Chapter 10.2, Problem 12HP

To determine

To calculate:The mean and the interquartile range of a real-world data set. Also to compare and describe any differences in the mean and the interquartile range for with and without the outliers.

Answer to Problem 12HP

For the real-world data set

The mean with the outlier (60) = 17.5.

The interquartile range with the outlier (60) = 13.

The mean without the outlier (60) = 12.7.

The interquartile range without the outlier (60) = 11.5.

Explanation of Solution

Given information:The real-world data set must have at least 8 values and one or more outliers.

Formula used:

For a given set of data

  $\begin{array}{l}\text{First quartile}=\text{median of the lower half of the data}\text{.}\hfill \\ \hfill \\ \text{Third quartile}=\text{median of the upper half of the data}\text{.}\hfill \\ \hfill \\ \text{Interquartile Range}=\text{Third quartile}-\text{First quartile}\hfill \end{array}$

The Mean of a given set of values is the average of the given set of values.

  $Mean=\frac{\text{sum}\text{of}\text{values}}{\text{number}\text{of}\text{values}}$

The Median of a given set of values is the middle value of the ordered set.

  $Median=\left\{\begin{array}{l}{\left(\frac{n+1}{2}\right)}^{th}\text{term,}\text{if}n\text{is}\text{odd}\\ \\ \frac{{\left(\frac{n}{2}\right)}^{th}\text{term}+{\left(\frac{n}{2}+1\right)}^{th}\text{term}}{2},\text{if}n\text{is}\text{even}\end{array}\right\}$

Calculation:

The real-world data set is given below

Number of Cafe in different regions of a state are shown below

	Region 1	Region2	Region 3	Region 4	Region 5	Region 6	Region 7	Region 8	Region 9	Region 10
Number of Cafe	4	6	8	10	13	15	16	21	22	60

First the outliers are found in the given data, for that the interquartile range is required. So the outliers and the interquartile range are given by:

  $\begin{array}{c}\text{First quartile}=\text{median of the lower half of the data}\\ \text{=median}\left\{4,6,8,10,13\right\}\\ ={\left(\frac{5+1}{2}\right)}^{th}term=3^{rd}term\\ =8\\ \\ \text{Third quartile}=\text{median of the upper half of the data}\\ \text{=median}\left\{15,16,21,22,50\right\}\\ ={\left(\frac{5+1}{2}\right)}^{th}term=3^{rd}term\\ =21\\ \\ \text{Interquartile Range}=\text{Third quartile}-\text{First quartile}\\ \text{=21}-8=13\end{array}$

  $\begin{array}{l}\text{To}\text{find}\text{the}\text{outlier},\text{Mutltipy}\text{Interquartile}\text{Range}\text{by}\text{1}\text{.5}\\ \text{=1}\text{.5}\times \text{13=19}\text{.5}\\ \\ \text{then}\\ 8-19.5=-11.5\\ 21+19.5=40.5\\ \text{Since}\text{60}\text{is}\text{the}\text{only}\text{value}\text{that}\text{lies}\text{outside}\text{40}\text{.5}\text{therefore}\text{it}\text{is}\text{the}\text{outlier}\\ \text{Outlier}=60\end{array}$

The mean and the interquartile range with the outlier are

  $\begin{array}{c}Mean=\frac{\text{sum}\text{of}\text{values}}{\text{number}\text{of}\text{values}}=\frac{4+6+8+10+13+15+16+21+22+60}{10}\\ Mean=17.5\mathrm{.........}\left(1\right)\\ \\ \text{Interquartile Range}=\text{Third quartile}-\text{First quartile}\\ \text{=21}-8=13\mathrm{........}\left(2\right)\end{array}$

The mean and the interquartile range without the outlier are

  $\begin{array}{c}Mean=\frac{\text{sum}\text{of}\text{values}}{\text{number}\text{of}\text{values}}=\frac{4+6+8+10+13+15+16+21+22}{9}\\ Mean=12.7\mathrm{........}\left(3\right)\\ \\ \text{First quartile}=\text{median of the lower half of the data}\\ \text{=median}\left\{4,6,8,10\right\}\\ =\frac{{\left(\frac{4}{2}\right)}^{th}term+{\left(\frac{4}{2}+1\right)}^{th}term}{2}=\frac{2^{nd}term+3^{rd}term}{2}\\ =\frac{6+8}{2}=7\\ \\ \text{Third quartile}=\text{median of the upper half of the data}\\ \text{=median}\left\{15,16,21,22\right\}\\ =\frac{{\left(\frac{4}{2}\right)}^{th}term+{\left(\frac{4}{2}+1\right)}^{th}term}{2}=\frac{2^{nd}term+3^{rd}term}{2}\\ =\frac{16+21}{2}=18.5\\ \\ \text{Interquartile Range}=\text{Third quartile}-\text{First quartile}\\ \text{=18}\text{.5}-7=11.5\mathrm{.........}\left(4\right)\\ \end{array}$

By comparing equation (1) & (2) it can be observed that the mean has decreased by $17.5-12.7=4.8$ .

By comparing equation (3) & (4) it can be observed that the interquartile range has decreased by $13-11.5=1.5$ .

Hence for the real-world data set

The mean with the outlier(60) =17.5.

The interquartile range with the outlier (60) = 13.

The mean without the outlier (60) = 12.7.

The interquartile range without the outlier (60) = 11.5.