Chapter 10.3, Problem 15E

Solution

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

The five-number summary of a data set is $\left\{10,33,42,48,73\right\}$ . The outlier are $10\text{ and }73$ .

Given information :

The stem and leaf plot is:

$1$	$07$
$2$	$139$
$3$	$023335$
$4$	$0112224458888$
$5$	$3567$
$6$	$7$
$7$	$3$

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 $30$ . So, the median will be the mean of the $15th$ and $16th$ number. So, the median is,

  $\begin{array}{c}\text{Median}=\left(\frac{42+42}{2}\right)\\ =42\end{array}$

  $Q_1$ will be the half of the first $15$ numbers. So,

  $Q_1=33$

  $Q_3$ will be the half of the first $15$ numbers. So,

  $Q_3=48$

So,

  $\begin{array}{c}\text{Minimum value}=10\\ Q_1=33\\ \text{Median}=42\\ Q_3=48\\ \text{Maximum value}=73\end{array}$

Therefore, the five-number summary of a data set is $\left\{10,33,42,48,73\right\}$ .

The Inter-Quartile Range is,

  $\begin{array}{c}IQR=48-33\\ =15\end{array}$

Now, the upper outlier is,

  $\begin{array}{c}{Q}_3+1.5IQR=48+1.5\left(15\right)\\ =48+22.5\\ =70.5\end{array}$

The lower outlier is,

  $\begin{array}{c}{Q}_1-1.5IQR=33-1.5\left(15\right)\\ =33-22.5\\ =10.5\end{array}$

Here, $10<10.5$ and $73>70.5$ .

Therefore, the outliers are $10\text{ and }73$ .