ECON 350 - CLASSICAL ECONOMICS Descriptive statistics Set 1

ECON 350 - CLASSICAL

ECONOMICS - Descriptive statistics

Question Bank - Set 1

Liberty University

Question 1

Question

A researcher collected data on the heights (in inches) of 50 students at a uni-

versity. The heights are summarized in the following descriptive statistics:



Mean height: 65 inches



Standard deviation: 3 inches

Assuming that the heights are normally distributed, calculate the z-score for a

student who is 68 inches tall.

Solution

Step 1: Calculate the z-score using the formula

z=x−µ

σ,

where xis the individual data point, µis the mean, and σis the standard

deviation.

Step 2: Substituting the given values into the formula, we get

z=68 −65

Step 3: Calculate the z-score:

z=3

3= 1.

Step 4: Therefore, the z-score for a student who is 68 inches tall is 1.

Question 2

Question

Consider a dataset with the following values:

15, 21, 19, 25, 16, 20, 22, 18, 27, 24

Find the sample mean, sample variance, standard deviation, and range for

this dataset.

Solution

To find the sample mean, sample variance, standard deviation, and range for

the dataset, follow these steps:

Step 1: Calculate the sample mean The sample mean (¯x) is calculated

using the formula:

¯x=Pn

i=1 xi

where xirepresents the individual data points and nis the number of data

points.

Calculating the sample mean:

¯x=15 + 21 + 19 + 25 + 16 + 20 + 22 + 18 + 27 + 24

10 =207

10 = 20.7

Step 2: Calculate the sample variance The sample variance (s2) is

calculated using the formula:

s2=Pn

i=1(xi−¯x)2

n−1

Calculating the sample variance:

s2=(15 −20.7)2+ (21 −20.7)2+. . . + (24 −20.7)2

9=140.1

9≈15.5667

Step 3: Calculate the standard deviation The standard deviation (s)

is the square root of the sample variance:

s=√s2=√15.5667 ≈3.9465

Step 4: Calculate the range The range is the difference between the

maximum and minimum values in the dataset. In this case, the range is:

Range = 27 −15 = 12

Therefore, the sample mean is 20.7, the sample variance is approximately

15.5667, the standard deviation is approximately 3.9465, and the range is 12.

Question 3

Question

Let Xbe a random variable representing the heights of students in a university.

The sample mean and sample standard deviation are calculated as 67 inches

and 4 inches, respectively. A student claims that at least 90

Solution

Step 1: Let’s set up the null and alternative hypotheses. - Null Hypothesis, H0:

The mean height of students in the university is 60 inches or less. - Alternative

Hypothesis, H1: The mean height of students in the university is greater than

60 inches.

Step 2: Calculate the Z-score for the sample mean using the formula:

Z=¯

X−µ0

√n

where ¯

Xis the sample mean, µ0is the hypothesized population mean under

the null hypothesis, σis the population standard deviation, and nis the sample

size.

Step 3: Given that ¯

X= 67 inches, σ= 4 inches, µ0= 60 inches, and

assuming the sample size is sufficiently large, we have:

Z=67 −60

√n

Step 4: Since the sample size is not given, we cannot calculate the exact value

of Z. However, we can make use of the fact that Z-scores can be converted to

percentages for normally distributed data.

Step 5: For a standard normal distribution, 90

Step 6: Compare the calculated Z-score to 1.28 to determine if we can reject

the null hypothesis. If the Z-score is greater than 1.28, we conclude that the

claim that at least 90

Question 4

Question

A researcher collects data on the heights of 100 students in a university. The

mean height is 65 inches with a standard deviation of 3 inches. If the heights are

normally distributed, what percentage of students are between 62 inches and 68

inches tall?

Solution

Step 1: Calculate the Z-scores for 62 inches and 68 inches using the formula:

Z=X−µ

σ, where Xis the height, µis the mean, and σis the standard deviation.

For 62 inches: Z62 =62−65

3=−1

For 68 inches: Z68 =68−65

3= 1

Step 2: Use a standard normal distribution table or calculator to find the

area under the curve between these two Z-scores. P(−1< Z < 1) = P(Z <

1) −P(Z < −1)

Step 3: Look up the values for P(Z < 1) and P(Z < −1) in the standard

normal distribution table. P(Z < 1) ≈0.8413 and P(Z < −1) ≈0.1587

Step 4: Calculate the percentage of students between 62 inches and 68 inches

by subtracting the two probabilities found in Step 3. P(−1< Z < 1) =

0.8413 −0.1587 = 0.6826

Therefore, approximately 68.26

Question 5

Question

A researcher collects data on the amount of time (in minutes) that students

spend studying each day. The following data represents the study times for a

sample of 15 students:

5,3,7,10,6,2,8,1,4,9,12,11,15,13,14

Calculate the mean, median, mode, range, variance, and standard deviation

for this data set.

Solution

Step 1: Calculate the mean. The mean is calculated by summing all the values

in the data set and dividing by the number of values.

Mean = 5+3+7+10+6+2+8+1+4+9+12+11+15+13+14

15 =110

15 = 7.33

Step 2: Calculate the median. To find the median, we first need to arrange

the data in ascending order:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

Since there are 15 data points, the median is the middle value, which is 8.

Step 3: Calculate the mode. The mode is the value that appears most

frequently in the data set. In this case, there is no mode as each value appears

only once.

Step 4: Calculate the range. The range is the difference between the maxi-

mum and minimum values in the data set.

Range = 15 −1 = 14

Step 5: Calculate the variance. The variance is a measure of how spread

out the values in the data set are from the mean. It is calculated by finding the

average of the squared differences between each value and the mean.

Variance = (5 −7.33)2+ (3 −7.33)2+. . . + (14 −7.33)2

Variance = 38.77 + 19.11 + . . . + 39.11

15 =308.02

15 = 20.54

Step 6: Calculate the standard deviation. The standard deviation is the

square root of the variance.

Standard Deviation = √20.54 ≈4.53

Therefore, for the given data set, the mean is 7.33, median is 8, mode is

none, range is 14, variance is approximately 20.54, and standard deviation is

approximately 4.53.

Question 6

Question

A researcher is studying the relationship between hours spent studying per week

and final exam scores for a group of university students. After collecting data

from 50 students, the researcher calculated the following descriptive statistics

for the two variables:

Hours Studied (hours) Final Exam Score

Mean 15 hours

Standard Deviation 3 hours

Assume that the relationship between hours studied and exam score follows

a linear pattern. If the researcher wants to predict the final exam score for a

student who studies 18 hours per week, what would be the predicted final exam

score?

Solution

Step 1: Calculate the correlation coefficient between hours studied and final

exam scores using the formula:

r=Cov(X, Y )

σX·σY

Where ris the correlation coefficient, Cov(X, Y ) is the covariance between

hours studied and final exam scores, and σXand σYare the standard deviations

of hours studied and final exam scores, respectively.

Step 2: Given that the correlation coefficient is r=Cov(X,Y )

σX·σY, rearrange the

formula to solve for Cov(X, Y ):

Cov(X, Y ) = r·σX·σY

Step 3: Substitute the given values into the formula:

Cov(X, Y ) = r·σX·σY= 0 (for a linear relationship)

Step 4: Calculate the regression equation for predicting final exam scores

based on hours studied:

Y=a+bX

Where Yis the predicted final exam score, Xis the number of hours studied,

and aand bare constants to be determined.

Step 5: Since the relationship between the two variables is linear, the regres-

sion equation simplifies to:

Y=a+bX =¯

Y+rσY

σX(X−¯

Step 6: Substitute the given values into the regression equation:

Y= 15 + 0.5(18 −15) = 15 + 0.5×3 = 15 + 1.5 = 16.5

Therefore, the predicted final exam score for a student who studies 18 hours

per week is 16.5.

Question 7

Question

Let X={3,3,5,6,9,10,11,12}be a set of data points. Calculate the five-

number summary for the data set X.

Solution

Step 1: Arrange the data set in ascending order.

