Chapter 10.2, Problem 4IP

To determine

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

Answer to Problem 4IP

The answer is

  $\text{Range}=80,\text{Inter-quartile}\text{range}=15,\text{Variance}=571.522\text{and}\text{Standard}\text{deviation}=23.91$

and the outlier is 150

Explanation of Solution

Given information:Sales of computer games.

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 89, 90, 80, 100, 92, 104, and 150.

Arranging the data in ascending order

80, 89, 90, 92, 100, 104, 150.

Now, the range of the given data is

  $\text{Range}=150-80=70$

Also, first and third quartile are

  $\begin{array}{l}{\text{Q}}_{\text{1}}={\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{7+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{8}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={(}^2\text{observation}\\ {\text{Q}}_1=89\end{array}$

and

  $\begin{array}{l}{\text{Q}}_{\text{3}}=3{\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{7+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{8}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}={(}^6\text{observation}\\ {\text{Q}}_3=104\end{array}$

Thus, Inter-quartile range is $Q_3-Q_1=104-89=15$

Calculation of Mean and Variance

  $\begin{array}{l}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{observations}}{\text{Number}\text{of}\text{observations}}\\ \text{Mean}=\frac{89+90+80+100+92+104+150}{7}\\ \text{Mean}=\frac{705}{7}\\ \text{Mean}=100.714\\ \text{Variance}=\frac{{\displaystyle \sum _{i=1}^n{(x_i)}^2}}{n}-{\overline{x}}^2,\text{where}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{{(80)}^2+{(89)}^2+{(90)}^2+{(92)}^2+{(100)}^2+{(104)}^2+{(150)}^2}{7}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{6400+7921+8100+8464+10000+10816+22500}{7}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{74201}{7}=10600.143\\ \text{Now,}\\ \text{Variance}=10600.143-{(}^{100.143}\\ \text{Variance}=10600.143-10028.621\\ \text{Variance}=571.522\end{array}$

Calculation of standard deviation

  $\begin{array}{l}\text{Standard Deviation}=\sqrt{\text{Variance}}\\ \text{Standard}\text{Deviation}=\sqrt{571.522}\\ \text{Standard}\text{Deviation}=23.91\end{array}$

Calculation of Outliers

  $\begin{array}{l}\text{Data}\text{Value}<{\text{Q}}_{\text{1}}-1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data}\text{Value}<89-1.5\times 15\\ \text{Data}\text{Value}<89-22.5\\ \text{Data}\text{Value}<66.5\end{array}$

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

  $\begin{array}{l}\text{Data}\text{Value}>{\text{Q}}_{\text{3}}+1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data Value}>104+1.5\times 15\\ \text{Data}\text{Value}>104+22.5\\ \text{Data}\text{Value}>126.5\end{array}$

Since, $150>126.5$

Therefore, the outlier is 150.

Hence, the required measure of variability is

  $\text{Range}=80,\text{Inter-quartile}\text{range}=15,\text{Variance}=571.522\text{and}\text{Standard}\text{deviation}=23.91$and the outlier is 150