Chapter 10.3, Problem 16E

Solution

To calculate : The five-number summary of the given data and find the outliers.

The five-number summary of a data set is $\left\{80,105,110.5,116,134\right\}$ . The outlier are $80\text{ and }134$

Given information :

The stem and leaf plot is

$8$	$099$
$9$	$5$
$10$	$0567799$
$11$	$013455667$
$12$	$469$
$13$	$4$

Formula Used:

The five-number summary of a data set is the collection $\left\{\text{minimum, }{Q}_1\text{, median, }{Q}_3\text{, maximum}\right\}$ .

A number in a data set, which falls below $Q_1-1.5\left(\text{IQR}\right)$ or above $Q_3+1.5\left(\text{IQR}\right)$ are the outliers.

First quartile is $Q_1$ , third quartile is $Q_3$ , and IQR is $Q_3-Q_1$ .

Calculation :

Consider the given stem-leaf plot.

Here, total of frequency is $24$ . So, the median will be the mean of the $12th$ and $13th$ number. So, the median is,

  $\begin{array}{c}\text{Median}=\left(\frac{110+111}{2}\right)\\ =110.5\end{array}$

  $Q_1$ will be the half of the first $12$ numbers. So,

  $Q_1=105$

  $Q_3$ will be the half of the first $12$ numbers. So,

  $Q_3=116$

So,

  $\begin{array}{c}\text{Minimum value}=80\\ Q_1=105\\ \text{Median}=110.5\\ Q_3=116\\ \text{Maximum value}=134\end{array}$

Therefore, the five-number summary of a data set is $\left\{80,105,110.5,116,134\right\}$ .

The Inter-Quartile Range is,

  $\begin{array}{c}IQR=116-105\\ =11\end{array}$

Now, the upper outlier is,

  $\begin{array}{c}{Q}_3+1.5IQR=116+1.5\left(11\right)\\ =116+16.5\\ =132.5\end{array}$

The lower outlier is,

  $\begin{array}{c}{Q}_1-1.5IQR=105-1.5\left(11\right)\\ =105-16.5\\ =88.5\end{array}$

Here, $80<88.5$ and $134>132.5$ .

Therefore, the outliers are $80\text{ and }134$ .