Chapter 13.2, Problem 36E

To determine

To calculate: The standard deviation of the data set, $9.2,10.6,7.2,4.3,7.0$ .

Answer to Problem 36E

The standard deviation of the data set, $9.2,10.6,7.2,4.3,7.0$ is $2.14$ .

Explanation of Solution

Given information:

The data set, $9.2,10.6,7.2,4.3,7.0$ .

Formula used:

For the data set $\left\{x_1,x_2,x_3,\dots x_n\right\}$ that consists of n data points,mean is denoted by $\overline{x}$ and mathematically expressed as, $\overline{x}=\frac{x_1+x_2+\cdots x_n}{n}$ .

For the data set $\left\{x_1,x_2,x_3,\dots x_n\right\}$ that consists of n data points, standard deviation is denoted by $\sigma$ and mathematically expressed as, $\sigma =\sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}-{\overline{x}}^2}$ , where $\overline{x}$ is the mean of the data.

Calculation:

Consider the data set, $9.2,10.6,7.2,4.3,7.0$ . There are five data points.

Recall that for the data set $\left\{x_1,x_2,x_3,\dots x_n\right\}$ that consists of n data points,mean is denoted by $\overline{x}$ and mathematically expressed as, $\overline{x}=\frac{x_1+x_2+\cdots x_n}{n}$ .

For the provided data set, mean is evaluated as,

  $\begin{array}{c}\overline{x}=\frac{x_1+x_2+\cdots x_n}{n}\\ =\frac{9.2+10.6+7.2+4.3+7.0}{5}\\ =\frac{383}{50}\\ =7.66\end{array}$

Therefore, mean of the data is $7.66$ .

Next, recall that for the data set $\left\{x_1,x_2,x_3,\dots x_n\right\}$ that consists of n data points, standard deviation is denoted by $\sigma$ and mathematically expressed as, $\sigma =\sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}-{\overline{x}}^2}$ , where $\overline{x}$ is the mean of the data.

For the provided data set, standard deviation is evaluated as,

  $\begin{array}{c}\sigma =\sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}-{\overline{x}}^2}\\ =\sqrt{\frac{9.2+10.6+7.2+4.3+7.0}{5}-{\left(7.66\right)}^2}\\ =\sqrt{4.5904}\\ =2.14\end{array}$

Thus, the standard deviation of the data set, $9.2,10.6,7.2,4.3,7.0$ is $2.14$ approximately.