X={3,3,5,6,9,10,11,12}

Step 2: Find the minimum value. The minimum value is 3.

Step 3: Find the maximum value. The maximum value is 12.

Step 4: Find the median. Since n= 8 is even, the median is the average of

the two middle numbers.

Median = 6+9

2= 7.5

Step 5: Find the lower quartile (Q1) and upper quartile (Q3).

Q1 = 3+5

2= 4

Q3 = 10 + 11

2= 10.5

Step 6: Construct the five-number summary. The five-number summary for

the data set Xis: Minimum = 3, Lower Quartile = 4, Median = 7.5, Upper

Quartile = 10.5, Maximum = 12.

Question 8

Question

A researcher collected the following data on the number of hours students study

per week: 5, 6, 7, 9, 10, 12, 14, 14, 15, 18. Find the mean, median, mode, range,

variance, and standard deviation of the data.

Solution

Step 1: Calculate the Mean To find the mean, we sum up all the data points

and divide by the total number of data points.

Mean = 5+6+7+9+10+12+14+14+15+18

Mean = 110

10 = 11

Step 2: Calculate the Median To find the median, we first need to order

the data: 5, 6, 7, 9, 10, 12, 14, 14, 15, 18. Since we have an even number of

data points, the median is the average of the two middle values. Median =

10+12

2= 11

Step 3: Calculate the Mode The mode is the value that appears most

frequently in the data set. In this case, the mode is 14, as it appears twice,

which is more than any other value.

Step 4: Calculate the Range The range is the difference between the

maximum and minimum values in the data set. Range = 18 - 5 = 13

Step 5: Calculate the Variance The formula for variance is: V ar(X) =

P(Xi−¯

X)2

n, where ¯

Xis the mean, Xiare the data points, and nis the total

number of data points. Now, we substitute the values into the formula:

V ar(X) = (5 −11)2+ (6 −11)2+... + (18 −11)2

V ar(X) = 72+25+16+4+1+1+9+9+16+49

V ar(X) = 202

10 = 20.2

Step 6: Calculate the Standard Deviation The standard deviation is

the square root of the variance. Standard deviation = √20.2≈4.49

Therefore, the mean is 11, the median is 11, the mode is 14, the range is 13,

the variance is 20.2, and the standard deviation is approximately 4.49.

Question 9

Question

Suppose a researcher collected the following data on the weights of 12 students

(in kg):

62,64,68,70,72,75,76,78,80,82,85,90

Calculate the five-number summary for this data set.

Solution

To find the five-number summary for the given data set, we need to determine

the minimum, first quartile (Q1), median (Q2), third quartile (Q3), and maxi-

mum values.

Step 1: Arrange the data in ascending order

62,64,68,70,72,75,76,78,80,82,85,90

Step 2: Find the median (Q2) Since there are 12 data points, the median

is the average of the 6th and 7th data points:

Q2 = 75 + 76

2= 75.5

Step 3: Find the first quartile (Q1) The first quartile is the median of

the lower half of the data set:

Q1 = 68 + 70

2= 69

Step 4: Find the third quartile (Q3) The third quartile is the median

of the upper half of the data set:

Q3 = 82 + 85

2= 83.5

Step 5: Find the minimum and maximum values The minimum value

is 62 and the maximum value is 90.

Five-number Summary: Minimum = 62

First Quartile (Q1) = 69

Median (Q2) = 75.5

Third Quartile (Q3) = 83.5

Maximum = 90

Therefore, the five-number summary for the given data set is: {62, 69, 75.5,

83.5, 90}.

Question 10

Question

Suppose we have collected the following data on the monthly salaries (in dollars)

of 10 employees at a software company:

$4000,$4200,$3800,$4300,$4100,$3900,$3700,$4500,$4000,$4800

Calculate the mean, median, mode, range, variance, and standard deviation

of the data.

Solution

Step 1: To calculate the mean, add all the values together and divide by the

total number of values.

Mean = 4000 + 4200 + 3800 + 4300 + 4100 + 3900 + 3700 + 4500 + 4000 + 4800

10 = 4100

Step 2: To find the median, first rearrange the data from least to greatest:

3700,3800,3900,4000,4000,4100,4200,4300,4500,4800

Since we have an even number of values, the median is the average of the middle

two values, which are 4000 and 4100.

Median = 4000 + 4100

2= 4050

Step 3: To determine the mode, identify the value that occurs most fre-

quently. The mode in this case is

4000 because it appears twice, which is more

than any other value.

Step 4: The range is found by subtracting the minimum value from the

maximum value.

Range = 4800 −3700 = 1100

Step 5: To calculate the variance, first find the squared difference between

each data point and the mean, then divide by the total number of data points.

Variance = (4000 −4100)2+ (4200 −4100)2+. . . + (4800 −4100)2

=10000 + 1002+. . . + 70000

=160000

10 = 16000

Step 6: Finally, the standard deviation is the square root of the variance.

Standard Deviation = √16000 = 40

Question 11

Question

Let Xbe a random variable representing the number of hours students sleep per

night in a university. A sample of 50 students showed the following statistics:

mean = 6 hours, standard deviation = 1.5 hours. Calculate the coefficient of

variation for the sample. Round your answer to two decimal places.

Solution

Step 1: Calculate the coefficient of variation using the formula:

Coefficient of Variation = Standard Deviation

Mean ×100%

Step 2: Substitute the values of the standard deviation and mean into the

formula:

Coefficient of Variation = 1.5

6×100%

Step 3: Calculate the coefficient of variation:

Coefficient of Variation = (0.25) ×100% = 25%

Therefore, the coefficient of variation for the sample is 25

Question 12

Question

Let’s consider a dataset with five observations: 12, 18, 22, 26, and x. If the

mean of the dataset is 20 and the standard deviation is 5, find the value of x.

Solution

Step 1: Recall that the mean of a dataset is calculated by summing all the

values and dividing by the total number of values. Additionally, the standard

deviation measures the variability or dispersion of a dataset from the mean.

Step 2: To find the mean of the dataset with five observations, we sum all

the values (12, 18, 22, 26, and x) and divide by 5. Thus:

12 + 18 + 22 + 26 + x

5= 20

Step 3: Simplifying the equation:

12 + 18 + 22 + 26 + x= 20 ×5

78 + x= 100

Step 4: Therefore, x= 100 −78 = 22.

Step 5: Now, let’s calculate the standard deviation of the dataset by finding

the squared differences between each value and the mean, summing these values,

dividing by the total number of observations, and then taking the square root.

We are given that the standard deviation is 5.

Step 6: The formula for standard deviation is:

r(12 −20)2+ (18 −20)2+ (22 −20)2+ (26 −20)2+ (22 −20)2

5= 5

Step 7: Simplifying the equation:

r64+4+4+36+4

5= 5

r112

5= 5

Step 8: Solving for this we get:

√22.4=5

5≈5

Step 9: Therefore, the value of x= 22 satisfies both the mean and standard

deviation conditions given.

Question 13

Question

Let Xand Ybe two random variables with the following joint probability

distribution:

X Y P (X, Y )

0 1 0.1

0 2 0.2

1 1 0.3

1 2 0.4

Calculate the marginal probability distributions of Xand Y.

Solution

Step 1: To find the marginal probability distribution of X, we sum the proba-

bilities of all possible values of Yfor each value of X.

For X= 0:

P(X= 0) = P(X= 0, Y = 1) + P(X= 0, Y = 2) = 0.1+0.2=0.3

For X= 1:

P(X= 1) = P(X= 1, Y = 1) + P(X= 1, Y = 2) = 0.3+0.4=0.7

Therefore, the marginal probability distribution of Xis:

X P (X)

0 0.3

1 0.7

Step 2: To find the marginal probability distribution of Y, we sum the

probabilities of all possible values of Xfor each value of Y.

