   # Screw Length (cm) Pipe Diameter (in.) 2.55 1.25 2.45 1.18 2.55 1.22 2.35 1.15 2.60 1.17 2.40 1.19 2.30 1.22 2.40 1.18 2.50 1.17 2.50 1.25 Determine the average, variance, and standard deviation for the following parts. The measured values are given in the accompanying table. ### Engineering Fundamentals: An Intro...

5th Edition
Saeed Moaveni
Publisher: Cengage Learning
ISBN: 9781305084766

#### Solutions

Chapter
Section ### Engineering Fundamentals: An Intro...

5th Edition
Saeed Moaveni
Publisher: Cengage Learning
ISBN: 9781305084766
Chapter 19, Problem 8P
Textbook Problem
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## Screw Length (cm) Pipe Diameter (in.) 2.55 1.25 2.45 1.18 2.55 1.22 2.35 1.15 2.60 1.17 2.40 1.19 2.30 1.22 2.40 1.18 2.50 1.17 2.50 1.25 Determine the average, variance, and standard deviation for the following parts. The measured values are given in the accompanying table.

To determine

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

### Explanation of Solution

Given data:

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

 Screw Length (cm) Pipe diameter (in.) 2.55 1.25 2.45 1.18 2.55 1.22 2.35 1.15 2.6 1.17 2.4 1.19 2.3 1.22 2.4 1.18 2.5 1.17 2.5 1.25

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

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