Chapter 14, Problem 1E

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Answer 1

Question

Chapter 14, Problem 1E

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

Expert Solution & Answer

To determine

To find:

The mean, median, mode, and standard deviation using built-in functions.

Answer to Problem 1E

Solution:

The required answers are stated as follows.

Explanation of Solution

The given data is,

Prod. line A:15.94 15.98 15.94 16.16 15.86 15.86 15.90 15.88Prod. line B:15.96 15.94 16.02 16.10 15.92 16.00 15.96 16.02

The formula to calculate the mean of set of data is,

Mean(x¯)=x1+x2+...+xnn=∑i1nxin

Here, xi is the ith data value and n is the number of data values in the data set.

Substitute 15.94, 15.98 15.94,...,15.88 for x1,x2,x3,...,x8 respectively in the formula.

Here number of data in the given data set is 8, substitute 8 for n in the formula.

Mean(x¯)=x1+x2+...+xnn=15.94+15.98+15.94+16.16+15.86+15.86+15.90+15.888≈15.94

Mean of A is 15.94.

Substitute 15.96, 15.94, 16.02,......16.02 for x1,x2,x3,...,x8 respectively in the formula.

Mean(x¯)=x1+x2+...+xnn=15.96+15.94+16.02+16.10+15.92+16.00+15.96+16.028≈15.99

Mean of B is 15.99.

The formula to calculate the median of set of ordered data is,

Median={(n+12)th value, n=odd12[(n2)th value+(n+12)th value], n=even

n is the number of data values in the data set.

Arrange the given data in increasing order.

Prod. line A:15.86 15.86 15.88 15.90 15.94 15.94 15.98 16.16Prod. line B:15.92 15.94 15.96 15.96 16.00 16.02 16.02 16.10

Number of values in data set is 8, it is even.

Substitute 8 for n in the formula for even number of data set A.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.90+15.94]

Simplify further,

Median=12[15.90+15.94]=15.92

The median of A is 15.92.

Substitute 8 for n in the formula for even number of data set B.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.96+16.00]

Simplify further,

Median=12[15.96+16.00]=15.98

The median of B is 15.98.

The most frequently occurred data is the mode.

Here, each of the data 15.86 in A occur 2 times.

Mode of the given data set A are 15.86.

Here, each of the data 15.96 in B occur 2 times.

Mode of the given data set B are 15.96.

Since, the mean value of set B is closer to diameter 16 mm. Hence, set B is better than set A.

MATLAB Code:

A = [15.94, 15.98, 15.94, 16.16, 15.86, 15.86, 15.90, 15.88];

B = [15.96, 15.94, 16.02, 16.10, 15.92, 16.00, 15.96, 16.02];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

mode_A = mode(A)

mode_B = mode(B)

stdev_A = std(A)

stdev_B = std(B)

Save the MATLAB script with the name, Chapter14_54793_14_1E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 1

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 2

Therefore, the required program is stated above.

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Answer 2

Question

Chapter 14, Problem 1E

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

Expert Solution & Answer

To determine

To find:

The mean, median, mode, and standard deviation using built-in functions.

Answer to Problem 1E

Solution:

The required answers are stated as follows.

Explanation of Solution

The given data is,

Prod. line A:15.94 15.98 15.94 16.16 15.86 15.86 15.90 15.88Prod. line B:15.96 15.94 16.02 16.10 15.92 16.00 15.96 16.02

The formula to calculate the mean of set of data is,

Mean(x¯)=x1+x2+...+xnn=∑i1nxin

Here, xi is the ith data value and n is the number of data values in the data set.

Substitute 15.94, 15.98 15.94,...,15.88 for x1,x2,x3,...,x8 respectively in the formula.

Here number of data in the given data set is 8, substitute 8 for n in the formula.

Mean(x¯)=x1+x2+...+xnn=15.94+15.98+15.94+16.16+15.86+15.86+15.90+15.888≈15.94

Mean of A is 15.94.

Substitute 15.96, 15.94, 16.02,......16.02 for x1,x2,x3,...,x8 respectively in the formula.