For Y= 1:

P(Y= 1) = P(X= 0, Y = 1) + P(X= 1, Y = 1) = 0.1+0.3 = 0.4

For Y= 2:

P(Y= 2) = P(X= 0, Y = 2) + P(X= 1, Y = 2) = 0.2+0.4 = 0.6

Therefore, the marginal probability distribution of Yis:

Y P (Y)

1 0.4

2 0.6

Question 14

Question

Consider a dataset with the following values: 15, 18, 20, 22, 25, 28, 30, 33, 35,

38, 40. Calculate the mean, median, mode, variance, and standard deviation of

this dataset.

Solution

Let’s calculate the mean, median, mode, variance, and standard deviation step

by step.

Step 1: Calculate the mean The mean is calculated by summing up all

the values in the dataset and dividing by the total number of values. Mean =

15 + 18 + 20 + 22 + 25 + 28 + 30 + 33 + 35 + 38 + 40

Mean = 304

Mean = 27.64 (rounded to two decimal places)

Step 2: Calculate the median Since the dataset has 11 values, the median

will be the middle value when the data is arranged in ascending order. Arranging

the dataset in ascending order: 15, 18, 20, 22, 25, 28, 30, 33, 35, 38, 40

Median = 28

Step 3: Calculate the mode The mode is the value that appears most

frequently in the dataset. In this dataset, there is no repeated value, so there is

no mode.

Step 4: Calculate the variance The variance is calculated by finding

the average of the squared differences between each data point and the mean.

Variance = 1

11[(15 −27.64)2+ (18 −27.64)2+... + (40 −27.64)2]

Variance = 1

11[148.2296 + 86.3396 + ... + 153.4596]

Variance = 1

11[1100.11]

Variance = 100 (rounded to two decimal places)

Step 5: Calculate the standard deviation The standard deviation is

the square root of the variance. Standard deviation = √100

Standard deviation = 10

Therefore, the mean is 27.64, the median is 28, the mode does not exist, the

variance is 100, and the standard deviation is 10 for the given dataset.

Question 15

Question

Let X={3,5,7,11,13}and Y={4,6,8,12,15}be two sets of data. Calculate

the covariance between Xand Y.

Solution

To calculate the covariance between Xand Y, we will use the formula:

Cov(X, Y ) = 1

i=1

(xi−¯

X)(yi−¯

where xiand yiare the data points in sets Xand Y, respectively, and ¯

and ¯

Yare the means of sets Xand Y, respectively.

Step 1: Calculate the means of Xand Y.



Mean of X:

X=3+5+7+11+13

5=39

5= 7.8



Mean of Y:

Y=4+6+8+12+15

5=45

5= 9

Step 2: Calculate the covariance using the formula.

Cov(X, Y ) = 1

5[(3 −7.8)(4 −9) + (5 −7.8)(6 −9) + (7 −7.8)(8 −9)

+ (11 −7.8)(12 −9) + (13 −7.8)(15 −9)]

5[(−4.8)(−5) + (−2.8)(−3) + (−0.8)(−1) + (3.2)(3) + (5.2)(6)]

5[24 + 8.4+0.8+9.6 + 31.2]

5×74 = 14.8

Therefore, the covariance between Xand Yis 14.8.

Question 16

Question

A researcher is investigating the relationship between the number of hours stu-

dents spend studying per week and their final exam grades. The data collected

from a sample of 30 students showed a mean study time of 15 hours per week

with a standard deviation of 4 hours. The mean exam grade was 80 with a

standard deviation of 10. Determine the covariance between study time and

exam grades.

Solution

Step 1: Recall that the formula for covariance between two variables Xand Y

is given by:

Cov(X, Y ) = 1

i=1

(Xi−¯

X)(Yi−¯

Step 2: Calculate the covariance using the given data.

Given: Mean study time ( ¯

X) = 15 hours Standard deviation of study time

= 4 hours Mean exam grade ( ¯

Y) = 80 Standard deviation of exam grade = 10

Step 3: First, let’s find the variance for both study time and exam grades.

The variance of study time (σ2

X) is calculated as:

σ2

X= (Standard deviation of study time)2= 42= 16

The variance of exam grades (σ2

Y) is calculated as:

σ2

Y= (Standard deviation of exam grades)2= 102= 100

Step 4: Next, find the covariance using the formula.

Cov(X, Y ) = 1

i=1

(Xi−¯

X)(Yi−¯

Substitute the given values:

Cov(X, Y ) = 1

i=1

(Xi−15)(Yi−80)

Step 5: Since we don’t have the actual data points, we can’t calculate the

exact covariance without them. However, given the means and standard devia-

tions, we can determine the general relationship.

If the covariance is positive, it indicates a positive relationship between study

time and exam grades. If negative, it indicates a negative relationship. And

if the covariance is zero, it indicates no linear relationship between the two

variables.

Therefore, the covariance between study time and exam grades can’t be

determined without the actual data points.

Question 17

Question

A researcher collected data on the test scores of 100 university students. The

mean test score was 75 with a standard deviation of 8. Assuming the distribution

is approximately normal, calculate the following:

1. the z-score for a test score of 82

2. the percentage of students who scored below 65

3. the percentage of students who scored between 70 and 80

Solution

1. To calculate the z-score for a test score of 82, we use the formula:

z=x−¯x

where: x= 82 (test score), ¯x= 75 (mean test score), σ= 8 (standard deviation).

Substitute the values into the formula:

z=82 −75

8=7

8= 0.875

So, the z-score for a test score of 82 is 0.875.

2. To find the percentage of students who scored below 65, we need to find

the z-score for 65 and then use a z-table to find the corresponding percentage.

The z-score for 65 can be calculated as:

z=65 −75

8=−10

8=−1.25

From the z-table, the percentage of students who scored below a z-score of

-1.25 is approximately 0.1056, or 10.56

3. To find the percentage of students who scored between 70 and 80, we

need to find the z-scores for 70 and 80, and then calculate the area between

them using the z-table. The z-scores for 70 and 80 are:

z70 =70 −75

8=−5

8=−0.625

z80 =80 −75

8=5

8= 0.625

Using the z-table, the area between -0.625 and 0.625 is approximately 0.2357,

or 23.57

Therefore, the percentage of students who scored between 70 and 80 is 23.57

Question 18

Question

Suppose we have a dataset representing the weights (in kg) of participants in two

different exercise programs. The mean weight for participants in Program A is

75 kg with a standard deviation of 5 kg, while the mean weight for participants

in Program B is 80 kg with a standard deviation of 4 kg.

Given that the dataset for Program A has a normal distribution and the

dataset for Program B has a skewed distribution, discuss the implications of

these differences in the context of conducting hypothesis tests and making in-

ferences about the populations these programs represent.

Solution

Step 1: Normal Distribution in Program A When the dataset for Program A has

a normal distribution, we can make certain assumptions about the population

and conduct parametric hypothesis tests with greater confidence. The mean

weight and standard deviation provide valuable information about the central

tendency and variability of the weights in Program A. We can use these parame-

ters to calculate confidence intervals and perform hypothesis tests like t-tests or

ANOVA with reasonable accuracy, assuming that the sample is representative

of the population.

Step 2: Skewed Distribution in Program B In contrast, when the dataset for

Program B has a skewed distribution, it indicates that the data may not meet

the assumptions of normality. In this case, conducting parametric hypothesis

tests without verifying the distribution assumptions could lead to misleading

results. Skewed data can affect the accuracy of inferential statistics, such as

confidence intervals and hypothesis tests, based on assumptions of normality. It

may be necessary to use non-parametric tests or transformations on the data to

address the skewness before making inferences about the population.

Step 3: Implications for Hypothesis Testing The difference in distributions

between Program A and Program B highlights the importance of considering

the underlying assumptions when conducting hypothesis tests. For Program

A, parametric tests can be applied with a higher level of confidence, leveraging

the information provided by the mean and standard deviation. However, for

Program B, the skewed distribution requires careful consideration and possibly

the use of alternative statistical methods to ensure the validity of the results.

In conclusion, understanding the distribution of the data is crucial for mak-

ing accurate inferences about the populations represented by the exercise pro-

grams. While a normal distribution in Program A allows for more straightfor-

ward analysis, a skewed distribution in Program B necessitates a more cautious

approach to hypothesis testing and statistical inference.

