Chapter 10.2, Problem 6IP

To determine

To Find: To find the measures of variability and outliers if any of the given data.

Answer to Problem 6IP

The answer is

  $\text{Range}=22,\text{Inter-quartile}\text{range}=7.75,\text{Variance}=39.61\text{and}\text{Standard}\text{deviation}=6.29$

And no outlier exists for the given set of data.

Explanation of Solution

Given information:Number of volunteer hours per month.

Formula used:

  $\begin{array}{l}\text{For}\text{Range}=\text{Highest}\text{value}-\text{Lowest}\text{value}\\ \text{For Inter-Quartile Range}={\text{Q}}_{\text{3}}-{\text{Q}}_{\text{1}}\\ \text{For Variance}=\frac{{\displaystyle \sum _{i=1}^n{(x_i)}^2}}{n}-{\overline{x}}^2\\ \text{For Standard Deviation}=\sqrt{\text{Variance}}\\ \text{For}\text{Outliers}:\text{Data}\text{Value}<\text{}{\text{Q}}_{\text{1}}-\text{1}\text{.5(IQR)}\text{or}\text{Data}\text{Value}>{\text{Q}}_{\text{3}}+\text{1}\text{.5(IQR)}\end{array}$

Calculation:

Since, the given data is 38, 43, 36, 37, 32, 37, 29, and 51.

Arranging the data in ascending order

29, 32, 36, 37, 37, 38, 43, 51

Now, the range of the given data is

  $\text{Range}=51-29=22$

Also, first and third quartile are

  $\begin{array}{l}{\text{Q}}_{\text{1}}={\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{8+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{9}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={(}^{2.25}\text{observation}\\ {\text{Q}}_1=2^{\text{nd}}\text{observation}+0.25(3^{\text{rd}}\text{observation}-2^{\text{nd}}\text{observation})\\ {\text{Q}}_{\text{1}}=32+0.25(36-32)\\ {\text{Q}}_{\text{1}}=32+2=34\end{array}$

and

  $\begin{array}{l}{\text{Q}}_{\text{3}}=3{\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{8+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{9}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}={(}^{6.75}\text{observation}\\ {\text{Q}}_3=6^{\text{th}}\text{observation}+0.75(7^{\text{th}}\text{observation}-6^{\text{th}}\text{observation})\\ {\text{Q}}_{\text{3}}=38+0.75(43-38)\\ {\text{Q}}_{\text{3}}=38+3.75=41.75\end{array}$

Thus, Inter-quartile range is $Q_3-Q_1=41.75-34=7.75$

Calculation of Mean and Variance

  $\begin{array}{l}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{observations}}{\text{Number}\text{of}\text{observations}}\\ \text{Mean}=\frac{29+32+36+37+37+38+43+51}{8}\\ \text{Mean}=\frac{303}{8}\\ \text{Mean}=37.875\\ \text{Variance}=\frac{{\displaystyle \sum _{i=1}^n{(x_i)}^2}}{n}-{\overline{x}}^2,\text{where}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{{(29)}^2+{(32)}^2+{(36)}^2+{(37)}^2+{(37)}^2+{(38)}^2+{(43)}^2+{(51)}^2}{8}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{841+1024+1296+1369+1369+1444+1849+2601}{8}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{11793}{8}=1474.125\\ \text{Now,}\\ \text{Variance}=1474.125-{(}^{37.875}\\ \text{Variance}=1474.125-1434.516\\ \text{Variance}=39.61\end{array}$

Calculation of standard deviation

  $\begin{array}{l}\text{Standard Deviation}=\sqrt{\text{Variance}}\\ \text{Standard}\text{Deviation}=\sqrt{39.61}\\ \text{Standard}\text{Deviation}=6.29\end{array}$

Calculation of Outliers

  $\begin{array}{l}\text{Data}\text{Value}<{\text{Q}}_{\text{1}}-1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data}\text{Value}<34-1.5\times 7.75\\ \text{Data}\text{Value}<34-11.625\\ \text{Data}\text{Value}<22.375\end{array}$

Which is not exist in the given set of data, also

  $\begin{array}{l}\text{Data}\text{Value}>{\text{Q}}_{\text{3}}+1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data Value}>41.75+1.5\times 7.75\\ \text{Data}\text{Value}>41.75+11.625\\ \text{Data}\text{Value}>53.375\end{array}$

No value exists in the given data

Hence, the required measure of variability is

  $\text{Range}=22,\text{Inter-quartile}\text{range}=7.75,\text{Variance}=39.61\text{and}\text{Standard}\text{deviation}=6.29$

And no outlier exists for the given set of data.