Mean(x¯)=x1+x2+...+xnn=15.96+15.94+16.02+16.10+15.92+16.00+15.96+16.028≈15.99

Mean of B is 15.99.

The formula to calculate the median of set of ordered data is,

Median={(n+12)th value, n=odd12[(n2)th value+(n+12)th value], n=even

n is the number of data values in the data set.

Arrange the given data in increasing order.

Prod. line A:15.86 15.86 15.88 15.90 15.94 15.94 15.98 16.16Prod. line B:15.92 15.94 15.96 15.96 16.00 16.02 16.02 16.10

Number of values in data set is 8, it is even.

Substitute 8 for n in the formula for even number of data set A.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.90+15.94]

Simplify further,

Median=12[15.90+15.94]=15.92

The median of A is 15.92.

Substitute 8 for n in the formula for even number of data set B.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.96+16.00]

Simplify further,

Median=12[15.96+16.00]=15.98

The median of B is 15.98.

The most frequently occurred data is the mode.

Here, each of the data 15.86 in A occur 2 times.

Mode of the given data set A are 15.86.

Here, each of the data 15.96 in B occur 2 times.

Mode of the given data set B are 15.96.

Since, the mean value of set B is closer to diameter 16 mm. Hence, set B is better than set A.

MATLAB Code:

A = [15.94, 15.98, 15.94, 16.16, 15.86, 15.86, 15.90, 15.88];

B = [15.96, 15.94, 16.02, 16.10, 15.92, 16.00, 15.96, 16.02];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

mode_A = mode(A)

mode_B = mode(B)

stdev_A = std(A)

stdev_B = std(B)

Save the MATLAB script with the name, Chapter14_54793_14_1E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 1

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 2

Therefore, the required program is stated above.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 14, Problem 1E

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

Expert Solution & Answer

To determine

To find:

The mean, median, mode, and standard deviation using built-in functions.

Answer to Problem 1E

Solution:

The required answers are stated as follows.

Explanation of Solution

The given data is,

Prod. line A:15.94 15.98 15.94 16.16 15.86 15.86 15.90 15.88Prod. line B:15.96 15.94 16.02 16.10 15.92 16.00 15.96 16.02

The formula to calculate the mean of set of data is,

Mean(x¯)=x1+x2+...+xnn=∑i1nxin

Here, xi is the ith data value and n is the number of data values in the data set.

Substitute 15.94, 15.98 15.94,...,15.88 for x1,x2,x3,...,x8 respectively in the formula.

Here number of data in the given data set is 8, substitute 8 for n in the formula.

Mean(x¯)=x1+x2+...+xnn=15.94+15.98+15.94+16.16+15.86+15.86+15.90+15.888≈15.94

Mean of A is 15.94.

Substitute 15.96, 15.94, 16.02,......16.02 for x1,x2,x3,...,x8 respectively in the formula.

Mean(x¯)=x1+x2+...+xnn=15.96+15.94+16.02+16.10+15.92+16.00+15.96+16.028≈15.99

Mean of B is 15.99.

The formula to calculate the median of set of ordered data is,

Median={(n+12)th value, n=odd12[(n2)th value+(n+12)th value], n=even

n is the number of data values in the data set.

Arrange the given data in increasing order.

Prod. line A:15.86 15.86 15.88 15.90 15.94 15.94 15.98 16.16Prod. line B:15.92 15.94 15.96 15.96 16.00 16.02 16.02 16.10

Number of values in data set is 8, it is even.

Substitute 8 for n in the formula for even number of data set A.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.90+15.94]

Simplify further,

Median=12[15.90+15.94]=15.92

The median of A is 15.92.

Substitute 8 for n in the formula for even number of data set B.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.96+16.00]

Simplify further,

Median=12[15.96+16.00]=15.98

The median of B is 15.98.

The most frequently occurred data is the mode.

Here, each of the data 15.86 in A occur 2 times.

Mode of the given data set A are 15.86.

Here, each of the data 15.96 in B occur 2 times.

Mode of the given data set B are 15.96.

