Chapter 10.2, Problem 2GP

To determine

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

Answer to Problem 2GP

The answer is

  $\text{Range}=14,\text{Inter-quartile}\text{range}=9,\text{Variance}=1161.143\text{and}\text{Standard}\text{deviation}=34.08$

and no outlier exists of the given set of data.

Explanation of Solution

Given information:Animal life spans in years.

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 6, 7, 12, 15, 15, 16, and 20.

Now, the range of the given data is

  $\text{Range}=20-6=14$

Also, first and third quartile are

  $\begin{array}{l}{\text{Q}}_{\text{1}}={\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{7+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{8}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={(}^2\text{observation}\\ {\text{Q}}_1=7\end{array}$

and

  $\begin{array}{l}{\text{Q}}_{\text{3}}=3{\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{7+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{8}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}={(}^6\text{observation}\\ {\text{Q}}_3=16\end{array}$

Thus, Inter-quartile range is $Q_3-Q_1=16-7=9$

Calculation of Mean and Variance

  $\begin{array}{l}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{observations}}{\text{Number}\text{of}\text{observations}}\\ \text{Mean}=\frac{6+7+12+15+15+16+20}{7}\\ \text{Mean}=\frac{91}{7}\\ \text{Mean}=13\\ \text{Variance}=\frac{{\displaystyle \sum _{i=1}^n{(x_i)}^2}}{n}-{\overline{x}}^2,\text{where}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{{(6)}^2+{(7)}^2+{(12)}^2+{(15)}^2+{(15)}^2+{(16)}^2+{(20)}^2}{7}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{36+49+144+225+225+256+400}{7}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{1335}{7}=992.143\\ \text{Now,}\\ \text{Variance}=992.143-{(}^{13}\\ \text{Variance}=992.143-169\\ \text{Variance}=1161.143\end{array}$

Calculation of standard deviation

  $\begin{array}{l}\text{Standard Deviation}=\sqrt{\text{Variance}}\\ \text{Standard}\text{Deviation}=\sqrt{1161.143}\\ \text{Standard}\text{Deviation}=34.08\end{array}$

Calculation of Outliers

  $\begin{array}{l}\text{Data}\text{Value}<{\text{Q}}_{\text{1}}-1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data}\text{Value}<7-1.5\times 9\\ \text{Data}\text{Value}<7-13.5\\ \text{Data}\text{Value}<-6.5\end{array}$

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

  $\begin{array}{l}\text{Data}\text{Value}>{\text{Q}}_{\text{3}}+1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data Value}>16+1.5\times 9\\ \text{Data}\text{Value}>16+13.5\\ \text{Data}\text{Value}>\text{29}\text{.5}\end{array}$

Not exist in the given data

Hence, the required measure of variability is

  $\text{Range}=14,\text{Inter-quartile}\text{range}=9,\text{Variance}=1161.143\text{and}\text{Standard}\text{deviation}=34.08$

And no outlier exists of the given set of data.