Chapter SH, Problem 13.3.3EP

To determine

To find:the range, interquartile range, and any outliers for the given data

Answer to Problem 13.3.3EP

The range is $39$ , the inter quartile range is $20.5$ and here outliers are $\text{none }$

Explanation of Solution

Given:

Ages of people in a restaurant:

Stem	Leaf
2	0 1 1 2 4 7 9
3	3 3 6 8 8 8
4	2 4 5 7 9 9
5	2 9

Calculation:

Range:

Range is the difference between greatest and smallest value.

  $\begin{array}{l}\text{Range}=59-20\\ =39\end{array}$

Hence, the range is 39

Inter quartile range:

Find the upper quartile and lower quartile.

Here median is 38 .

Divide the data into two halves.

Median of lower half is

  $LQ=\frac{24+27}{2}=25.5$

Median of upper half is

  $UQ=\frac{45+47}{2}=46$

Inter quartile range is

  $UQ-LQ=46-25.5=20.5$

Hence, the inter quartile range is 20.5

Outlier

Data that are more than 1.5 times the value of the interquartile range beyond the quartiles is called outlier.

Here inter quartile range is 20.5 .

  $20.5\times 1.5=30.75$

Now subtract 30.75 from the lower quartile and add to upper quartile.

  $\text{LQ}-30.75=-5.25$

  $\text{UQ}+30.75=76.75$

Thus the data which is not in the range of -5.25 to 76.75 is outlier.

Here outliers are $\text{none }$

Conclusion:

Therefore, the range is $39$ , the inter quartile range is $20.5$ and here outliers are $\text{none }$