Thinking Like an Engineer: An Active Learning Approach (4th Edition)
Thinking Like an Engineer: An Active Learning Approach (4th Edition)
4th Edition
ISBN: 9780134639673
Author: Elizabeth A. Stephan, David R. Bowman, William J. Park, Benjamin L. Sill, Matthew W. Ohland
Publisher: PEARSON
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Chapter 14, Problem 2ICA

For the following pressure data, recorded in units of pound-force per square inch, answer the following questions

Chapter 14, Problem 2ICA, For the following pressure data, recorded in units of pound-force per square inch, answer the

a. What is the mean of the data?

b. What is the median of the data?

c. What is the variance of the data?

d. What is the standard deviation of the data?

(a)

Expert Solution
Check Mark
To determine

Calculate the mean of the data provided.

Answer to Problem 2ICA

The mean of the data provided is 23.3 psi.

Explanation of Solution

Formula used:

Write the expression for mean.

Mean= Sum of all the data providedTotal number of data (1)

Calculation:

Find the sum of all the data provided.

Sum of all the data provided=(3psi+42psi+6psi+45psi+30psi+18psi+9psi+3psi+54psi)=210psi

Substitute 210psi for Sum of all the data provided and 9 for Total number of data in equation (1) to calculate the mean.

Mean=210psi9=23.3 psi

Conclusion:

Thus, the mean of the data provided is 23.3 psi.

(b)

Expert Solution
Check Mark
To determine

Calculate the median of the data provided.

Answer to Problem 2ICA

The median of the data provided is 18psi_.

Explanation of Solution

Calculation:

Median of the data set can be found out by arranging the data in ascending order and choosing the center number.

Arranging in ascending order yields the series.

3psi,3psi,6psi,9psi,18psi,30psi,42psi,45psi,54psi

The center value in an ascending order series is 18psi so the median is 18psi.

Conclusion:

Thus, the median of the data provided is 18psi_.

(c)

Expert Solution
Check Mark
To determine

Calculate the variance of the data provided.

Answer to Problem 2ICA

The variance of the data provided is 395.5psi2_.

Explanation of Solution

Formula used:

Write the expression for variance.

Variance=1N1×i=1N(MeanDatai)2 (2)

Calculation:

Substitute 9 for N in equation (2) to expand the expression.

Variance=191×i=19(MeanDatai)2=18((MeanData1)2+(MeanData2)2+(MeanData3)2+(MeanData4)2+(MeanData5)2+(MeanData6)2+(MeanData7)2+(MeanData8)2+(MeanData9)2)

Substitute 23.3 psi for Mean, 3 psi for Data1, 42 psi for Data2, 6 psi for Data3, 45 psi for Data4, 30 psi for Data5, 18 psi for Data6, 9 psi for Data7, 3 psi for Data8 and 54 psi for Data9 to calculate the variance.

Variance=18×((23.3psi3psi)2+(23.3psi42psi)2+(23.3psi6psi)2+(23.3psi45psi)2+(23.3psi30psi)2+(23.3psi18psi)2+(23.3psi9psi)2+(23.3psi3psi)2+(23.3psi54psi)2)=18×(412.09psi2+349.69psi2+299.29psi2+470.89psi2+44.89psi2+28.09psi2+204.49psi2+412.09psi2+942.49psi2)=395.5psi2

Conclusion:

Thus, the variance of the data provided is 395.5psi2_.

(d)

Expert Solution
Check Mark
To determine

Calculate the standard deviation of the data provided.

Answer to Problem 2ICA

The standard deviation of the data provided is 19.88psi_.

Explanation of Solution

Formula used:

Write the expression for standard deviation.

Standard deviation=Variance (3)

Calculation:

Substitute 395.5psi2 for Variance in equation (3) to obtain the value of standard deviation.

Standard deviation=395.5psi2=19.88psi

Conclusion:

Thus, the standard deviation of the data provided is 19.88psi_.

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