Question 19

Question

Suppose you have a dataset of 1000 observations on a variable. After calcu-

lating the sample mean and sample standard deviation, you decide to remove

all observations that are more than 3 standard deviations away from the mean.

Calculate the new sample mean and standard deviation after removing these

observations.

Solution

Let’s denote the original dataset as X={x1, x2, . . . , x1000}.

Step 1: Calculate the sample mean (¯x) and sample standard deviation (s)

of the original dataset. The sample mean is given by:

¯x=1

i=1

where nis the number of observations.

The sample standard deviation is given by:

s=sPn

i=1(xi−¯x)2

n−1

Step 2: Remove observations more than 3 standard deviations away from

the mean. Let X′be the new dataset after removing these observations.

Step 3: Calculate the new sample mean and sample standard deviation.

The new sample mean is given by:

¯x′=1

i=1

x′

where mis the number of observations in X′.

The new sample standard deviation is given by:

s′=sPm

i=1(x′

i−¯x′)2

m−1

Step 4: Finally, calculate the new sample mean and standard deviation.

Question 20

Question

Consider the following data set:

12,15,18,20,22,25,28,30,45,50

Calculate the mean, median, mode, range, variance, and standard deviation

for this data set.

Solution

Step 1: Calculate the mean

Mean = 1

i=1

10 (12 + 15 + 18 + 20 + 22 + 25 + 28 + 30 + 45 + 50)

=265

= 26.5

Step 2: Calculate the median Since the data set is already in ascending

order: In this case, the median is the average of the 5th and 6th numbers.

Median = 22 + 25

2= 23.5

Step 3: Calculate the mode The mode is the number that appears most

frequently in the data set. Since no number repeats, there is no mode in this

data set.

Step 4: Calculate the range

Range = Max −Min = 50 −12 = 38

Step 5: Calculate the variance

Variance = 1

i=1

(xi−Mean)2

10[(12 −26.5)2+ (15 −26.5)2+. . . + (50 −26.5)2]

10[472.25 + 133.225 + . . . + 627.5625]

10 ×892.85

= 89.285

Step 6: Calculate the standard deviation

Standard Deviation = √Variance = √89.285 ≈9.46

Therefore, the mean is 26.5, median is 23.5, mode does not exist, range is 38,

variance is 89.285, and standard deviation is approximately 9.46 for the given

data set.

Question 21

Question

Consider a dataset with 20 observations. The mean of the dataset is 45 and the

standard deviation is 5. If one observation is changed from 30 to 60, how does

this affect the mean and standard deviation?

Solution

Let’s denote the original dataset as Xand the new dataset as X′. We will first

calculate the mean and standard deviation of the original dataset X, and then

find the mean and standard deviation of the new dataset X′after changing one

observation.

Step 1: Calculate the mean and standard deviation of the original dataset

X: Given: n= 20, ¯

X= 45, s= 5.

Step 1a: Calculate the sum of the observations in dataset X:

Sum of observations = n×¯

X= 20 ×45 = 900

Step 1b: Calculate the sum of squares of deviations from the mean:

SS =

i=1

(Xi−¯

X)2=n×s2= 20 ×52= 500

Step 1c: Calculate the variance of dataset X:

Variance = SS

n=500

20 = 25

Step 1d: Calculate the standard deviation of dataset X:

Standard deviation = √Variance = √25 = 5

Therefore, the original dataset Xhas a mean of 45 and a standard deviation

of 5.

Step 2: Calculate the mean and standard deviation of the new dataset X′

after changing one observation: We will change one observation from 30 to 60

in dataset X.

Step 2a: Calculate the sum of observations in dataset X′: The new sum of

observations is 900 −30 + 60 = 930.

Step 2b: Calculate the new mean ¯

X′:

X′=New sum of observations

n=930

20 = 46.5

Step 2c: Calculate the new sum of squares of deviations from the mean SS′:

Since only one observation changed, we can calculate SS′as follows:

SS′=SS −(Xnew −Xold)2= 500 −(60 −30)2= 500 −900 = −400

Step 2d: Calculate the new variance of dataset X′:

Variance of X′=SS′

n=−400

20 =−20

Step 2e: Calculate the new standard deviation of dataset X′:

Standard deviation of X′=√Variance of X′=√−20 (Note: standard deviation cannot be negative)

Hence, changing the observation from 30 to 60 increases the mean from 45

to 46.5, and the standard deviation is no longer a meaningful measure due to

the negative variance.

Question 22

Question

The following dataset represents the weights of 12 individuals:

176,182,196,200,205,192,187,210,195,185,190,198.

Calculate the mean, median, mode, range, variance, and standard deviation of

the dataset.

Solution

Step 1: Calculate the mean. Step 2: Calculate the median. Step 3: Calculate

the mode. Step 4: Calculate the range. Step 5: Calculate the variance. Step 6:

Calculate the standard deviation.

Step 1: To calculate the mean, we sum up all the weights and divide by the

total number of weights. Mean = 176+182+196+200+205+192+187+210+195+185+190+198

12 =

2316

12 = 193.

Step 2: To calculate the median, we first arrange the weights in ascending

order: 176, 182, 185, 187, 190, 192, 195, 196, 198, 200, 205, 210. The median is

the middle value, so in this case, the median weight is 193.

Step 3: The mode is the value that appears most frequently in the dataset.

In this case, since all the weights are different, there is no mode.

Step 4: To calculate the range, we subtract the smallest weight from the

largest weight. Range = 210 - 176 = 34.

Step 5: To calculate the variance, first calculate the mean squared deviation

for each weight and then find the average of these squared deviations. Variance

=(176−193)2+(182−193)2+...+(198−193)2

12 =2892

12 = 241.33.

Step 6: To calculate the standard deviation, simply take the square root of

the variance. Standard deviation = √241.33 ≈15.53.

Question 23

Question

Let’s say we have a dataset with the following values for a variable X: 9, 12, 11,

14, 13, 10, 9, 12, 11, 15. Calculate the sample variance for variable X.

Solution

Step 1: Find the mean ( ¯

X) of the data.

X=9+12+11+14+13+10+9+12+11+15

10 =116

10 = 11.6

Step 2: Calculate the squared distance from the mean for each data point.

XX−¯

X(X−¯

X)2

9 9 −11.6 = −2.6 (−2.6)2= 6.76

12 12 −11.6 = 0.4 (0.4)2= 0.16

11 11 −11.6 = −0.6 (−0.6)2= 0.36

14 14 −11.6 = 2.4 (2.4)2= 5.76

13 13 −11.6 = 1.4 (1.4)2= 1.96

10 10 −11.6 = −1.6 (−1.6)2= 2.56

9 9 −11.6 = −2.6 (−2.6)2= 6.76

12 12 −11.6 = 0.4 (0.4)2= 0.16

11 11 −11.6 = −0.6 (−0.6)2= 0.36

15 15 −11.6 = 3.4 (3.4)2= 11.56

Step 3: Sum the squared differences and divide by the sample size minus 1

to get the sample variance.

s2=6.76 + 0.16 + 0.36 + 5.76 + 1.96 + 2.56 + 6.76 + 0.16 + 0.36 + 11.56

10 −1

s2=36.56

9= 4.06

Therefore, the sample variance of variable X is 4.06.

Question 24

Question

A university conducts a survey on the number of hours students sleep per night.

The data collected is as follows: 5, 6, 7, 8, 9, 10, 11. Calculate the mean,

median, mode, range, variance, and standard deviation of the data.

Solution

Step 1: Calculate the mean The mean of a set of data is calculated by sum-

ming all the values and dividing by the total number of values. Mean =

5+6+7+8+9+10+11

7Mean = 56

7Mean = 8

Step 2: Calculate the median To find the median, we first need to arrange

the data in ascending order: 5, 6, 7, 8, 9, 10, 11. Since there are 7 values, the

median is the 4th value, which is 8.