Since, the mean value of set B is closer to diameter 16 mm. Hence, set B is better than set A.

MATLAB Code:

A = [15.94, 15.98, 15.94, 16.16, 15.86, 15.86, 15.90, 15.88];

B = [15.96, 15.94, 16.02, 16.10, 15.92, 16.00, 15.96, 16.02];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

mode_A = mode(A)

mode_B = mode(B)

stdev_A = std(A)

stdev_B = std(B)

Save the MATLAB script with the name, Chapter14_54793_14_1E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 1

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 2

Therefore, the required program is stated above.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 14, Problem 1E

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

Expert Solution & Answer

To determine

To find:

The mean, median, mode, and standard deviation using built-in functions.

Answer to Problem 1E

Solution:

The required answers are stated as follows.

Explanation of Solution

The given data is,

Prod. line A:15.94 15.98 15.94 16.16 15.86 15.86 15.90 15.88Prod. line B:15.96 15.94 16.02 16.10 15.92 16.00 15.96 16.02

The formula to calculate the mean of set of data is,

Mean(x¯)=x1+x2+...+xnn=∑i1nxin

Here, xi is the ith data value and n is the number of data values in the data set.

Substitute 15.94, 15.98 15.94,...,15.88 for x1,x2,x3,...,x8 respectively in the formula.

Here number of data in the given data set is 8, substitute 8 for n in the formula.

Mean(x¯)=x1+x2+...+xnn=15.94+15.98+15.94+16.16+15.86+15.86+15.90+15.888≈15.94

Mean of A is 15.94.

Substitute 15.96, 15.94, 16.02,......16.02 for x1,x2,x3,...,x8 respectively in the formula.

Mean(x¯)=x1+x2+...+xnn=15.96+15.94+16.02+16.10+15.92+16.00+15.96+16.028≈15.99

Mean of B is 15.99.

The formula to calculate the median of set of ordered data is,

Median={(n+12)th value, n=odd12[(n2)th value+(n+12)th value], n=even

n is the number of data values in the data set.

Arrange the given data in increasing order.

Prod. line A:15.86 15.86 15.88 15.90 15.94 15.94 15.98 16.16Prod. line B:15.92 15.94 15.96 15.96 16.00 16.02 16.02 16.10

Number of values in data set is 8, it is even.

Substitute 8 for n in the formula for even number of data set A.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.90+15.94]

Simplify further,

Median=12[15.90+15.94]=15.92

The median of A is 15.92.

Substitute 8 for n in the formula for even number of data set B.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.96+16.00]

Simplify further,

Median=12[15.96+16.00]=15.98

The median of B is 15.98.

The most frequently occurred data is the mode.

Here, each of the data 15.86 in A occur 2 times.

Mode of the given data set A are 15.86.

Here, each of the data 15.96 in B occur 2 times.

Mode of the given data set B are 15.96.

Since, the mean value of set B is closer to diameter 16 mm. Hence, set B is better than set A.

MATLAB Code:

A = [15.94, 15.98, 15.94, 16.16, 15.86, 15.86, 15.90, 15.88];

B = [15.96, 15.94, 16.02, 16.10, 15.92, 16.00, 15.96, 16.02];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

mode_A = mode(A)

mode_B = mode(B)

stdev_A = std(A)

stdev_B = std(B)

Save the MATLAB script with the name, Chapter14_54793_14_1E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 1

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 2

Therefore, the required program is stated above.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Expert Solution & Answer

To determine

To find:

The mean, median, mode, and standard deviation using built-in functions.

Answer to Problem 1E

Solution:

The required answers are stated as follows.

Explanation of Solution

The given data is,

Prod. line A:15.94 15.98 15.94 16.16 15.86 15.86 15.90 15.88Prod. line B:15.96 15.94 16.02 16.10 15.92 16.00 15.96 16.02

The formula to calculate the mean of set of data is,

Mean(x¯)=x1+x2+...+xnn=∑i1nxin

Here, xi is the ith data value and n is the number of data values in the data set.

