Chapter 12.3, Problem 2CB

To determine

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

Answer to Problem 2CB

  $\begin{array}{l}\text{Data set 1: Mean}=100,\text{ Standard deviation}=7.23\\ \text{Data set 2: Mean}=95.833,\text{ Standard deviation}=4.81\end{array}$

Explanation of Solution

Given information:

Two data sets are given as below −

  $\begin{array}{l}\text{Data set 1: 102,98,103,86,101,110}\\ \text{Data set 2: 90,89,100,97,102,97}\end{array}$

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 −

  $\begin{array}{l}\text{Data set 1: 102,98,103,86,101,110}\\ \text{Data set 2: 90,89,100,97,102,97}\end{array}$

For data set 1:

  $\text{Mean}=\frac{102+98+103+86+101+110}{6}=100$

$x$	$\overline{x}$	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$
$102$	$100$	$2$	$4$
$98$	$100$	$-2$	$4$
$103$	$100$	$3$	$9$
$86$	$100$	$-14$	$196$
$101$	$100$	$1$	$1$
$110$	$100$	$10$	$100$
			$\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}=\frac{4+4+9+196+1+100}{6}=52.33$
			$\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}=\sqrt{52.33}=7.23$

Similarly for data set $2$ :

  $\text{Mean}=\frac{90+89+100+97+102+97}{6}=95.833$

$x$	$\overline{x}$	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$
$90$	$95.833$	$-5.833$	$34$
$89$	$95.833$	$-6.833$	$46.689$
$100$	$95.833$	$4.167$	$17.363$
$97$	$95.833$	$1.167$	$1.361$
$102$	$95.833$	$6.167$	$38.031$
$97$	$95.833$	$1.167$	$1.361$
			$\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}=\frac{138.805}{6}=23.13$
			$\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}=\sqrt{23.13}=4.81$

Hence, for the given data sets −

  $\begin{array}{l}\text{Data set 1: Mean}=100,\text{ Standard deviation}=7.23\\ \text{Data set 2: Mean}=95.833,\text{ Standard deviation}=4.81\end{array}$