Step 3: Calculate the mode The mode is the value that appears most fre-

quently in the data. In this case, there is no mode as each value appears only

once.

Step 4: Calculate the range The range is the difference between the highest

and lowest values in the data. Range = 11 - 5 = 6

Step 5: Calculate the variance The variance is a measure of how spread out

the values in the data set are from the mean. Variance = (5−8)2+(6−8)2+(7−8)2+(8−8)2+(9−8)2+(10−8)2+(11−8)2

Variance = 22

7Variance ≈3.14

Step 6: Calculate the standard deviation The standard deviation is the

square root of the variance. Standard deviation = √3.14 Standard deviation ≈

1.77

Therefore, the mean is 8, the median is 8, there is no mode, the range is 6,

the variance is approximately 3.14, and the standard deviation is approximately

1.77.

Question 25

Question

In a study on the academic performance of students, the following scores were

obtained: 72, 85, 63, 91, 78, 60, 88, 82, 69, 75. Find the mean, median, mode,

range, variance, and standard deviation of these scores.

Solution

Step 1: Calculate the mean. Step 2: Find the median. Step 3: Determine the

mode. Step 4: Compute the range. Step 5: Calculate the variance. Step 6:

Find the standard deviation.

Step 1: Calculate the mean The mean is the sum of all values divided

by the number of values.

Mean = 72 + 85 + 63 + 91 + 78 + 60 + 88 + 82 + 69 + 75

10 =763

10 = 76.3

Step 2: Find the median To find the median, first arrange the scores

in ascending order: {60, 63, 69, 72, 75, 78, 82, 85, 88, 91}. Since there are

10 values, the median is the average of the 5th and 6th values. Median =

75+78

2=153

2= 76.5.

Step 3: Determine the mode In this set of scores, there is no mode since

all values occur only once.

Step 4: Compute the range The range is the difference between the

largest and smallest values. Range = 91 - 60 = 31.

Step 5: Calculate the variance The variance is the average of the squared

differences from the mean. Variance = (72−76.3)2+(85−76.3)2+...+(75−76.3)2

10 . Vari-

ance = (−4.3)2+8.72+...+(−1.3)2

10 =18.49+75.69+...+1.69

10 = 98.41.

Step 6: Find the standard deviation The standard deviation is the

square root of the variance. Standard deviation = √98.41 ≈9.92.

Question 26

Question

The scores of 20 students on a statistics exam are as follows: 78, 85, 92, 64, 70,

88, 76, 85, 90, 72, 68, 79, 83, 80, 87, 75, 81, 86, 74, 82. Calculate the mean,

median, mode, range, variance, and standard deviation of the scores.

Solution

Step 1: Calculate the Mean The mean is calculated by summing up all the scores

and dividing by the total number of scores. Mean = 78+85+92+64+70+88+76+85+90+72+68+79+83+80+87+75+81+86+74+82

Mean = 1600

20 = 80

Step 2: Calculate the Median To find the median, we first need to arrange

the scores in ascending order. 64, 68, 70, 72, 74, 75, 76, 78, 79, 80, 81, 82, 83,

85, 85, 86, 87, 88, 90, 92 Since we have 20 scores, the median is the average of

the 10th and 11th scores. Median = 80+81

2= 80.5

Step 3: Calculate the Mode The mode is the score that appears most fre-

quently. The mode of the given scores is 85.

Step 4: Calculate the Range The range is calculated by subtracting the

smallest score from the largest score. Range = 92 - 64 = 28

Step 5: Calculate the Variance The variance is calculated by finding the

average of the squared differences between each score and the mean. Variance

=(78−80)2+(85−80)2+...+(82−80)2

Variance = 256+25+144+...+4

Variance = 866

20 = 43.3

Step 6: Calculate the Standard Deviation The standard deviation is the

square root of the variance. Standard Deviation = √43.3 = 6.58

Therefore, the mean score is 80, median score is 80.5, mode is 85, range is

28, variance is 43.3, and standard deviation is 6.58.

Question 27

Question

A researcher conducted a study on the relationship between the number of hours

spent studying and final exam scores of 30 university students. The following

data was collected:

Hours Studied (X) Exam Score (Y)

5 65

9 78

6 70

3 50

8 85

Calculate the mean, median, mode, variance, and standard deviation for

both the hours studied and exam scores.

Solution

Step 1: Mean, Median, and Mode for Hours Studied



Mean = 5+9+6+3+8

5=31

5= 6.2



To find the median, first arrange the data in ascending order: 3,5,6,8,9.

Since there are 5 data points, the median is the middle value, which is 6.



There is no mode for this dataset as all values occur only once.

Step 2: Variance and Standard Deviation for Hours Studied



Calculate the variance using the formula:

Variance = Pn

i=1(Xi−Mean)2

n−1

Var(X) = (5 −6.2)2+ (9 −6.2)2+ (6 −6.2)2+ (3 −6.2)2+ (8 −6.2)2

5−1

Var(X) = 10.56 + 8.84 + 0.04 + 9.64 + 3.24

4=32.32

4= 8.08



Calculate the standard deviation:

SD(X) = pVar(X) = √8.08 ≈2.8446

Step 3: Mean, Median, and Mode for Exam Scores



Mean = 65+78+70+50+85

5=348

5= 69.6



To find the median, first arrange the data in ascending order: 50,65,70,78,85.

Since there are 5 data points, the median is the middle value, which is 70.



There is no mode for this dataset as all values occur only once.

Step 4: Variance and Standard Deviation for Exam Scores



Calculate the variance using the formula:

Variance = Pn

i=1(Yi−Mean)2

n−1

Var(Y) = (65 −69.6)2+ (78 −69.6)2+ (70 −69.6)2+ (50 −69.6)2+ (85 −69.6)2

5−1

Var(Y) = 20.16 + 72.36 + 0.16 + 399.36 + 291.6

4=783.64

4= 195.91



Calculate the standard deviation:

SD(Y) = pVar(Y) = √195.91 ≈13.9907

Question 28

Question

Suppose you have a dataset with the following values: 14, 17, 20, 23, 26, 29, 32,

35, 38, 41. Calculate the mean, median, mode, variance, standard deviation,

range, and interquartile range for this dataset.

Solution

Step 1: Calculate the Mean

Mean = 14 + 17 + 20 + 23 + 26 + 29 + 32 + 35 + 38 + 41

=275

10 = 27.5

Step 2: Calculate the Median Since the dataset is already in ascending

order, the median is the middle value:

Median = 26

Step 3: Calculate the Mode Since there are no repeated values, there is

no mode for this dataset.

Step 4: Calculate the Variance

Variance = 1

i=1

(xi−¯x)2

10 (14 −27.5)2+ (17 −27.5)2+. . . + (41 −27.5)2

10 [171.25 + 121 + 72.25 + . . . + 182.25]

10 ×634.5 = 63.45

Step 5: Calculate the Standard Deviation

Standard Deviation = √Variance = √63.45 ≈7.97

Step 6: Calculate the Range

Range = Max −Min = 41 −14 = 27

Step 7: Calculate the Interquartile Range (IQR) To find the first and

third quartiles: - Q1 = 0.25 * (n + 1)th term = 0.25 * 11th term = 2.75, which

is between the 2nd and 3rd data points, so Q1 = 20. - Q3 = 0.75 * (n + 1)th

term = 0.75 * 11th term = 8.25, which is between the 8th and 9th data points,

so Q3 = 38.

IQR = Q3−Q1 = 38 −20 = 18

Therefore, the mean is 27.5, the median is 26, there is no mode, the variance

is 63.45, the standard deviation is approximately 7.97, the range is 27, and the

interquartile range is 18.

Question 29

Question

In a study of heights of individuals, the following data was collected:

165,171,178,162,175,183,168,170,176,169,172,180,174,179

Calculate the mean, median, mode, variance, and standard deviation for the

given data.

Solution

Step 1: Calculate the Mean



To calculate the mean, we first add up all the data points and then divide

by the total number of data points.



