Chapter 13.2, Problem 29E

To determine

To calculate: The mean $\left(\overline{x}\right)$ , variance $\left(v\right)$ and standard deviation $\left(\sigma \right)$ of the data set, $42,50,61,47,56,68$ .

Answer to Problem 29E

For the data set, $42,50,61,47,56,68$ , the mean $\overline{x}=54$ , variance $v=76.33$ and standard deviation $\sigma =8.74$ .

Explanation of Solution

Given information:

The data set, $42,50,61,47,56,68$ .

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, $42,50,61,47,56,68$ . There are six 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,

  $\begin{array}{c}\overline{x}=\frac{x_1+x_2+\cdots x_n}{n}\\ =\frac{42+50+61+47+56+68}{6}\\ =54\end{array}$

Therefore, mean of the data is 54.

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,

  $\begin{array}{c}v=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\cdots {\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{{\left(42-54\right)}^2+{\left(50-54\right)}^2+{\left(61-54\right)}^2+{\left(47-54\right)}^2+{\left(56-54\right)}^2+{\left(68-54\right)}^2}{6}\\ =\frac{458}{6}\\ =76.33\end{array}$

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,

  $\begin{array}{c}\sigma =\sqrt{v}\\ =\sqrt{76.33}\\ =8.74\end{array}$

Thus, for the data set, $42,50,61,47,56,68$ , the mean $\overline{x}=54$ , variance $v=76.33$ and standard deviation $\sigma =8.74$ .