Chapter 12.3, Problem 3CB

To determine

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

Answer to Problem 3CB

  $\begin{array}{l}\text{Data set 1: Mean}=9.7,\text{ Standard deviation}=1.08\\ \text{Data set 2: Mean}=9.8,\text{ Standard deviation}=1.25\end{array}$

Explanation of Solution

Given information:

Two data sets are given as below −

  $\begin{array}{l}\text{Data set 1: 8}\text{.2,11}\text{.6,8}\text{.7,10}\text{.6,9}\text{.4,10}\text{.1,9}\text{.3}\\ \text{Data set 2: 9}\text{.3,10}\text{.2,8}\text{.1,12}\text{.3,8}\text{.7,9}\text{.9,10}\text{.1}\end{array}$

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 −

  $\begin{array}{l}\text{Data set 1: 8}\text{.2,11}\text{.6,8}\text{.7,10}\text{.6,9}\text{.4,10}\text{.1,9}\text{.3}\\ \text{Data set 2: 9}\text{.3,10}\text{.2,8}\text{.1,12}\text{.3,8}\text{.7,9}\text{.9,10}\text{.1}\end{array}$

For data set 1:

  $\text{Mean}=\frac{8.2+11.6+8.7+10.6+9.4+10.1+9.3}{7}=9.7$

$x$	$\overline{x}$	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$
$8.2$	$9.7$	$-1.5$	$2.25$
$11.6$	$9.7$	$1.9$	$3.71$
$8.7$	$9.7$	$-1.0$	$1.0$
$10.6$	$9.7$	$0.9$	$0.81$
$9.4$	$9.7$	$-0.3$	$0.09$
$10.1$	$9.7$	$0.4$	$0.16$
$9.3$	$9.7$	$-0.4$	$0.16$
			$\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}=\frac{2.25+3.71+1.00+0.81+0.09+0.16+0.16}{7}=1.1685$
			$\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}=\sqrt{1.1685}=1.08$

Similarly for data set $2$ :

  $\text{Mean}=\frac{9.3+10.2+8.1+12.3+8.7+9.9+10.1}{7}=9.8$

$x$	$\overline{x}$	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$
$9.3$	$9.8$	$0.5$	$0.25$
$10.2$	$9.8$	$0.4$	$0.16$
$8.1$	$9.8$	$-1.7$	$2.89$
$12.3$	$9.8$	$2.5$	$6.25$
$8.7$	$9.8$	$-1.1$	$1.21$
$9.9$	$9.8$	$0.1$	$0.01$
$10.1$	$9.8$	$0.3$	$0.09$
			$\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}=\frac{10.86}{7}=1.551$
			$\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}=\sqrt{1.551}=1.25$

Hence, for the given data sets −

  $\begin{array}{l}\text{Data set 1: Mean}=9.7,\text{ Standard deviation}=1.08\\ \text{Data set 2: Mean}=9.8,\text{ Standard deviation}=1.25\end{array}$