Sum of all data points = 165 + 171 + 178 + 162 + 175 + 183 + 168 + 170 +

176 + 169 + 172 + 180 + 174 + 179 = 2477.



Number of data points = 14.



Mean = 2477

14 = 177.

Step 2: Calculate the Median



To calculate the median, we first arrange the data points in ascending

order: 162,165,168,169,170,171,172,174,175,176,178,179,180,183.



Since we have an odd number of data points, the median is the middle

value, which is 172.

Step 3: Calculate the Mode



The mode is the value that appears most frequently in the data.



In this case, there is no mode as all data points appear only once.

Step 4: Calculate the Variance



To calculate the variance, we first find the squared difference between each

data point and the mean, then sum these squared differences, and finally

divide by the total number of data points.



Squared differences: (165 −177)2,(171 −177)2, ..., (179 −177)2.



Sum of squared differences = 168.



Variance = 168

14−1= 14.

Step 5: Calculate the Standard Deviation



The standard deviation is the square root of the variance.



Standard deviation = √14 ≈3.74.

Question 30

Question

Let X={12,8,15,7,10,19,13,6,11,14}be a set of data points. Calculate the

interquartile range (IQR) of the data set X.

Solution

To find the interquartile range (IQR) of a data set, we need to first find the first

and third quartiles.

Step 1: Find the median (Q2) of the data set. Arrange the data set

Xin ascending order:

6,7,8,10,11,12,13,14,15,19

Since there are 10 data points, the median is the average of the 5th and 6th

data points:

Median = 11 + 12

2= 11.5

Step 2: Find the median of the lower half of the data set as the

first quartile (Q1). The lower half of the data set is {6,7,8,10,11}. Since

there are 5 data points in the lower half, the median is the 3rd data point:

Q1=8

Step 3: Find the median of the upper half of the data set as the

third quartile (Q3). The upper half of the data set is {12,13,14,15,19}.

Since there are 5 data points in the upper half, the median is the 3rd data

point:

Q3 = 14

Step 4: Calculate the interquartile range (IQR). The interquartile

range is given by:

IQR = Q3−Q1 = 14 −8=6

Therefore, the interquartile range (IQR) of the data set Xis 6.

Question 2

Question

Consider a dataset with the following values:

15, 21, 19, 25, 16, 20, 22, 18, 27, 24

Find the sample mean, sample variance, standard deviation, and range for

this dataset.

Solution

To find the sample mean, sample variance, standard deviation, and range for

the dataset, follow these steps:

Step 1: Calculate the sample mean The sample mean (¯x) is calculated

using the formula:

¯x=Pn

i=1 xi

where xirepresents the individual data points and nis the number of data

points.

Calculating the sample mean:

¯x=15 + 21 + 19 + 25 + 16 + 20 + 22 + 18 + 27 + 24

10 =207

10 = 20.7

Step 2: Calculate the sample variance The sample variance (s2) is

calculated using the formula:

s2=Pn

i=1(xi−¯x)2

n−1

Calculating the sample variance:

s2=(15 −20.7)2+ (21 −20.7)2+. . . + (24 −20.7)2

9=140.1

9≈15.5667

Step 3: Calculate the standard deviation The standard deviation (s)

is the square root of the sample variance:

s=√s2=√15.5667 ≈3.9465

Step 4: Calculate the range The range is the difference between the

maximum and minimum values in the dataset. In this case, the range is:

Range = 27 −15 = 12

Therefore, the sample mean is 20.7, the sample variance is approximately

15.5667, the standard deviation is approximately 3.9465, and the range is 12.

Question 3

Question

Let Xbe a random variable representing the heights of students in a university.

The sample mean and sample standard deviation are calculated as 67 inches

and 4 inches, respectively. A student claims that at least 90

Solution

Step 1: Let’s set up the null and alternative hypotheses. - Null Hypothesis, H0:

The mean height of students in the university is 60 inches or less. - Alternative

Hypothesis, H1: The mean height of students in the university is greater than

60 inches.

Step 2: Calculate the Z-score for the sample mean using the formula:

Z=¯

X−µ0

√n

where ¯

Xis the sample mean, µ0is the hypothesized population mean under

the null hypothesis, σis the population standard deviation, and nis the sample

size.

Step 3: Given that ¯

X= 67 inches, σ= 4 inches, µ0= 60 inches, and

assuming the sample size is sufficiently large, we have:

Z=67 −60

√n

Step 4: Since the sample size is not given, we cannot calculate the exact value

of Z. However, we can make use of the fact that Z-scores can be converted to

percentages for normally distributed data.

Step 5: For a standard normal distribution, 90

Step 6: Compare the calculated Z-score to 1.28 to determine if we can reject

the null hypothesis. If the Z-score is greater than 1.28, we conclude that the

claim that at least 90

Question 4

Question

A researcher collects data on the heights of 100 students in a university. The

mean height is 65 inches with a standard deviation of 3 inches. If the heights are

normally distributed, what percentage of students are between 62 inches and 68

inches tall?

Solution

Step 1: Calculate the Z-scores for 62 inches and 68 inches using the formula:

Z=X−µ

σ, where Xis the height, µis the mean, and σis the standard deviation.

For 62 inches: Z62 =62−65

3=−1

For 68 inches: Z68 =68−65

3= 1

Step 2: Use a standard normal distribution table or calculator to find the

area under the curve between these two Z-scores. P(−1< Z < 1) = P(Z <

1) −P(Z < −1)

Step 3: Look up the values for P(Z < 1) and P(Z < −1) in the standard

normal distribution table. P(Z < 1) ≈0.8413 and P(Z < −1) ≈0.1587

Step 4: Calculate the percentage of students between 62 inches and 68 inches

by subtracting the two probabilities found in Step 3. P(−1< Z < 1) =

0.8413 −0.1587 = 0.6826

Therefore, approximately 68.26

Question 5

Question

A researcher collects data on the amount of time (in minutes) that students

spend studying each day. The following data represents the study times for a

sample of 15 students:

5,3,7,10,6,2,8,1,4,9,12,11,15,13,14

Calculate the mean, median, mode, range, variance, and standard deviation

for this data set.

Solution

Step 1: Calculate the mean. The mean is calculated by summing all the values

in the data set and dividing by the number of values.

Mean = 5+3+7+10+6+2+8+1+4+9+12+11+15+13+14

15 =110

15 = 7.33

Step 2: Calculate the median. To find the median, we first need to arrange

the data in ascending order:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

Since there are 15 data points, the median is the middle value, which is 8.

Step 3: Calculate the mode. The mode is the value that appears most

frequently in the data set. In this case, there is no mode as each value appears

only once.

Step 4: Calculate the range. The range is the difference between the maxi-

mum and minimum values in the data set.

Range = 15 −1 = 14

Step 5: Calculate the variance. The variance is a measure of how spread

out the values in the data set are from the mean. It is calculated by finding the

average of the squared differences between each value and the mean.

Variance = (5 −7.33)2+ (3 −7.33)2+. . . + (14 −7.33)2

Variance = 38.77 + 19.11 + . . . + 39.11

15 =308.02

15 = 20.54

Step 6: Calculate the standard deviation. The standard deviation is the

square root of the variance.

Standard Deviation = √20.54 ≈4.53

Therefore, for the given data set, the mean is 7.33, median is 8, mode is

none, range is 14, variance is approximately 20.54, and standard deviation is

approximately 4.53.

Question 6

Question

A researcher is studying the relationship between hours spent studying per week

and final exam scores for a group of university students. After collecting data

from 50 students, the researcher calculated the following descriptive statistics

for the two variables:

Hours Studied (hours) Final Exam Score

Mean 15 hours

Standard Deviation 3 hours

Assume that the relationship between hours studied and exam score follows

a linear pattern. If the researcher wants to predict the final exam score for a

student who studies 18 hours per week, what would be the predicted final exam

score?

Solution

Step 1: Calculate the correlation coefficient between hours studied and final

exam scores using the formula:

r=Cov(X, Y )

σX·σY

Where ris the correlation coefficient, Cov(X, Y ) is the covariance between

hours studied and final exam scores, and σXand σYare the standard deviations

of hours studied and final exam scores, respectively.

