Chapter 10.2, Problem 1GP

To determine

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

Answer to Problem 1GP

The answer is

  $\text{Range}=145,\text{Inter-quartile}\text{range}=72.5,\text{Variance}=1896.484\text{and}\text{Standard}\text{deviation}=43.55$

And no outliers exist of the given set of data.

Explanation of Solution

Given information:Number of minutes spent in bike riding

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 120, 80, 170, 100, 120, 110, 180, and 35.

Arranging the given data in ascending order, then

35, 80, 100, 110, 120, 120, 170, 180.

Now, the range of the given data is

  $\text{Range}=180-35=145$

Also, first and third quartile are

  $\begin{array}{l}{\text{Q}}_{\text{1}}={\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{8+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{9}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={(}^{2.25}\text{observation}\\ {\text{Q}}_1=2^{\text{nd}}\text{observation}+0.25(3^{\text{rd}}\text{observation}-2^{\text{nd}}\text{observation})\\ {\text{Q}}_{\text{1}}=80+0.25(100-80)\\ {\text{Q}}_{\text{1}}=80+5=85\end{array}$

and

  $\begin{array}{l}{\text{Q}}_{\text{3}}=3{\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{8+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{9}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}={(}^{6.75}\text{observation}\\ {\text{Q}}_3=6^{\text{th}}\text{observation}+0.75(7^{\text{th}}\text{observation}-6^{\text{th}}\text{observation})\\ {\text{Q}}_{\text{3}}=120+0.75(170-120)\\ {\text{Q}}_{\text{3}}=120+37.5=157.5\end{array}$

Thus, Inter-quartile range is $Q_3-Q_1=157.5-85=72.5$

Calculation of Mean and Variance

  $\begin{array}{l}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{observations}}{\text{Number}\text{of}\text{observations}}\\ \text{Mean}=\frac{120+80+170+100+120+110+180+35}{8}\\ \text{Mean}=\frac{915}{8}\\ \text{Mean}=114.375\\ \text{Variance}=\frac{{\displaystyle \sum _{i=1}^n{(x_i)}^2}}{n}-{\overline{x}}^2,\text{where}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{{(120)}^2+{(80)}^2+{(170)}^2+{(100)}^2+{(120)}^2+{(110)}^2+{(180)}^2+{(35)}^2}{8}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{14400+6400+28900+10000+14400+12100+32400+1225}{8}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=14978.125\\ \text{Now,}\\ \text{Variance}=14978.125-{(}^{114.375}\\ \text{Variance}=14978.125-13081.641\\ \text{Variance}=1896.484\end{array}$

Calculation of standard deviation

  $\begin{array}{l}\text{Standard Deviation}=\sqrt{\text{Variance}}\\ \text{Standard}\text{Deviation}=\sqrt{1896.484}\\ \text{Standard}\text{Deviation}=43.55\end{array}$

Calculation of Outliers

  $\begin{array}{l}\text{Data}\text{Value}<{\text{Q}}_{\text{1}}-1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data}\text{Value}<85-1.5\times 72.5\\ \text{Data}\text{Value}<85-108.75\\ \text{Data}\text{Value}<-23.75\end{array}$

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

  $\begin{array}{l}\text{Data}\text{Value}>{\text{Q}}_{\text{3}}+1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data Value}>157.5+1.5\times 72.5\\ \text{Data}\text{Value}>157.5+108.75\\ \text{Data}\text{Value}>\text{266}\text{.25}\end{array}$

Not exist in the given data

Hence, the required measure of variability is

  $\text{Range}=145,\text{Inter-quartile}\text{range}=72.5,\text{Variance}=1896.484\text{and}\text{Standard}\text{deviation}=43.55$

And no outlier exists of the given set of data.