Chapter 19, Problem 8P

To determine

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

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.)
2.55	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,

$\overline{x}=\frac{x_1+x_2+x_3+\mathrm{............}+x_{n-1}+x_n}{n}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$ (1)

Here,

$\overline{x}$ is the mean,

$x_i$ 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,

$v=\frac{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}$ (2)

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

$s=\sqrt{\frac{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}}$ (3)

Calculation:

Calculation for Screw length:

Substitute all the value of screw length for $x_i$ up to the range $n$, and $10$ for $n$ in equation (1) to calculate mean $\left(\overline{x}\right)$,

$\begin{array}{c}\overline{x}=\frac{2.55+2.45+2.55+2.35+2.60+2.40+2.30+2.40+2.50+2.50}{10}\\ =\frac{24.6}{10}\\ \overline{x}=2.46\end{array}$

Substitute all the value of screw length for $x_i$ up to the range $n$, $2.46$ for $\overline{x}$, and $10$ for $n$ in equation (2) to find variance $\left(v\right)$,

$\begin{array}{c}v=\frac{\begin{array}{l}{\left(2.55-2.46\right)}^2+{\left(2.45-2.46\right)}^2+{\left(2.55-2.46\right)}^2+{\left(2.35-2.46\right)}^2+{\left(2.60-2.46\right)}^2+\\ {\left(2.40-2.46\right)}^2+{\left(2.30-2.46\right)}^2+{\left(2.40-2.46\right)}^2+{\left(2.50-2.46\right)}^2+{\left(2.50-2.46\right)}^2\end{array}}{10-1}\\ =\frac{0.084}{9}\\ v=0.00933\end{array}$

Substitute all the value of screw length for $x_i$ up to the range $n$, $2.46$ for $\overline{x}$, and $10$ for $n$ in equation (3) to find standard deviation $\left(s\right)$,

$s=\sqrt{\frac{\begin{array}{l}{\left(2.55-2.46\right)}^2+{\left(2.45-2.46\right)}^2+{\left(2.55-2.46\right)}^2+{\left(2.35-2.46\right)}^2+{\left(2.60-2.46\right)}^2+\\ {\left(2.40-2.46\right)}^2+{\left(2.30-2.46\right)}^2+{\left(2.40-2.46\right)}^2+{\left(2.50-2.46\right)}^2+{\left(2.50-2.46\right)}^2\end{array}}{10-1}}$

$\begin{array}{c}s=\sqrt{\frac{0.084}{9}}\\ s=\sqrt{0.00933}\\ s=0.09661\end{array}$

Calculation for pipe diameter:

Substitute all the value of pipe diameter for $x_i$ up to the range $n$, and $10$ for $n$ in equation (1) to calculate mean $\left(\overline{x}\right)$,

$\begin{array}{c}\overline{x}=\frac{1.25+1.18+1.22+1.15+1.17+1.19+1.22+1.18+1.17+1.25}{10}\\ =\frac{11.98}{10}\\ \overline{x}=1.198\end{array}$

Substitute all the value of pipe diameter for $x_i$ up to the range $n$, $1.198$ for $\overline{x}$, and $10$ for $n$ in equation (2) to find variance $\left(v\right)$,

$\begin{array}{c}v=\frac{\begin{array}{l}{\left(1.25-1.198\right)}^2+{\left(1.18-1.198\right)}^2+{\left(1.22-1.198\right)}^2+{\left(1.15-1.198\right)}^2+{\left(1.17-1.198\right)}^2+\\ {\left(1.19-1.198\right)}^2+{\left(1.22-1.198\right)}^2+{\left(1.18-1.198\right)}^2+{\left(1.17-1.198\right)}^2+{\left(1.25-1.198\right)}^2\end{array}}{10-1}\\ =\frac{0.01096}{9}\\ v=0.00122\end{array}$

Substitute all the value of pipe diameter for $x_i$ up to the range $n$, $1.198$ for $\overline{x}$, and $10$ for $n$ in equation (3) to find standard deviation $\left(s\right)$,

$\begin{array}{c}s=\sqrt{\frac{\begin{array}{l}{\left(1.25-1.198\right)}^2+{\left(1.18-1.198\right)}^2+{\left(1.22-1.198\right)}^2+{\left(1.15-1.198\right)}^2+{\left(1.17-1.198\right)}^2+\\ {\left(1.19-1.198\right)}^2+{\left(1.22-1.198\right)}^2+{\left(1.18-1.198\right)}^2+{\left(1.17-1.198\right)}^2+{\left(1.25-1.198\right)}^2\end{array}}{10-1}}\\ =\sqrt{\frac{0.01096}{9}}\\ =\sqrt{0.00122}\\ s=0.0349\end{array}$

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.