Chapter 15.2, Problem 18WE

To determine

Identify the affect on mean and the standard deviation.

Answer to Problem 18WE

Each will be doubled.

Explanation of Solution

Given information:

If each score in a set of scores was multiplied by $2$ points.

Calculation:

The mean is the average of the numbers.

  $\begin{array}{l}M=\frac{Sumofthescores}{Numberofscores}\\ =\frac{x_1+x_2+\mathrm{....}+x_n}{n}\end{array}$

If each score in the data set is multiplied by $2$ , then the mean of this new data set is given by:

  $\begin{array}{l}{M}^{\text{'}}=\frac{2\left(x_1\right)+2\left(x_2\right)+\mathrm{...}+2\left(x_n\right)}{n}\\ =\frac{2\left(x_1+x_2+\mathrm{...}+x_n\right)}{n}\\ =2\frac{x_1+x_2+\mathrm{...}+x_n}{n}\\ =2M\end{array}$

Hence, the mean of the new data set is twice the mean of the original data set. Hence if each score in a set of scores was multiplied by $2$ , then the mean will be doubled.

The standard deviation, $\sigma$ , for a set of $n$ numbers: $x_1,x_2,\mathrm{....},x_n,$ whose mean is $M$ is given by:

  $\sigma =\sqrt{\frac{{\left(x_1-M\right)}^2+{\left(x_2-M\right)}^2+\mathrm{.......}+{\left(x_n-M\right)}^2}{n}}$

If each of the scores is multiplied by $2$ , the mean of the new data set will be $2M$ and hence the standard deviation of the new data set will be:

  $\begin{array}{l}{\sigma }^{\text{'}}=\sqrt{\frac{{\left(2x_1-2M\right)}^2+{\left(2x_2-2M\right)}^2+\mathrm{.......}+{\left(2x_n-2M\right)}^2}{n}}\\ =\sqrt{\frac{4\left[{\left(x_1-M\right)}^2+{\left(x_2-M\right)}^2+\mathrm{...}+{\left(x_n-M\right)}^2\right]}{n}}\\ =2\sqrt{\frac{{\left(x_1-M\right)}^2+{\left(x_2-M\right)}^2+\mathrm{...}+{\left(x_n-M\right)}^2}{n}}\\ =2\sigma \end{array}$

Hence the standard deviation of the new data set is twice the standard deviation of the original data set.

Hence if each score in a set of scores was multiplied by $2$ , then both the mean and standard deviation will be doubled.