Engineering Fundamentals
Engineering Fundamentals
6th Edition
ISBN: 9780357112144
Author: Saeed Moaveni
Publisher: MISC PUBS
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Chapter 19, Problem 8P
To determine

Calculate the mean, standard deviation, and variance for the given measured values.

Expert Solution & Answer
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Answer to Problem 8P

The mean, variance, and standard deviation for the given values of screw length are 2.46, 0.00933, and 0.09661 respectively.

The mean, variance, and standard deviation for the given values of pipe diameter are 1.198, 0.00122, and 0.03490 respectively.

Explanation of Solution

Given data:

The given measured values of screw length and pipe diameter are shown below.

Screw Length (cm)Pipe diameter (in.)
2.551.25
2.451.18
2.551.22
2.351.15
2.61.17
2.41.19
2.31.22
2.41.18
2.51.17
2.51.25

The total number of measured values, n=10.

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

x¯=x1+x2+x3+............+xn1+xnn=1ni=1nxi (1)

Here,

x¯ is the mean,

xi is the data points,

n is the number of data points.

From equation 19.5 in the textbook, the formula to find the variance is,

v=i=1n(xix¯)2n1 (2)

From equation 19.6 in the textbook, the formula to find standard deviation is,

s=i=1n(xix¯)2n1 (3)

Calculation:

Calculation for Screw length:

Substitute all the value of screw length for xi up to the range n, and 10 for n in equation (1) to calculate mean (x¯),

x¯=2.55+2.45+2.55+2.35+2.60+2.40+2.30+2.40+2.50+2.5010=24.610x¯=2.46

Substitute all the value of screw length for xi up to the range n, 2.46 for x¯, and 10 for n in equation (2) to find variance (v),

v=(2.552.46)2+(2.452.46)2+(2.552.46)2+(2.352.46)2+(2.602.46)2+(2.402.46)2+(2.302.46)2+(2.402.46)2+(2.502.46)2+(2.502.46)2101=0.0849v=0.00933

Substitute all the value of screw length for xi up to the range n, 2.46 for x¯, and 10 for n in equation (3) to find standard deviation (s),

s=(2.552.46)2+(2.452.46)2+(2.552.46)2+(2.352.46)2+(2.602.46)2+(2.402.46)2+(2.302.46)2+(2.402.46)2+(2.502.46)2+(2.502.46)2101

s=0.0849s=0.00933s=0.09661

Calculation for pipe diameter:

Substitute all the value of pipe diameter for xi up to the range n, and 10 for n in equation (1) to calculate mean (x¯),

x¯=1.25+1.18+1.22+1.15+1.17+1.19+1.22+1.18+1.17+1.2510=11.9810x¯=1.198

Substitute all the value of pipe diameter for xi up to the range n, 1.198 for x¯, and 10 for n in equation (2) to find variance (v),

v=(1.251.198)2+(1.181.198)2+(1.221.198)2+(1.151.198)2+(1.171.198)2+(1.191.198)2+(1.221.198)2+(1.181.198)2+(1.171.198)2+(1.251.198)2101=0.010969v=0.00122

Substitute all the value of pipe diameter for xi up to the range n, 1.198 for x¯, and 10 for n in equation (3) to find standard deviation (s),

s=(1.251.198)2+(1.181.198)2+(1.221.198)2+(1.151.198)2+(1.171.198)2+(1.191.198)2+(1.221.198)2+(1.181.198)2+(1.171.198)2+(1.251.198)2101=0.010969=0.00122s=0.0349

Therefore, the mean, variance, and standard deviation for the given values of screw length are 2.46, 0.00933, and 0.09661 respectively, and the mean, variance, and standard deviation for the given values of pipe diameter are 1.198, 0.00122, and 0.03490 respectively.

Conclusion:

Thus, the mean, variance, and standard deviation for the given values of screw length are 2.46, 0.00933, and 0.09661 respectively, and the mean, variance, and standard deviation for the given values of pipe diameter are 1.198, 0.00122, and 0.03490 respectively.

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