Step 2: Given that the correlation coefficient is r=Cov(X,Y )

σX·σY, rearrange the

formula to solve for Cov(X, Y ):

Cov(X, Y ) = r·σX·σY

Step 3: Substitute the given values into the formula:

Cov(X, Y ) = r·σX·σY= 0 (for a linear relationship)

Step 4: Calculate the regression equation for predicting final exam scores

based on hours studied:

Y=a+bX

Where Yis the predicted final exam score, Xis the number of hours studied,

and aand bare constants to be determined.

Step 5: Since the relationship between the two variables is linear, the regres-

sion equation simplifies to:

Y=a+bX =¯

Y+rσY

σX(X−¯

Step 6: Substitute the given values into the regression equation:

Y= 15 + 0.5(18 −15) = 15 + 0.5×3 = 15 + 1.5 = 16.5

Therefore, the predicted final exam score for a student who studies 18 hours

per week is 16.5.

Question 7

Question

Let X={3,3,5,6,9,10,11,12}be a set of data points. Calculate the five-

number summary for the data set X.

Solution

Step 1: Arrange the data set in ascending order.

X={3,3,5,6,9,10,11,12}

Step 2: Find the minimum value. The minimum value is 3.

Step 3: Find the maximum value. The maximum value is 12.

Step 4: Find the median. Since n= 8 is even, the median is the average of

the two middle numbers.

Median = 6+9

2= 7.5

Step 5: Find the lower quartile (Q1) and upper quartile (Q3).

Q1 = 3+5

2= 4

Q3 = 10 + 11

2= 10.5

Step 6: Construct the five-number summary. The five-number summary for

the data set Xis: Minimum = 3, Lower Quartile = 4, Median = 7.5, Upper

Quartile = 10.5, Maximum = 12.

Question 8

Question

A researcher collected the following data on the number of hours students study

per week: 5, 6, 7, 9, 10, 12, 14, 14, 15, 18. Find the mean, median, mode, range,

variance, and standard deviation of the data.

Solution

Step 1: Calculate the Mean To find the mean, we sum up all the data points

and divide by the total number of data points.

Mean = 5+6+7+9+10+12+14+14+15+18

Mean = 110

10 = 11

Step 2: Calculate the Median To find the median, we first need to order

the data: 5, 6, 7, 9, 10, 12, 14, 14, 15, 18. Since we have an even number of

data points, the median is the average of the two middle values. Median =

10+12

2= 11

Step 3: Calculate the Mode The mode is the value that appears most

frequently in the data set. In this case, the mode is 14, as it appears twice,

which is more than any other value.

Step 4: Calculate the Range The range is the difference between the

maximum and minimum values in the data set. Range = 18 - 5 = 13

Step 5: Calculate the Variance The formula for variance is: V ar(X) =

P(Xi−¯

X)2

n, where ¯

Xis the mean, Xiare the data points, and nis the total

number of data points. Now, we substitute the values into the formula:

V ar(X) = (5 −11)2+ (6 −11)2+... + (18 −11)2

V ar(X) = 72+25+16+4+1+1+9+9+16+49

V ar(X) = 202

10 = 20.2

Step 6: Calculate the Standard Deviation The standard deviation is

the square root of the variance. Standard deviation = √20.2≈4.49

Therefore, the mean is 11, the median is 11, the mode is 14, the range is 13,

the variance is 20.2, and the standard deviation is approximately 4.49.

Question 9

Question

Suppose a researcher collected the following data on the weights of 12 students

(in kg):

62,64,68,70,72,75,76,78,80,82,85,90

Calculate the five-number summary for this data set.

Solution

To find the five-number summary for the given data set, we need to determine

the minimum, first quartile (Q1), median (Q2), third quartile (Q3), and maxi-

mum values.

Step 1: Arrange the data in ascending order

62,64,68,70,72,75,76,78,80,82,85,90

Step 2: Find the median (Q2) Since there are 12 data points, the median

is the average of the 6th and 7th data points:

Q2 = 75 + 76

2= 75.5

Step 3: Find the first quartile (Q1) The first quartile is the median of

the lower half of the data set:

Q1 = 68 + 70

2= 69

Step 4: Find the third quartile (Q3) The third quartile is the median

of the upper half of the data set:

Q3 = 82 + 85

2= 83.5

Step 5: Find the minimum and maximum values The minimum value

is 62 and the maximum value is 90.

Five-number Summary: Minimum = 62

First Quartile (Q1) = 69

Median (Q2) = 75.5

Third Quartile (Q3) = 83.5

Maximum = 90

Therefore, the five-number summary for the given data set is: {62, 69, 75.5,

83.5, 90}.

Question 10

Question

Suppose we have collected the following data on the monthly salaries (in dollars)

of 10 employees at a software company:

$4000,$4200,$3800,$4300,$4100,$3900,$3700,$4500,$4000,$4800

Calculate the mean, median, mode, range, variance, and standard deviation

of the data.

Solution

Step 1: To calculate the mean, add all the values together and divide by the

total number of values.

Mean = 4000 + 4200 + 3800 + 4300 + 4100 + 3900 + 3700 + 4500 + 4000 + 4800

10 = 4100

Step 2: To find the median, first rearrange the data from least to greatest:

3700,3800,3900,4000,4000,4100,4200,4300,4500,4800

Since we have an even number of values, the median is the average of the middle

two values, which are 4000 and 4100.

Median = 4000 + 4100

2= 4050

Step 3: To determine the mode, identify the value that occurs most fre-

quently. The mode in this case is

4000 because it appears twice, which is more

than any other value.

Step 4: The range is found by subtracting the minimum value from the

maximum value.

Range = 4800 −3700 = 1100

Step 5: To calculate the variance, first find the squared difference between

each data point and the mean, then divide by the total number of data points.

Variance = (4000 −4100)2+ (4200 −4100)2+. . . + (4800 −4100)2

=10000 + 1002+. . . + 70000

=160000

10 = 16000

Step 6: Finally, the standard deviation is the square root of the variance.

Standard Deviation = √16000 = 40

Question 11

Question

Let Xbe a random variable representing the number of hours students sleep per

night in a university. A sample of 50 students showed the following statistics:

mean = 6 hours, standard deviation = 1.5 hours. Calculate the coefficient of

variation for the sample. Round your answer to two decimal places.

Solution

Step 1: Calculate the coefficient of variation using the formula:

Coefficient of Variation = Standard Deviation

Mean ×100%

Step 2: Substitute the values of the standard deviation and mean into the

formula:

Coefficient of Variation = 1.5

6×100%

Step 3: Calculate the coefficient of variation:

Coefficient of Variation = (0.25) ×100% = 25%

Therefore, the coefficient of variation for the sample is 25

Question 12

Question

Let’s consider a dataset with five observations: 12, 18, 22, 26, and x. If the

mean of the dataset is 20 and the standard deviation is 5, find the value of x.

Solution

Step 1: Recall that the mean of a dataset is calculated by summing all the

values and dividing by the total number of values. Additionally, the standard

deviation measures the variability or dispersion of a dataset from the mean.

Step 2: To find the mean of the dataset with five observations, we sum all

the values (12, 18, 22, 26, and x) and divide by 5. Thus:

12 + 18 + 22 + 26 + x

5= 20

Step 3: Simplifying the equation:

12 + 18 + 22 + 26 + x= 20 ×5

78 + x= 100

Step 4: Therefore, x= 100 −78 = 22.

Step 5: Now, let’s calculate the standard deviation of the dataset by finding

the squared differences between each value and the mean, summing these values,

dividing by the total number of observations, and then taking the square root.

We are given that the standard deviation is 5.

Step 6: The formula for standard deviation is:

r(12 −20)2+ (18 −20)2+ (22 −20)2+ (26 −20)2+ (22 −20)2

5= 5

Step 7: Simplifying the equation:

r64+4+4+36+4

5= 5

r112

5= 5

Step 8: Solving for this we get:

√22.4=5

5≈5

Step 9: Therefore, the value of x= 22 satisfies both the mean and standard

deviation conditions given.

