FINDING THE EXPONENTIAL DISTRIBUTION WITH
MEAN
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
1. The probability that the machine lasts more than 7 years.
1. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
1. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
1. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
1. The probability that a randomly selected student is taller than 180 cm.
1. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
1. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
1. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
1. Exactly 2 items are defective.
1. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
1. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
1. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
2. The probability that the machine lasts more than 7 years.
2. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
2. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
2. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
2. The probability that a randomly selected student is taller than 180 cm.
2. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
2. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
2. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
2. Exactly 2 items are defective.
2. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
2. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
2. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
3. The probability that the machine lasts more than 7 years.
3. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
3. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
3. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
3. The probability that a randomly selected student is taller than 180 cm.
3. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
3. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
3. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
3. Exactly 2 items are defective.
3. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
3. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
3. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
4. The probability that the machine lasts more than 7 years.
4. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
4. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
4. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
4. The probability that a randomly selected student is taller than 180 cm.
4. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
4. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
4. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
4. Exactly 2 items are defective.
4. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
4. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
4. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
5. The probability that the machine lasts more than 7 years.
5. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
5. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
5. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
5. The probability that a randomly selected student is taller than 180 cm.
5. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
5. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
5. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
5. Exactly 2 items are defective.
5. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
5. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
5. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
6. The probability that the machine lasts more than 7 years.
6. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
6. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
6. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
6. The probability that a randomly selected student is taller than 180 cm.
6. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
6. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
6. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
6. Exactly 2 items are defective.
6. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
6. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
6. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
7. The probability that the machine lasts more than 7 years.
7. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
7. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
7. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
7. The probability that a randomly selected student is taller than 180 cm.
7. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
7. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
7. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
7. Exactly 2 items are defective.
7. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
7. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
7. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
8. The probability that the machine lasts more than 7 years.
8. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
8. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
8. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
8. The probability that a randomly selected student is taller than 180 cm.
8. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
8. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
8. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
8. Exactly 2 items are defective.
8. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
8. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
8. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
9. The probability that the machine lasts more than 7 years.
9. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
9. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
9. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
9. The probability that a randomly selected student is taller than 180 cm.
9. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
9. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
9. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
9. Exactly 2 items are defective.
9. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
9. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
9. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
10. The probability that the machine lasts more than 7 years.
10. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
10. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
10. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
10. The probability that a randomly selected student is taller than 180 cm.
10. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
10. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
10. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
10. Exactly 2 items are defective.
10. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
10. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
10. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
11. The probability that the machine lasts more than 7 years.
11. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
11. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
11. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
11. The probability that a randomly selected student is taller than 180 cm.
11. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
11. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
11. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
11. Exactly 2 items are defective.
11. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
11. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
11. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
12. The probability that the machine lasts more than 7 years.
12. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
12. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
12. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
12. The probability that a randomly selected student is taller than 180 cm.
12. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
12. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
12. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
12. Exactly 2 items are defective.
12. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
12. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
12. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
13. The probability that the machine lasts more than 7 years.
13. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
13. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
13. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
13. The probability that a randomly selected student is taller than 180 cm.
13. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
13. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
13. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
13. Exactly 2 items are defective.
13. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
13. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
13. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
14. The probability that the machine lasts more than 7 years.
14. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
14. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
14. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
14. The probability that a randomly selected student is taller than 180 cm.
14. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
14. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
14. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
14. Exactly 2 items are defective.
14. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
14. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
14. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
15. The probability that the machine lasts more than 7 years.
15. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
15. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
15. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
15. The probability that a randomly selected student is taller than 180 cm.
15. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
15. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
15. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
15. Exactly 2 items are defective.
15. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
15. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
15. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
16. The probability that the machine lasts more than 7 years.
16. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
16. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
16. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
16. The probability that a randomly selected student is taller than 180 cm.
16. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
16. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
16. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
16. Exactly 2 items are defective.
16. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
16. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
16. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
17. The probability that the machine lasts more than 7 years.
17. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
17. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
17. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
17. The probability that a randomly selected student is taller than 180 cm.
17. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
17. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
17. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
17. Exactly 2 items are defective.
17. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
17. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
17. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
18. The probability that the machine lasts more than 7 years.
18. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
18. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
18. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
18. The probability that a randomly selected student is taller than 180 cm.
18. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
18. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
18. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
18. Exactly 2 items are defective.
18. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
18. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
18. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
19. The probability that the machine lasts more than 7 years.
19. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
19. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
19. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
19. The probability that a randomly selected student is taller than 180 cm.
19. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
19. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
19. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
19. Exactly 2 items are defective.
19. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
19. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
19. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
20. The probability that the machine lasts more than 7 years.
20. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
20. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
20. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
