Chapter 13.2, Problem 26E

To determine

To calculate: The mean $\left(\overline{x}\right)$ , variance $\left(v\right)$ and standard deviation $\left(\sigma \right)$ of the data set, $2,12,4,7,5$ .

Answer to Problem 26E

For the data set, $2,12,4,7,5$ , the mean $\overline{x}=6$ , variance $v=11.6$ and standard deviation $\sigma =3.41$ .

Explanation of Solution

Given information:

The data set, $2,12,4,7,5$ .

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, variance is denoted by $v$ and mathematically expressed as, $v=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\cdots {\left(x_n-\overline{x}\right)}^2}{n}$ , where $\overline{x}$ is the mean of the data.

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, $\sigma =\sqrt{v}$ , where $v$ is the variance of the data.

Calculation:

Consider the data set, $2,12,4,7,5$ . 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,

  $\begin{array}{c}\overline{x}=\frac{x_1+x_2+\cdots x_n}{n}\\ =\frac{2+12+4+7+5}{5}\\ =\frac{30}{5}\\ =6\end{array}$

Therefore, mean of the data is 6.

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

For the provided data set, variance is evaluated as,

  $\begin{array}{c}v=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\cdots {\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{{\left(2-6\right)}^2+{\left(12-6\right)}^2+{\left(4-6\right)}^2+{\left(7-6\right)}^2+{\left(5-6\right)}^2}{4}\\ =\frac{58}{5}\\ =11.6\end{array}$

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, $\sigma =\sqrt{v}$ , where $v$ is the variance of the data.

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

  $\begin{array}{c}\sigma =\sqrt{v}\\ =\sqrt{11.6}\\ =3.41\end{array}$

Thus, for the data set, $2,12,4,7,5$ , the mean $\overline{x}=6$ , variance $v=11.6$ and standard deviation $\sigma =3.41$ .