Chapter 12.3, Problem 1CB

To determine

To find: The mean and standard deviation of the given data set.

Answer to Problem 1CB

  $\begin{array}{l}\text{Data set 1: Mean}=6,\text{ Standard deviation}=3.16\\ \text{Data set 2: Mean}=6.875,\text{ Standard deviation}=3.06\end{array}$

Explanation of Solution

Given information:

Two data sets are given as below −

  $\begin{array}{l}\text{Data set 1: 4,8,5,12,3,9,5,2}\\ \text{Data set 2: 5,9,11,4,6,11,2,7}\end{array}$

Concepts used:

Standard deviation for a data set is given as −

  $\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x_1}\right)}^2}}{n}};\text{ where }x=\text{value in the data set, }\overline{x}=\text{mean of the data set,}\\ n=\text{number of values in the data set}\end{array}$

Calculation:

Given data sets are −

  $\begin{array}{l}\text{Data set 1: 4,8,5,12,3,9,5,2}\\ \text{Data set 2: 5,9,11,4,6,11,2,7}\end{array}$

For data set 1:

  $\text{Mean}=\frac{4+8+5+12+3+9+5+2}{8}=6$

$x$	$\overline{x}$	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$
$4$	$6$	$-2$	$4$
$8$	$6$	$2$	$4$
$5$	$6$	$1$	$1$
$12$	$6$	$6$	$36$
$3$	$6$	$-3$	$9$
$9$	$6$	$3$	$9$
$5$	$6$	$1$	$1$
$2$	$6$	$4$	$16$
			$\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}=\frac{4+4+1+36+9+9+1+16}{8}=10$
			$\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}=\sqrt{10}=3.16$

Similarly for data set $2$ :

  $\text{Mean}=\frac{5+9+11+4+6+11+2+7}{8}=6.875$

$x$	$\overline{x}$	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$
$5$	$6.875$	$-1.875$	$3.5156$
$9$	$6.875$	$2.125$	$4.5156$
$11$	$6.875$	$4.125$	$17.0156$
$4$	$6.875$	$-2.875$	$8.2656$
$6$	$6.875$	$-0.875$	$0.7656$
$11$	$6.875$	$4.125$	$17.0156$
$2$	$6.875$	$-4.875$	$23.7656$
$7$	$6.875$	$0.125$	$0.0156$
			$\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}=9.359375$
			$\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}=\sqrt{9.359375}=3.06$

Hence, for the given data sets −

  $\begin{array}{l}\text{Data set 1: Mean}=6,\text{ Standard deviation}=3.16\\ \text{Data set 2: Mean}=6.875,\text{ Standard deviation}=3.06\end{array}$