20. The probability that a randomly selected student is taller than 180 cm.
20. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
20. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
20. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
20. Exactly 2 items are defective.
20. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
20. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
20. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
21. The probability that the machine lasts more than 7 years.
21. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
21. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
21. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
21. The probability that a randomly selected student is taller than 180 cm.
21. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
21. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
21. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
21. Exactly 2 items are defective.
21. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
21. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
21. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
22. The probability that the machine lasts more than 7 years.
22. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
22. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
22. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
22. The probability that a randomly selected student is taller than 180 cm.
22. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
22. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
22. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
22. Exactly 2 items are defective.
22. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
22. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
22. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
23. The probability that the machine lasts more than 7 years.
23. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
23. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
23. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
23. The probability that a randomly selected student is taller than 180 cm.
23. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
23. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
23. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
23. Exactly 2 items are defective.
23. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
23. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
23. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
24. The probability that the machine lasts more than 7 years.
24. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
24. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
24. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
24. The probability that a randomly selected student is taller than 180 cm.
24. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
24. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
24. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
24. Exactly 2 items are defective.
24. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
24. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
24. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
25. The probability that the machine lasts more than 7 years.
25. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
25. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
25. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
25. The probability that a randomly selected student is taller than 180 cm.
25. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
25. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
25. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
25. Exactly 2 items are defective.
25. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
25. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
25. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
26. The probability that the machine lasts more than 7 years.
26. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
26. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
26. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
26. The probability that a randomly selected student is taller than 180 cm.
26. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
26. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
26. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
26. Exactly 2 items are defective.
26. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
26. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
26. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
27. The probability that the machine lasts more than 7 years.
27. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
27. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
27. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
27. The probability that a randomly selected student is taller than 180 cm.
27. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
27. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
27. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
27. Exactly 2 items are defective.
27. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
27. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
27. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
28. The probability that the machine lasts more than 7 years.
28. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
28. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
28. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
28. The probability that a randomly selected student is taller than 180 cm.
28. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
28. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
28. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
28. Exactly 2 items are defective.
28. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
28. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
28. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
29. The probability that the machine lasts more than 7 years.
29. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
29. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
29. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
29. The probability that a randomly selected student is taller than 180 cm.
29. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
29. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
29. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
29. Exactly 2 items are defective.
29. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
29. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
29. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
30. The probability that the machine lasts more than 7 years.
30. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
30. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
30. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
30. The probability that a randomly selected student is taller than 180 cm.
30. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
30. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
30. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
30. Exactly 2 items are defective.
30. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
30. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
30. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
31. The probability that the machine lasts more than 7 years.
31. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
31. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
31. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
31. The probability that a randomly selected student is taller than 180 cm.
31. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
31. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
31. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
31. Exactly 2 items are defective.
31. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
31. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
31. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
32. The probability that the machine lasts more than 7 years.
32. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
32. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
32. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
32. The probability that a randomly selected student is taller than 180 cm.
32. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
32. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
32. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
32. Exactly 2 items are defective.
32. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
32. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
32. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
33. The probability that the machine lasts more than 7 years.
33. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
33. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
33. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
33. The probability that a randomly selected student is taller than 180 cm.
33. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
33. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
33. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
33. Exactly 2 items are defective.
33. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
33. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
33. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
34. The probability that the machine lasts more than 7 years.
34. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
34. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
34. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
34. The probability that a randomly selected student is taller than 180 cm.
34. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
34. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
34. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
34. Exactly 2 items are defective.
34. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
34. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
34. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
35. The probability that the machine lasts more than 7 years.
35. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
35. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
35. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
35. The probability that a randomly selected student is taller than 180 cm.
35. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
35. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
35. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
35. Exactly 2 items are defective.
35. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
35. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
35. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
36. The probability that the machine lasts more than 7 years.
36. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
36. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
36. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
36. The probability that a randomly selected student is taller than 180 cm.
36. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
36. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
36. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
36. Exactly 2 items are defective.
36. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
36. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
36. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
37. The probability that the machine lasts more than 7 years.
37. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
37. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
37. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
37. The probability that a randomly selected student is taller than 180 cm.
37. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
37. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
37. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
37. Exactly 2 items are defective.
37. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
37. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
37. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
38. The probability that the machine lasts more than 7 years.
38. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
38. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
38. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
38. The probability that a randomly selected student is taller than 180 cm.
38. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
38. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
38. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
38. Exactly 2 items are defective.
38. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
38. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
38. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
39. The probability that the machine lasts more than 7 years.
39. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
39. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
39. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
39. The probability that a randomly selected student is taller than 180 cm.
39. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
39. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
39. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
39. Exactly 2 items are defective.
39. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
39. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
39. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
40. The probability that the machine lasts more than 7 years.
40. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
40. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
40. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
40. The probability that a randomly selected student is taller than 180 cm.
40. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
40. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
40. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
40. Exactly 2 items are defective.
40. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
40. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
40. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
41. The probability that the machine lasts more than 7 years.
41. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
41. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
41. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
41. The probability that a randomly selected student is taller than 180 cm.
41. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
41. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
41. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
41. Exactly 2 items are defective.
41. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
41. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
41. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
42. The probability that the machine lasts more than 7 years.
42. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
42. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
42. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
42. The probability that a randomly selected student is taller than 180 cm.
42. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
42. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
42. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
42. Exactly 2 items are defective.
42. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
42. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
42. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
43. The probability that the machine lasts more than 7 years.
43. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
43. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
43. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
43. The probability that a randomly selected student is taller than 180 cm.
43. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
43. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
43. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
43. Exactly 2 items are defective.
43. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
43. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
43. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
44. The probability that the machine lasts more than 7 years.
44. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
44. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
44. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
44. The probability that a randomly selected student is taller than 180 cm.
44. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
44. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
44. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
44. Exactly 2 items are defective.
44. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
44. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
44. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
45. The probability that the machine lasts more than 7 years.
45. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
45. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
45. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
45. The probability that a randomly selected student is taller than 180 cm.
45. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
45. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
45. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
45. Exactly 2 items are defective.
45. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
45. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
45. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
46. The probability that the machine lasts more than 7 years.
46. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
46. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
46. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
46. The probability that a randomly selected student is taller than 180 cm.
46. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
46. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
46. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
46. Exactly 2 items are defective.
46. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
46. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
46. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
47. The probability that the machine lasts more than 7 years.
47. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
47. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
47. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
47. The probability that a randomly selected student is taller than 180 cm.
47. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
47. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
47. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
47. Exactly 2 items are defective.
47. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
47. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
47. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
48. The probability that the machine lasts more than 7 years.
48. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
48. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
48. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
48. The probability that a randomly selected student is taller than 180 cm.
48. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
48. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
48. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
48. Exactly 2 items are defective.
48. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
48. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
48. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
49. The probability that the machine lasts more than 7 years.
49. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
49. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
49. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
49. The probability that a randomly selected student is taller than 180 cm.
49. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
49. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
49. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
49. Exactly 2 items are defective.
49. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
49. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
49. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
50. The probability that the machine lasts more than 7 years.
50. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
50. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
50. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
50. The probability that a randomly selected student is taller than 180 cm.
50. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
50. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
50. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
50. Exactly 2 items are defective.
50. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
50. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
50. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
51. The probability that the machine lasts more than 7 years.
51. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
51. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
51. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
51. The probability that a randomly selected student is taller than 180 cm.
51. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
51. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
51. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
51. Exactly 2 items are defective.
51. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
51. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
51. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
52. The probability that the machine lasts more than 7 years.
52. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
52. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
52. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
52. The probability that a randomly selected student is taller than 180 cm.
52. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
52. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
52. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
52. Exactly 2 items are defective.
52. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
52. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
52. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
53. The probability that the machine lasts more than 7 years.
53. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
53. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
53. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
53. The probability that a randomly selected student is taller than 180 cm.
53. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
53. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
53. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
53. Exactly 2 items are defective.
53. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
53. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
53. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
54. The probability that the machine lasts more than 7 years.
54. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
54. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
54. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
54. The probability that a randomly selected student is taller than 180 cm.
54. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
54. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
54. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
54. Exactly 2 items are defective.
54. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
54. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
54. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
55. The probability that the machine lasts more than 7 years.
55. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
55. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
55. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
55. The probability that a randomly selected student is taller than 180 cm.
55. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
55. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
55. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
55. Exactly 2 items are defective.
55. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
55. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
55. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
56. The probability that the machine lasts more than 7 years.
56. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
56. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
56. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
56. The probability that a randomly selected student is taller than 180 cm.
56. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
56. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
56. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
56. Exactly 2 items are defective.
56. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
56. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
56. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
57. The probability that the machine lasts more than 7 years.
57. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
57. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
57. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
57. The probability that a randomly selected student is taller than 180 cm.
57. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
57. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
57. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
57. Exactly 2 items are defective.
57. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
57. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
57. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
58. The probability that the machine lasts more than 7 years.
58. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
58. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
58. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
58. The probability that a randomly selected student is taller than 180 cm.
58. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
58. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
58. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
58. Exactly 2 items are defective.
58. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
58. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
58. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
59. The probability that the machine lasts more than 7 years.
59. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
59. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
59. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
59. The probability that a randomly selected student is taller than 180 cm.
59. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
59. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
59. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
59. Exactly 2 items are defective.
59. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
59. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
59. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
60. The probability that the machine lasts more than 7 years.
60. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
60. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
60. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
60. The probability that a randomly selected student is taller than 180 cm.
60. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
60. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
60. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
60. Exactly 2 items are defective.
60. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
60. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
60. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
61. The probability that the machine lasts more than 7 years.
61. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
61. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
61. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
61. The probability that a randomly selected student is taller than 180 cm.
61. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
61. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
61. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
61. Exactly 2 items are defective.
61. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
61. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
61. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
62. The probability that the machine lasts more than 7 years.
62. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
62. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
62. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
62. The probability that a randomly selected student is taller than 180 cm.
62. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
62. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
62. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
62. Exactly 2 items are defective.
62. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
62. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
62. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
63. The probability that the machine lasts more than 7 years.
63. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
63. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
63. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
63. The probability that a randomly selected student is taller than 180 cm.
63. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
63. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
63. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
63. Exactly 2 items are defective.
63. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
63. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
63. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
64. The probability that the machine lasts more than 7 years.
64. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
64. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
64. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
64. The probability that a randomly selected student is taller than 180 cm.
64. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
64. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
64. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
64. Exactly 2 items are defective.
64. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
64. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
64. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
65. The probability that the machine lasts more than 7 years.
65. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
65. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
65. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
65. The probability that a randomly selected student is taller than 180 cm.
65. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
65. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
65. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
65. Exactly 2 items are defective.
65. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
65. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
65. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
66. The probability that the machine lasts more than 7 years.
66. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
66. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
66. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
66. The probability that a randomly selected student is taller than 180 cm.
66. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
66. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
66. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
66. Exactly 2 items are defective.
66. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
66. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
66. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
67. The probability that the machine lasts more than 7 years.
67. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
67. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
67. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
67. The probability that a randomly selected student is taller than 180 cm.
67. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
67. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
67. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
67. Exactly 2 items are defective.
67. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
67. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
67. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
68. The probability that the machine lasts more than 7 years.
68. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
68. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
68. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
68. The probability that a randomly selected student is taller than 180 cm.
68. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
68. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
68. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
68. Exactly 2 items are defective.
68. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
68. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
68. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
69. The probability that the machine lasts more than 7 years.
69. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
69. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
69. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
69. The probability that a randomly selected student is taller than 180 cm.
69. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
69. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
69. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
69. Exactly 2 items are defective.
69. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
69. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
69. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
70. The probability that the machine lasts more than 7 years.
70. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
70. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
70. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
70. The probability that a randomly selected student is taller than 180 cm.
70. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
70. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
70. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
70. Exactly 2 items are defective.
70. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
70. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
70. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
71. The probability that the machine lasts more than 7 years.
71. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
71. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
71. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
71. The probability that a randomly selected student is taller than 180 cm.
71. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
71. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
71. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
71. Exactly 2 items are defective.
71. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
71. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
71. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
72. The probability that the machine lasts more than 7 years.
72. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
72. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
72. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
72. The probability that a randomly selected student is taller than 180 cm.
72. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
72. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
72. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
72. Exactly 2 items are defective.
72. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
72. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
72. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
73. The probability that the machine lasts more than 7 years.
73. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
73. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
73. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
73. The probability that a randomly selected student is taller than 180 cm.
73. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
73. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
73. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
73. Exactly 2 items are defective.
73. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
73. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
73. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
74. The probability that the machine lasts more than 7 years.
74. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
74. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
74. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
74. The probability that a randomly selected student is taller than 180 cm.
74. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
74. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
74. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
74. Exactly 2 items are defective.
74. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
74. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
74. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
75. The probability that the machine lasts more than 7 years.
75. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
75. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
75. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
75. The probability that a randomly selected student is taller than 180 cm.
75. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
75. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
75. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
75. Exactly 2 items are defective.
75. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
75. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
75. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
76. The probability that the machine lasts more than 7 years.
76. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
76. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
76. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
76. The probability that a randomly selected student is taller than 180 cm.
76. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
76. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
76. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
76. Exactly 2 items are defective.
76. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
76. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
76. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
77. The probability that the machine lasts more than 7 years.
77. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
77. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
77. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
77. The probability that a randomly selected student is taller than 180 cm.
77. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
77. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
77. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
77. Exactly 2 items are defective.
77. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
77. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
77. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
Problem 1: Exponential Distribution
A machine has a lifetime (in years) that follows an exponential distribution with a mean of 5
years. Calculate:
78. The probability that the machine lasts more than 7 years.
78. The probability that the machine fails within the first 3 years.
Solution
Let
X
be the lifetime of the machine,
X
∼
Exponential
(
λ
)
where
λ=1
5
.
