Chapter 13, Problem 8CT

To determine

To calculate: The mean $\left(\overline{x}\right)$ , variance $\left(v\right)$ and standard deviation $\left(\sigma \right)$ of the data set, $0,1,2,3,4,5,6$ .

Answer to Problem 8CT

For the data set, $0,1,2,3,4,5,6$ , the mean $\overline{x}=3$ , variance $v=4$ and standard deviation $\sigma =2$ .

Explanation of Solution

Given information:

The data set, $0,1,2,3,4,5,6$ .

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, $0,1,2,3,4,5,6$ . There are seven 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,

  $\begin{array}{c}\overline{x}=\frac{x_1+x_2+\cdots x_n}{n}\\ =\frac{0+1+2+3+4+5+6}{7}\\ =\frac{6}{2}\\ =3\end{array}$

Therefore, mean of the data is 3.

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,

  $\begin{array}{c}v=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\cdots {\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{{\left(0-3\right)}^2+{\left(1-3\right)}^2+{\left(2-3\right)}^2+{\left(3-3\right)}^2+{\left(4-3\right)}^2+{\left(5-3\right)}^2+{\left(6-3\right)}^2}{7}\\ =\frac{28}{7}\\ =4\end{array}$

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,

  $\begin{array}{c}\sigma =\sqrt{v}\\ =\sqrt{4}\\ =2\end{array}$

Thus, for the data set, $0,1,2,3,4,5,6$ , the mean $\overline{x}=3$ , variance $v=4$ and standard deviation $\sigma =2$ .