Chapter 1.6, Problem 13E

a.

To determine

Verify that the mean value, variance, and standard deviation of the statistical population $y_1,\dots ,y_N$ are $\overline{y}=c_1+\overline{x}$, $S_y^2=S_x^2$, and $S_y=S_x$, respectively.

Explanation of Solution

Here, $y_i=c_1+x_i$.

The mean value of y is obtained as follows:

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}\\ =\frac{{\displaystyle \sum _{i=1}^n\left(c_1+x_i\right)}}{n}\\ =\frac{n{c}_1+{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =c_1+\overline{x}\end{array}$

The variance of y is obtained as follows:

$\begin{array}{c}{S}_y^2=\frac{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}{n-1}\\ =\frac{{\displaystyle \sum _{i=1}^n{\left(\left(c_1+x_i\right)-\left(c_1+\overline{x}\right)\right)}^2}}{n-1}\\ =\frac{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}\\ =S_x^2\end{array}$

The standard deviation of y is obtained as follows:

$\begin{array}{c}{S}_y=\sqrt{S_y^2}\\ =\sqrt{S_x^2}\\ =S_x\end{array}$

Hence, it is verified that the mean value, variance, and standard deviation of the statistical population $y_1,\dots ,y_N$ are $\overline{y}=c_1+\overline{x}$, $S_y^2=S_x^2$, and $S_y=S_x$, respectively.

b.

To determine

Verify that the mean value, variance, and standard deviation of the statistical population $y_1,\dots ,y_N$ are $\overline{y}=c_2\overline{x}$, $S_y^2=c_2^2{S}_x^2$, and $S_y=\left|c_2\right|S_x$, respectively.

Explanation of Solution

Here, $y_i=c_2{x}_i$.

The mean value of y is obtained as follows:

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}\\ =\frac{{\displaystyle \sum _{i=1}^n\left(c_2{x}_i\right)}}{n}\\ =\frac{n{c}_2{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =c_2\overline{x}\end{array}$

The variance of y is obtained as follows:

$\begin{array}{c}{S}_y^2=\frac{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}{n-1}\\ =\frac{{\displaystyle \sum _{i=1}^n{\left(c_2{x}_i-c_2\overline{x}\right)}^2}}{n-1}\\ =\frac{c_2^2{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}\\ =c_2^2{S}_x^2\end{array}$

The standard deviation of y is obtained as follows:

$\begin{array}{c}{S}_y=\sqrt{S_y^2}\\ =\sqrt{c_2^2{S}_x^2}\\ =\left|c_2\right|S_x\end{array}$

Hence, it is verified that the mean value, variance, and standard deviation of the statistical population $y_1,\dots ,y_N$ are $\overline{y}=c_2\overline{x}$, $S_y^2=c_2^2{S}_x^2$, and $S_y=\left|c_2\right|S_x$, respectively.

c.

To determine

Verify that the mean value, variance, and standard deviation of the statistical population $y_1,\dots ,y_N$ are $\overline{y}=c_1+c_2\overline{x}$, $S_y^2=c_2^2{S}_x^2$, and $S_y=\left|c_2\right|S_x$, respectively.

Explanation of Solution

Here, $y_i=c_1+c_2{x}_i$.

The mean value of y is obtained as follows:

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}\\ =\frac{{\displaystyle \sum _{i=1}^n\left(c_1+c_2{x}_i\right)}}{n}\\ =\frac{n{c}_1+c_2{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =c_1+c_2\overline{x}\end{array}$

The variance of y is obtained as follows:

$\begin{array}{c}{S}_y^2=\frac{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}{n-1}\\ =\frac{{\displaystyle \sum _{i=1}^n{\left(\left(c_1+c_2{x}_i\right)-\left(c_1+c_2\overline{x}\right)\right)}^2}}{n-1}\\ =\frac{c_2^2{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}\\ =c_2^2{S}_x^2\end{array}$

The standard deviation of y is obtained as follows:

$\begin{array}{c}{S}_y=\sqrt{S_y^2}\\ =\sqrt{c_2^2{S}_x^2}\\ =\left|c_2\right|S_x\end{array}$

Hence, it is verified that the mean value, variance, and standard deviation of the statistical population $y_1,\dots ,y_N$ are $\overline{y}=c_1+c_2\overline{x}$, $S_y^2=c_2^2{S}_x^2$, and $S_y=\left|c_2\right|S_x$, respectively.