Chapter 10.2, Problem 21STP

To determine

To calculate: Find the measures of variability and any outliers for the set of data.

Answer to Problem 21STP

Therefore, $\text{range}=68,\text{median}=46,\text{First Quartile}=27,\text{Third Quartile}=\text{46,Inter-Quartile range=19,Outlier}=\text{80}$ is the solution of data set.

Explanation of Solution

Given information: The given set of data is,

  $30,62,35,80,12,24,30,39,53,38$

Formula Used:

  $\begin{array}{l}\text{mean}=\frac{\text{sum}\text{of}\text{the}\text{data}}{\text{number}\text{of}\text{data}}\\ \text{median}={\left\{\frac{\left(n+1\right)}{2}\right\}}^{th}\text{term}\text{ for odd number of data}\\ \text{median}=\text{mean}\text{of}{\left\{\frac{\left(n\right)}{2}\right\}}^{th}\text{term}+{\left\{\frac{\left(n+1\right)}{2}\right\}}^{th}\text{term}\text{ for even number of data}\\ \text{mode}=\text{maximum}\text{occurence}\text{of}\text{particular}\text{data}\end{array}$

Calculation: The given data can be evaluated as,

  $\begin{array}{l}30,62,35,80,12,24,30,39,53,38\\ \text{Rearranging the data in ascending order,}\\ \text{12,24,30,30,35,38,39,53,62,80}\\ \text{range}=80-12=68\\ \text{median}=\frac{39+53}{2}\\ \text{median}=46\\ \text{12,24,30,30,35,38,39,53,62,80}\\ \text{First Quartile=}\frac{24+30}{2}\\ \text{First Quartile=}27\\ \text{Third Quartile=}\frac{39+53}{2}\\ \text{Third Quartile=46}\\ \text{Inter-Quartile range=46-27}\\ \text{Inter-Quartile range=19}\\ \text{Calculation for Outliers,}\\ \text{Inter-Quartile range}\times 1.5=19\times 1.5=28.5\\ \text{Determine Outliers,}\\ \text{27-28}\text{.5=}-1.5\\ 46+28.5=74.5\\ \text{Therefore, 80 is outlier}\text{.}\end{array}$

Therefore, $\text{range}=68,\text{median}=46,\text{First Quartile}=27,\text{Third Quartile}=\text{46,Inter-Quartile range=19,Outlier}=\text{80}$ is the solution of data set.