Substitute 15.94, 15.98 15.94,...,15.88 for x1,x2,x3,...,x8 respectively in the formula.

Here number of data in the given data set is 8, substitute 8 for n in the formula.

Mean(x¯)=x1+x2+...+xnn=15.94+15.98+15.94+16.16+15.86+15.86+15.90+15.888≈15.94

Mean of A is 15.94.

Substitute 15.96, 15.94, 16.02,......16.02 for x1,x2,x3,...,x8 respectively in the formula.

Mean(x¯)=x1+x2+...+xnn=15.96+15.94+16.02+16.10+15.92+16.00+15.96+16.028≈15.99

Mean of B is 15.99.

The formula to calculate the median of set of ordered data is,

Median={(n+12)th value, n=odd12[(n2)th value+(n+12)th value], n=even

n is the number of data values in the data set.

Arrange the given data in increasing order.

Prod. line A:15.86 15.86 15.88 15.90 15.94 15.94 15.98 16.16Prod. line B:15.92 15.94 15.96 15.96 16.00 16.02 16.02 16.10

Number of values in data set is 8, it is even.

Substitute 8 for n in the formula for even number of data set A.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.90+15.94]

Simplify further,

Median=12[15.90+15.94]=15.92

The median of A is 15.92.

Substitute 8 for n in the formula for even number of data set B.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.96+16.00]

Simplify further,

Median=12[15.96+16.00]=15.98

The median of B is 15.98.

The most frequently occurred data is the mode.

Here, each of the data 15.86 in A occur 2 times.

Mode of the given data set A are 15.86.

Here, each of the data 15.96 in B occur 2 times.

Mode of the given data set B are 15.96.

Since, the mean value of set B is closer to diameter 16 mm. Hence, set B is better than set A.

MATLAB Code:

A = [15.94, 15.98, 15.94, 16.16, 15.86, 15.86, 15.90, 15.88];

B = [15.96, 15.94, 16.02, 16.10, 15.92, 16.00, 15.96, 16.02];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

mode_A = mode(A)

mode_B = mode(B)

stdev_A = std(A)

stdev_B = std(B)

Save the MATLAB script with the name, Chapter14_54793_14_1E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 1

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 2

Therefore, the required program is stated above.

Answer 6

Expert Solution & Answer

To determine

To find:

The mean, median, mode, and standard deviation using built-in functions.

Answer to Problem 1E

Solution:

The required answers are stated as follows.

Explanation of Solution

The given data is,

Prod. line A:15.94 15.98 15.94 16.16 15.86 15.86 15.90 15.88Prod. line B:15.96 15.94 16.02 16.10 15.92 16.00 15.96 16.02

The formula to calculate the mean of set of data is,

Mean(x¯)=x1+x2+...+xnn=∑i1nxin

Here, xi is the ith data value and n is the number of data values in the data set.

Substitute 15.94, 15.98 15.94,...,15.88 for x1,x2,x3,...,x8 respectively in the formula.

Here number of data in the given data set is 8, substitute 8 for n in the formula.

Mean(x¯)=x1+x2+...+xnn=15.94+15.98+15.94+16.16+15.86+15.86+15.90+15.888≈15.94

Mean of A is 15.94.

Substitute 15.96, 15.94, 16.02,......16.02 for x1,x2,x3,...,x8 respectively in the formula.

Mean(x¯)=x1+x2+...+xnn=15.96+15.94+16.02+16.10+15.92+16.00+15.96+16.028≈15.99

Mean of B is 15.99.

The formula to calculate the median of set of ordered data is,

Median={(n+12)th value, n=odd12[(n2)th value+(n+12)th value], n=even

n is the number of data values in the data set.

Arrange the given data in increasing order.

Prod. line A:15.86 15.86 15.88 15.90 15.94 15.94 15.98 16.16Prod. line B:15.92 15.94 15.96 15.96 16.00 16.02 16.02 16.10

Number of values in data set is 8, it is even.

Substitute 8 for n in the formula for even number of data set A.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.90+15.94]

Simplify further,

Median=12[15.90+15.94]=15.92

The median of A is 15.92.