78. The probability that the machine lasts more than 7 years is:
P
(
X>7
)
= 1−F
(
7
)
= 1−
(
1−e− λ
⋅
7
)
=e− λ
⋅
7
P
(
X>7
)
=e−7/5 ≈0.3012
78. The probability that the machine fails within the first 3 years is:
P
(
X≤3
)
=F
(
3
)
= 1−e− λ
⋅
3= 1−e−3/5
P
(
X≤3
)
= 1−e−0.6≈0.4512
Problem 2: Normal Distribution
The heights of a group of students are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. Calculate:
78. The probability that a randomly selected student is taller than 180 cm.
78. The height that separates the shortest 25% from the rest.
Solution
Let
X
be the height of a student,
X
∼
N
(
170,102
)
.
78. The probability that a student is taller than 180 cm is:
P
(
X>180
)
= 1−P
(
X≤180
)
= 1−F
(
180
)
Converting to the standard normal variable
Z
:
Z=X− μ
σ=180−170
10 = 1
Using standard normal distribution tables:
P
(
Z≤1
)
≈0.8413
P
(
X>180
)
= 1−0.8413= 0.1587
78. To find the height that separates the shortest 25% from the rest, we need to find the 25th
percentile (
Q1
):
P
(
X≤Q1
)
= 0.25
Using standard normal distribution tables, we find
z0.25≈ − 0.674
. Converting back to the
original variable:
Q1=μ+z0.25
⋅
σ= 170+
(
−0.674
)
⋅
10= 170−6.74= 163.26cm
Problem 3: Binomial Distribution
In a quality control process, 5% of the items produced are defective. If a sample of 20 items is
selected, calculate the probability that:
78. Exactly 2 items are defective.
78. At most 3 items are defective.
Solution
Let
X
be the number of defective items in a sample of 20,
X
∼
Binomial
(
n=20,p= 0.05
)
.
78. The probability that exactly 2 items are defective is:
P
(
X= 2
)
=
(
20
2
)
(
0.05
)
2
(
0.95
)
18
P
(
X= 2
)
=20!
2!
(
20−2
)
!
(
0.05
)
2
(
0.95
)
18≈0.1887
78. The probability that at most 3 items are defective is:
P
(
X≤3
)
=P
(
X= 0
)
+P
(
X= 1
)
+P
(
X= 2
)
+P
(
X= 3
)
P
(
X= 0
)
=
(
20
0
)
(
0.05
)
0
(
0.95
)
20≈0.3585
P
(
X= 1
)
=
(
20
1
)
(
0.05
)
1
(
0.95
)
19≈0.3774
P
(
X= 2
)
≈0.1887
P
(
X= 3
)
=
(
20
3
)
(
0.05
)
3
(
0.95
)
17 ≈0.0628
P
(
X≤3
)
≈0.3585+0.3774+0.1887+0.0628= 0.9874
Problem 4: Chi-Square Goodness of Fit Test
A die is rolled 60 times, and the observed frequencies of the outcomes are as follows:
Observed frequencies of die rolls
Outcome
Frequency
1
8
2
10
3
9
4
11
5
13
6
9
Test the hypothesis that the die is fair at the 0.05 significance level.
Solution
The null hypothesis
H0
is that the die is fair, i.e., each outcome has an equal probability of
1
6
.
The expected frequency for each outcome if the die is fair:
Ei=60
6=10
The test statistic for the chi-square goodness of fit test is:
χ2=∑
i=1
6
(
Oi−Ei
)
2
Ei=
(
8−10
)
2
10 +
(
10−10
)
2
10 +
(
9−10
)
2
10 +
(
11−10
)
2
10 +
(
13−10
)
2
10 +
(
9−10
)
2
10
χ2=4
10 +0+ 1
10+1
10 +9
10 +1
10= 1.6
With
6−1= 5
degrees of freedom, the critical value from the chi-square distribution table at the
0.05 significance level is
χ0.05,5
2= 11.07
.
Since
1.6<11.07
, we do not reject the null hypothesis. There is no significant evidence to
suggest that the die is not fair.
