Chapter 10.2, Problem 8IP

To determine

To Find: To find the measures of variability of the given data.

Answer to Problem 8IP

The answer is

  $\text{Range}=45,\text{Inter-quartile}\text{range}=15.75,\text{Variance}=113.867\text{and}\text{Standard}\text{deviation}=10.671$

and the outlier does not exist in the given set of data.

Explanation of Solution

Given information:Scores in science test

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:

Arranging the given data in ascending order

64, 79, 82, 82, 85, 88, 88, 91, 91, 94, 97, 97, 97, 100, 100, 103, 106, 109.

Now, the range of the given data is

  $\text{Range}=109-64=45$

Also, first and third quartile are

  $\begin{array}{l}{\text{Q}}_{\text{1}}={\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{18+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{19}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={(}^{4.75}\text{observation}\\ {\text{Q}}_1=4^{\text{th}}\text{observation}+0.75(5^{\text{th}}\text{observation}-4^{\text{th}}\text{observation})\\ {\text{Q}}_{\text{1}}=82+0.75(85-82)\\ {\text{Q}}_{\text{1}}=82+2.25=84.25\end{array}$

and

  $\begin{array}{l}{\text{Q}}_{\text{3}}=3{\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{18+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{19}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}={(}^{14.25}\text{observation}\\ {\text{Q}}_3={14}^{\text{th}}\text{observation}+0.25({15}^{\text{th}}\text{observation}-{14}^{\text{th}}\text{observation})\\ {\text{Q}}_{\text{3}}=100+0.25(100-100)\\ {\text{Q}}_{\text{3}}=100+0=100\end{array}$

Thus, Inter-quartile range is $Q_3-Q_1=100-84.25=15.75$

Calculation of Mean and Variance

  $\begin{array}{l}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{observations}}{\text{Number}\text{of}\text{observations}}\\ \text{Mean}=\frac{\left(\begin{array}{l}64+79+82+82+85+88\\ +88+91+91+94+97+97\\ +97+100+100+103+106+109\end{array}\right)}{18}\\ \text{Mean}=\frac{1653}{18}\\ \text{Mean}=91.833\\ \text{Variance}=\frac{{\displaystyle \sum _{i=1}^n{(x_i)}^2}}{n}-{\overline{x}}^2,\text{where}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{\left[\begin{array}{l}{(}^{64}+{(}^{79}+{(}^{82}+{(}^{82}+{(}^{85}\\ +{(}^{88}+{(}^{88}+{(}^{91}+{(}^{91}+{(}^{94}\\ +{(}^{97}+{(}^{97}+{(}^{97}+{(}^{100}+{(}^{100}\\ +{(}^{103}+{(}^{106}+{(}^{109}\end{array}\right]}{18}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{\left(\begin{array}{l}4096+6241+6724+6724+7225\\ +7744+7744+8281+8281+8836\\ +9409+9409+9409+10000+10000\\ +10609+11236+11881\end{array}\right)}{18}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{153849}{18}=8547.167\\ \text{Now,}\\ \text{Variance}=8547.167-{(}^{91.833}\\ \text{Variance}=8547.167-8433.3\\ \text{Variance}=113.867\end{array}$

Calculation of standard deviation

  $\begin{array}{l}\text{Standard Deviation}=\sqrt{\text{Variance}}\\ \text{Standard}\text{Deviation}=\sqrt{113.867}\\ \text{Standard}\text{Deviation}=10.671\end{array}$

Calculation of Outliers

  $\begin{array}{l}\text{Data}\text{Value}<{\text{Q}}_{\text{1}}-1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data}\text{Value}<84.25-1.5\times 15.75\\ \text{Data}\text{Value}<84.25-23.625\\ \text{Data}\text{Value}<60.625\end{array}$

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

  $\begin{array}{l}\text{Data}\text{Value}>{\text{Q}}_{\text{3}}+1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data Value}>100+1.5\times 15.75\\ \text{Data}\text{Value}>100+23.625\\ \text{Data}\text{Value}>123.625\end{array}$

No value exists in the data

Hence, the required measure of variability is

  $\text{Range}=45,\text{Inter-quartile}\text{range}=15.75,\text{Variance}=113.867\text{and}\text{Standard}\text{deviation}=10.671$ and the outlier is not existing in the given set of data.