Substitute 8 for n in the formula for even number of data set B.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.96+16.00]

Simplify further,

Median=12[15.96+16.00]=15.98

The median of B is 15.98.

The most frequently occurred data is the mode.

Here, each of the data 15.86 in A occur 2 times.

Mode of the given data set A are 15.86.

Here, each of the data 15.96 in B occur 2 times.

Mode of the given data set B are 15.96.

Since, the mean value of set B is closer to diameter 16 mm. Hence, set B is better than set A.

MATLAB Code:

A = [15.94, 15.98, 15.94, 16.16, 15.86, 15.86, 15.90, 15.88];

B = [15.96, 15.94, 16.02, 16.10, 15.92, 16.00, 15.96, 16.02];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

mode_A = mode(A)

mode_B = mode(B)

stdev_A = std(A)

stdev_B = std(B)

Save the MATLAB script with the name, Chapter14_54793_14_1E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 1

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 2

Therefore, the required program is stated above.

Answer 7

Expert Solution & Answer

Answer 8

Expert Solution & Answer

Answer 9

To determine

To find:

The mean, median, mode, and standard deviation using built-in functions.

Answer 10

Answer to Problem 1E

Solution:

The required answers are stated as follows.

Answer 11

Explanation of Solution

The given data is,

Prod. line A:15.94 15.98 15.94 16.16 15.86 15.86 15.90 15.88Prod. line B:15.96 15.94 16.02 16.10 15.92 16.00 15.96 16.02

The formula to calculate the mean of set of data is,

Mean(x¯)=x1+x2+...+xnn=∑i1nxin

Here, xi is the ith data value and n is the number of data values in the data set.

Substitute 15.94, 15.98 15.94,...,15.88 for x1,x2,x3,...,x8 respectively in the formula.

Here number of data in the given data set is 8, substitute 8 for n in the formula.

Mean(x¯)=x1+x2+...+xnn=15.94+15.98+15.94+16.16+15.86+15.86+15.90+15.888≈15.94

Mean of A is 15.94.

Substitute 15.96, 15.94, 16.02,......16.02 for x1,x2,x3,...,x8 respectively in the formula.

Mean(x¯)=x1+x2+...+xnn=15.96+15.94+16.02+16.10+15.92+16.00+15.96+16.028≈15.99

Mean of B is 15.99.

The formula to calculate the median of set of ordered data is,

Median={(n+12)th value, n=odd12[(n2)th value+(n+12)th value], n=even

n is the number of data values in the data set.

Arrange the given data in increasing order.

Prod. line A:15.86 15.86 15.88 15.90 15.94 15.94 15.98 16.16Prod. line B:15.92 15.94 15.96 15.96 16.00 16.02 16.02 16.10

Number of values in data set is 8, it is even.

Substitute 8 for n in the formula for even number of data set A.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.90+15.94]

Simplify further,

Median=12[15.90+15.94]=15.92

The median of A is 15.92.

Substitute 8 for n in the formula for even number of data set B.

Median=12[(n2)th value+(n2+1)th value]=12[(82)th value+(82+1)th value]=12[4th value+5th value]=12[15.96+16.00]

Simplify further,

Median=12[15.96+16.00]=15.98

The median of B is 15.98.

The most frequently occurred data is the mode.

Here, each of the data 15.86 in A occur 2 times.

Mode of the given data set A are 15.86.

Here, each of the data 15.96 in B occur 2 times.

Mode of the given data set B are 15.96.

Since, the mean value of set B is closer to diameter 16 mm. Hence, set B is better than set A.

MATLAB Code:

A = [15.94, 15.98, 15.94, 16.16, 15.86, 15.86, 15.90, 15.88];

B = [15.96, 15.94, 16.02, 16.10, 15.92, 16.00, 15.96, 16.02];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

mode_A = mode(A)

mode_B = mode(B)

stdev_A = std(A)

stdev_B = std(B)

Save the MATLAB script with the name, Chapter14_54793_14_1E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 1

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 1E , additional homework tip 2

Therefore, the required program is stated above.