Chapter 14.3, Problem 6CFU

To determine

To calculate: The mean deviation and the standard deviation of $\left\{\$4.45,\$5.50,\$5.50,\$6.30,\$7.80,\$11.00,\$12.20,\$17.20\right\}.$

Answer to Problem 6CFU

The standard deviation is $4.10619$ and mean deviation is $=3.5421875$ .

Explanation of Solution

Given information:

  $\left\{\$4.45,\$5.50,\$5.50,\$6.30,\$7.80,\$11.00,\$12.20,\$17.20\right\}.$

Formula used:

A measure of variability associated with the arithmetic mean is the standard deviation.

  $\sigma =\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}$

The arithmetic mean of the absolute values of the deviations from the mean of a set of data is called the mean deviation.

  $MD=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left|X_i-\overline{X}\right|}$

Formula of Mean

  $\overline{X}=\frac{{\displaystyle \sum X}}{n}$ where X represents each of the values in the data set.

Calculation:

Consider the data ,

  $\$4.45,\$5.50,\$5.50,\$6.30,\$7.80,\$11.00,\$12.20,\$17.20$

So, $n=8$ .

The mean of the data is

  $\begin{array}{l}=\frac{4.45+5.50+5.50+6.30+7.80+11.00+12.20+17.20}{8}\\ =\frac{69.95}{8}\\ =8.74375\end{array}$

Therefore , the standard deviation is −

$X_i$	$\overline{X}$	$X_i-\overline{X}$	${\left(X_i-\overline{X}\right)}^2$
4.45	8.74375	$\begin{array}{l}=4.45-8.74375\\ =-4.29375\end{array}$	$\begin{array}{l}={\left(-4.29375\right)}^2\\ =18.4362\end{array}$
5.50	8.74375	$\begin{array}{l}=5.50-8.74375\\ =-3.24375\end{array}$	$\begin{array}{l}={\left(-3.24375\right)}^2\\ =10.5219\end{array}$
5.50	8.74375	$\begin{array}{l}=5.50-8.74375\\ =-3.24375\end{array}$	$\begin{array}{l}={\left(-3.24375\right)}^2\\ =10.5219\end{array}$
6.30	8.74375	$\begin{array}{l}=6.30-8.74375\\ =-2.44375\end{array}$	$\begin{array}{l}={\left(-2.44375\right)}^2\\ =5.9719\end{array}$
7.80	8.74375	$\begin{array}{l}=7.80-8.74375\\ =-0.94375\end{array}$	$\begin{array}{l}={\left(-0.94375\right)}^2\\ =0.8906\end{array}$
11.00	8.74375	$\begin{array}{l}=11.00-8.74375\\ =2.25625\end{array}$	$\begin{array}{l}={\left(2.25625\right)}^2\\ =5.0906\end{array}$
12.20	8.74375	$\begin{array}{l}=12.20-8.74375\\ =3.45625\end{array}$	$\begin{array}{l}={\left(3.45625\right)}^2\\ =11.9456\end{array}$
17.20	8.74375	$\begin{array}{l}=17.20-8.74375\\ =8.45625\end{array}$	$\begin{array}{l}={\left(8.45625\right)}^2\\ =71.5081\end{array}$
${{\displaystyle \sum _{i=1}^8\left(X_i-\overline{X}\right)}}^2$			$134.8868$

  $\begin{array}{l}=\sigma =\sqrt{\frac{1}{8}\left(134.8868\right)}\\ =\sigma =\sqrt{16.86085}\\ =\sigma =4.10619\end{array}$

The mean deviation is −

$X_i$	$\overline{X}$	$\left|X_i-\overline{X}\right|$
4.45	8.74375	$\begin{array}{l}=\left|4.45-8.74375\right|\\ =4.29375\end{array}$
5.50	8.74375	$\begin{array}{l}=\left|5.50-8.74375\right|\\ =3.24375\end{array}$
5.50	8.74375	$\begin{array}{l}=\left|5.50-8.74375\right|\\ =3.24375\end{array}$
6.30	8.74375	$\begin{array}{l}=\left|6.30-8.74375\right|\\ =2.44375\end{array}$
7.80	8.74375	$\begin{array}{l}=\left|7.80-8.74375\right|\\ =0.94375\end{array}$
11.00	8.74375	$\begin{array}{l}=\left|11.00-8.74375\right|\\ =2.25625\end{array}$
12.20	8.74375	$\begin{array}{l}=\left|12.20-8.74375\right|\\ =3.45625\end{array}$
17.20	8.74375	$\begin{array}{l}=\left|17.20-8.74375\right|\\ =8.45625\end{array}$
${\displaystyle \sum _{i=1}^8\left|X_i-\overline{X}\right|}$		$28.3375$

  $\begin{array}{l}MD=\frac{1}{8}\left(28.3375\right)\\ =3.5421875\end{array}$

Hence, the standard deviation is $4.10619$ and mean deviation is $=3.5421875$ .