Chapter 12.1, Problem 1PS

To determine

What happens to the mean and the variance when we add 7 to every output x and what are the new sample mean, the new expected, and new variance.

Answer to Problem 1PS

New mean gets added by 7 to the original mean and new variance is equals to the original variance.

Explanation of Solution

Given:

Mean and variance

Calculation:

Let $x_1,x_2,x_3,\dots ,x_N$ be the $N$ output then mean of the N outputs is given by $m=\frac{x_1+x_2+\mathrm{............}+x_N}{N}\mathrm{...................}(1)$ and

Variance of the N outputs is given by ${\sigma }^2=\frac{{\displaystyle \sum {\left(X+7-m\right)}^2}}{N}\mathrm{......................}(2)$, where $m$ is mean of the N outputs.

After adding 7 to each of output, then we get $x_1+7,x_2\text{+ 7,}{x}_3+7,\dots \dots \mathrm{..}{x}_N+7$

New mean $m_{{}^1}=\frac{x_1+7+x_2+7+\mathrm{............}+x_N+7}{N}$

$\begin{array}{l}=\frac{x_1+x_2+\mathrm{............}+x_N}{N}+\frac{7N}{N}\\ m_{{}^1}=m+7\end{array}$

Thus, new mean gets added by 7 to the original mean.

The new variance ${\sigma }^2=\frac{{\displaystyle \sum {\left(X+7-m\right)}^2}}{N}$

Thus, every mean gets added by 7 and distance remains same.

Thus new variance is equals to the original variance.