Find Probability Density Function -Statistics and Probability Problems.

Find Probability Density Function -Statistics and

Probability Problems

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

1. 𝑃(𝑋 ≤ 1)

2. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

1. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

2. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

3. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

4. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

1. Exactly 3 emails.

2. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

1. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

2. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null

hypothesis. There is no significant evidence to suggest

that the population variance is different from 15. Problem

1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

3. 𝑃(𝑋 ≤ 1)

4. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

5. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

6. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

7. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

8. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

9. Exactly 3 emails.

10. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

11. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

12. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

13. 𝑃(𝑋 ≤ 1)

14. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

15. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

16. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

17. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

18. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

19. Exactly 3 emails.

20. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

21. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

22. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

23. 𝑃(𝑋 ≤ 1)

24. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

25. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

26. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

27. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

28. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

29. Exactly 3 emails.

30. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

31. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

32. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

33. 𝑃(𝑋 ≤ 1)

34. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

35. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

36. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

37. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

38. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

39. Exactly 3 emails.

40. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

41. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

42. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

43. 𝑃(𝑋 ≤ 1)

44. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

45. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

46. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

47. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

48. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

49. Exactly 3 emails.

50. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

51. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

52. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

53. 𝑃(𝑋 ≤ 1)

54. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

55. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

56. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

57. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

58. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

59. Exactly 3 emails.

60. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

61. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

62. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

63. 𝑃(𝑋 ≤ 1)

64. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

65. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

66. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

67. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

68. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

69. Exactly 3 emails.

70. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

71. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

72. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

73. 𝑃(𝑋 ≤ 1)

74. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

75. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

76. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

77. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

78. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

79. Exactly 3 emails.

80. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

81. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

82. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

83. 𝑃(𝑋 ≤ 1)

84. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

85. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

86. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

87. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

88. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

89. Exactly 3 emails.

90. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

91. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

92. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

93. 𝑃(𝑋 ≤ 1)

94. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

95. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

96. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

97. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

98. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

99. Exactly 3 emails.

100. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

101. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

102. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

103. 𝑃(𝑋 ≤ 1)

104. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

105. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

106. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

107. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

108. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

109. Exactly 3 emails.

110. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

111. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

112. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

113. 𝑃(𝑋 ≤ 1)

114. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

115. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

116. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

117. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

118. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

119. Exactly 3 emails.

120. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

121. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

122. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

123. 𝑃(𝑋 ≤ 1)

124. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

125. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

126. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

127. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

128. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

129. Exactly 3 emails.

130. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

131. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

132. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

133. 𝑃(𝑋 ≤ 1)

134. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

135. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

136. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

137. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

138. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

139. Exactly 3 emails.

140. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

141. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

142. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

143. 𝑃(𝑋 ≤ 1)

144. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

145. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

146. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

147. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

148. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

149. Exactly 3 emails.

150. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

151. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

152. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

153. 𝑃(𝑋 ≤ 1)

154. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

155. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

156. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

157. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

158. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

159. Exactly 3 emails.

160. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

161. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

162. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

163. 𝑃(𝑋 ≤ 1)

164. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

165. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

166. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

167. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

168. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

169. Exactly 3 emails.

170. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

171. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

172. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

173. 𝑃(𝑋 ≤ 1)

174. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

175. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

176. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

177. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

178. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

179. Exactly 3 emails.

180. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

181. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

182. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

183. 𝑃(𝑋 ≤ 1)

184. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

185. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

186. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

187. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

188. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

189. Exactly 3 emails.

190. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

191. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

192. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

193. 𝑃(𝑋 ≤ 1)

194. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

195. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

196. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

197. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

198. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

199. Exactly 3 emails.

200. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

201. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

202. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

203. 𝑃(𝑋 ≤ 1)

204. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

205. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

206. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

207. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

208. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

209. Exactly 3 emails.

210. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

211. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

212. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

213. 𝑃(𝑋 ≤ 1)

214. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

215. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

216. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

217. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

218. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

219. Exactly 3 emails.

220. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

221. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

222. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

223. 𝑃(𝑋 ≤ 1)

224. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

225. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

226. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

227. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

228. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

229. Exactly 3 emails.

230. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

231. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

232. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

233. 𝑃(𝑋 ≤ 1)

234. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

235. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

236. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

237. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

238. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

239. Exactly 3 emails.

240. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

241. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

242. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

243. 𝑃(𝑋 ≤ 1)

244. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

245. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

246. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

247. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

248. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

249. Exactly 3 emails.

250. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

251. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

252. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

253. 𝑃(𝑋 ≤ 1)

254. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

255. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

256. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

257. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

258. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

259. Exactly 3 emails.

260. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

261. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

262. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

263. 𝑃(𝑋 ≤ 1)

264. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

265. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

266. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

267. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

268. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

269. Exactly 3 emails.

270. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

271. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

272. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

273. 𝑃(𝑋 ≤ 1)

274. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

275. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

276. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

277. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

278. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

279. Exactly 3 emails.

280. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

281. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

282. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

283. 𝑃(𝑋 ≤ 1)

284. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

285. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

286. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

287. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

288. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

289. Exactly 3 emails.

290. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

291. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

292. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

293. 𝑃(𝑋 ≤ 1)

294. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

295. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

296. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

297. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

298. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

299. Exactly 3 emails.

300. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

301. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

302. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

303. 𝑃(𝑋 ≤ 1)

304. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

305. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

306. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

307. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

308. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

309. Exactly 3 emails.

310. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

311. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

312. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

313. 𝑃(𝑋 ≤ 1)

314. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

315. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

316. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

317. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

318. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

319. Exactly 3 emails.

320. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

321. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

322. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

323. 𝑃(𝑋 ≤ 1)

324. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

325. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

326. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

327. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

328. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

329. Exactly 3 emails.

330. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

331. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

332. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

333. 𝑃(𝑋 ≤ 1)

334. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

335. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

336. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

337. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

338. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

339. Exactly 3 emails.

340. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

341. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

342. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

343. 𝑃(𝑋 ≤ 1)

344. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

345. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

346. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

347. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

348. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

349. Exactly 3 emails.

350. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

351. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

352. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

353. 𝑃(𝑋 ≤ 1)

354. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

355. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

356. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

357. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

358. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

359. Exactly 3 emails.

360. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

361. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

362. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

363. 𝑃(𝑋 ≤ 1)

364. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

365. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

366. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

367. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

368. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

369. Exactly 3 emails.

370. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

371. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

372. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

373. 𝑃(𝑋 ≤ 1)

374. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

375. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

376. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

377. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

378. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

379. Exactly 3 emails.

380. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

381. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

382. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

383. 𝑃(𝑋 ≤ 1)

384. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

385. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

386. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

387. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

388. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

389. Exactly 3 emails.

390. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

391. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

392. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

393. 𝑃(𝑋 ≤ 1)

394. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

395. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

396. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

397. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

398. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

399. Exactly 3 emails.

400. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

401. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

402. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

403. 𝑃(𝑋 ≤ 1)

404. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

405. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

406. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

407. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

408. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

409. Exactly 3 emails.

410. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

411. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

412. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

Problem 1: Probability Density Function

Given the probability density function (PDF) of a continuous random variable 𝑋 is 𝑓(𝑥)=

√2𝜋𝑒−𝑥2/2, find:

413. 𝑃(𝑋 ≤ 1)

414. 𝑃(−1 ≤ 𝑋 ≤ 1)

Solution

The given PDF 𝑓(𝑥)=1

√2𝜋𝑒−𝑥2/2 is the standard normal distribution, 𝑁(0,1).

415. To find 𝑃(𝑋 ≤ 1), we need to compute the cumulative distribution function (CDF) 𝐹(1):

𝑃(𝑋 ≤ 1)= 𝐹(1)= ∫ 1

√2𝜋

−∞ 𝑒−𝑥2/2 𝑑𝑥

416. Using standard normal distribution tables or a computational tool, we find:

𝑃(𝑋 ≤ 1)≈ 0.8413

417. To find 𝑃(−1 ≤ 𝑋 ≤ 1), we use the symmetry of the normal distribution:

𝑃(−1 ≤ 𝑋 ≤ 1)= 𝑃(𝑋 ≤ 1)−𝑃(𝑋 ≤ −1)

418. Since 𝑃(𝑋 ≤ −1)= 1−𝑃(𝑋 ≤ 1):

𝑃(−1 ≤ 𝑋 ≤ 1)= 2⋅𝑃(𝑋 ≤ 1)−1 ≈ 2⋅0.8413−1 = 0.6826

Problem 2: Poisson Distribution

Suppose the number of emails a person receives per day follows a Poisson distribution with an

average rate of 5 emails per day. Calculate the probability that on a given day, the person

receives:

419. Exactly 3 emails.

420. At most 2 emails.

Solution

Let 𝑋 be the number of emails received per day, 𝑋 ∼ Poisson(𝜆 = 5).

421. The probability of receiving exactly 3 emails is:

𝑃(𝑋 = 3)=𝑒−553

3! =𝑒−5 ⋅125

6≈ 0.1404

422. The probability of receiving at most 2 emails is:

𝑃(𝑋 ≤ 2)= 𝑃(𝑋 = 0)+𝑃(𝑋 = 1)+𝑃(𝑋 = 2)

𝑃(𝑋 = 0)=𝑒−550

0! = 𝑒−5 ≈ 0.0067

𝑃(𝑋 = 1)=𝑒−551

1! = 5𝑒−5 ≈ 0.0337

𝑃(𝑋 = 2)=𝑒−552

2! =25𝑒−5

2≈ 0.0842

𝑃(𝑋 ≤ 2)≈ 0.0067+0.0337+0.0842 = 0.1246

Problem 3: Binomial Distribution

Consider a scenario where a fair coin is flipped 10 times. What is the probability of getting

exactly 6 heads?

Solution

Let 𝑋 be the number of heads in 10 flips, 𝑋 ∼ Binomial(𝑛 = 10,𝑝 = 0.5).

𝑃(𝑋 = 6)=(10

6)(0.5)6(0.5)4=10!

6!4!(0.5)10

𝑃(𝑋 = 6)=210⋅(0.5)10 =210⋅1

1024 ≈ 0.2051

Problem 4: Chi-Square Distribution

A random sample of 15 observations from a normal population has a sample variance of 20.

Test the hypothesis that the population variance is 15 at the 0.05 significance level.

Solution

The test statistic is:

𝜒2=(𝑛−1)𝑠2

𝜎2=14⋅20

15 ≈18.67

With 𝑛−1 = 14 degrees of freedom, the critical values from the chi-square distribution table at

the 0.05 significance level are 𝜒0.025,14

2≈ 5.63 and 𝜒0.975,14

2≈26.12.

Since 5.63 <18.67 <26.12, we do not reject the null hypothesis. There is no significant

evidence to suggest that the population variance is different from 15.

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