Question 13

Question

Let Xand Ybe two random variables with the following joint probability

distribution:

X Y P (X, Y )

0 1 0.1

0 2 0.2

1 1 0.3

1 2 0.4

Calculate the marginal probability distributions of Xand Y.

Solution

Step 1: To find the marginal probability distribution of X, we sum the proba-

bilities of all possible values of Yfor each value of X.

For X= 0:

P(X= 0) = P(X= 0, Y = 1) + P(X= 0, Y = 2) = 0.1+0.2=0.3

For X= 1:

P(X= 1) = P(X= 1, Y = 1) + P(X= 1, Y = 2) = 0.3+0.4=0.7

Therefore, the marginal probability distribution of Xis:

X P (X)

0 0.3

1 0.7

Step 2: To find the marginal probability distribution of Y, we sum the

probabilities of all possible values of Xfor each value of Y.

For Y= 1:

P(Y= 1) = P(X= 0, Y = 1) + P(X= 1, Y = 1) = 0.1+0.3 = 0.4

For Y= 2:

P(Y= 2) = P(X= 0, Y = 2) + P(X= 1, Y = 2) = 0.2+0.4 = 0.6

Therefore, the marginal probability distribution of Yis:

Y P (Y)

1 0.4

2 0.6

Question 14

Question

Consider a dataset with the following values: 15, 18, 20, 22, 25, 28, 30, 33, 35,

38, 40. Calculate the mean, median, mode, variance, and standard deviation of

this dataset.

Solution

Let’s calculate the mean, median, mode, variance, and standard deviation step

by step.

Step 1: Calculate the mean The mean is calculated by summing up all

the values in the dataset and dividing by the total number of values. Mean =

15 + 18 + 20 + 22 + 25 + 28 + 30 + 33 + 35 + 38 + 40

Mean = 304

Mean = 27.64 (rounded to two decimal places)

Step 2: Calculate the median Since the dataset has 11 values, the median

will be the middle value when the data is arranged in ascending order. Arranging

the dataset in ascending order: 15, 18, 20, 22, 25, 28, 30, 33, 35, 38, 40

Median = 28

Step 3: Calculate the mode The mode is the value that appears most

frequently in the dataset. In this dataset, there is no repeated value, so there is

no mode.

Step 4: Calculate the variance The variance is calculated by finding

the average of the squared differences between each data point and the mean.

Variance = 1

11[(15 −27.64)2+ (18 −27.64)2+... + (40 −27.64)2]

Variance = 1

11[148.2296 + 86.3396 + ... + 153.4596]

Variance = 1

11[1100.11]

Variance = 100 (rounded to two decimal places)

Step 5: Calculate the standard deviation The standard deviation is

the square root of the variance. Standard deviation = √100

Standard deviation = 10

Therefore, the mean is 27.64, the median is 28, the mode does not exist, the

variance is 100, and the standard deviation is 10 for the given dataset.

Question 15

Question

Let X={3,5,7,11,13}and Y={4,6,8,12,15}be two sets of data. Calculate

the covariance between Xand Y.

Solution

To calculate the covariance between Xand Y, we will use the formula:

Cov(X, Y ) = 1

i=1

(xi−¯

X)(yi−¯

where xiand yiare the data points in sets Xand Y, respectively, and ¯

and ¯

Yare the means of sets Xand Y, respectively.

Step 1: Calculate the means of Xand Y.



Mean of X:

X=3+5+7+11+13

5=39

5= 7.8



Mean of Y:

Y=4+6+8+12+15

5=45

5= 9

Step 2: Calculate the covariance using the formula.

Cov(X, Y ) = 1

5[(3 −7.8)(4 −9) + (5 −7.8)(6 −9) + (7 −7.8)(8 −9)

+ (11 −7.8)(12 −9) + (13 −7.8)(15 −9)]

5[(−4.8)(−5) + (−2.8)(−3) + (−0.8)(−1) + (3.2)(3) + (5.2)(6)]

5[24 + 8.4+0.8+9.6 + 31.2]

5×74 = 14.8

Therefore, the covariance between Xand Yis 14.8.

Question 16

Question

A researcher is investigating the relationship between the number of hours stu-

dents spend studying per week and their final exam grades. The data collected

from a sample of 30 students showed a mean study time of 15 hours per week

with a standard deviation of 4 hours. The mean exam grade was 80 with a

standard deviation of 10. Determine the covariance between study time and

exam grades.

Solution

Step 1: Recall that the formula for covariance between two variables Xand Y

is given by:

Cov(X, Y ) = 1

i=1

(Xi−¯

X)(Yi−¯

Step 2: Calculate the covariance using the given data.

Given: Mean study time ( ¯

X) = 15 hours Standard deviation of study time

= 4 hours Mean exam grade ( ¯

Y) = 80 Standard deviation of exam grade = 10

Step 3: First, let’s find the variance for both study time and exam grades.

The variance of study time (σ2

X) is calculated as:

σ2

X= (Standard deviation of study time)2= 42= 16

The variance of exam grades (σ2

Y) is calculated as:

σ2

Y= (Standard deviation of exam grades)2= 102= 100

Step 4: Next, find the covariance using the formula.

Cov(X, Y ) = 1

i=1

(Xi−¯

X)(Yi−¯

Substitute the given values:

Cov(X, Y ) = 1

i=1

(Xi−15)(Yi−80)

Step 5: Since we don’t have the actual data points, we can’t calculate the

exact covariance without them. However, given the means and standard devia-

tions, we can determine the general relationship.

If the covariance is positive, it indicates a positive relationship between study

time and exam grades. If negative, it indicates a negative relationship. And

if the covariance is zero, it indicates no linear relationship between the two

variables.

Therefore, the covariance between study time and exam grades can’t be

determined without the actual data points.

Question 17

Question

A researcher collected data on the test scores of 100 university students. The

mean test score was 75 with a standard deviation of 8. Assuming the distribution

is approximately normal, calculate the following:

1. the z-score for a test score of 82

2. the percentage of students who scored below 65

3. the percentage of students who scored between 70 and 80

Solution

1. To calculate the z-score for a test score of 82, we use the formula:

z=x−¯x

where: x= 82 (test score), ¯x= 75 (mean test score), σ= 8 (standard deviation).

Substitute the values into the formula:

z=82 −75

8=7

8= 0.875

So, the z-score for a test score of 82 is 0.875.

2. To find the percentage of students who scored below 65, we need to find

the z-score for 65 and then use a z-table to find the corresponding percentage.

The z-score for 65 can be calculated as:

z=65 −75

8=−10

8=−1.25

From the z-table, the percentage of students who scored below a z-score of

-1.25 is approximately 0.1056, or 10.56

3. To find the percentage of students who scored between 70 and 80, we

need to find the z-scores for 70 and 80, and then calculate the area between

them using the z-table. The z-scores for 70 and 80 are:

z70 =70 −75

8=−5

8=−0.625

z80 =80 −75

8=5

8= 0.625

Using the z-table, the area between -0.625 and 0.625 is approximately 0.2357,

or 23.57

Therefore, the percentage of students who scored between 70 and 80 is 23.57

Question 18

Question

Suppose we have a dataset representing the weights (in kg) of participants in two

different exercise programs. The mean weight for participants in Program A is

75 kg with a standard deviation of 5 kg, while the mean weight for participants

in Program B is 80 kg with a standard deviation of 4 kg.

Given that the dataset for Program A has a normal distribution and the

dataset for Program B has a skewed distribution, discuss the implications of

these differences in the context of conducting hypothesis tests and making in-

ferences about the populations these programs represent.

Solution

Step 1: Normal Distribution in Program A When the dataset for Program A has

a normal distribution, we can make certain assumptions about the population

and conduct parametric hypothesis tests with greater confidence. The mean

weight and standard deviation provide valuable information about the central