Chapter 10.3, Problem 13E

Solution

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

The five-number summary of a data set is $\left\{1,2,5,5,8\right\}$ . There is no outlier.

Given information :

The data set is:

Megapixels	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency	$2$	$2$	$0$	$1$	$12$	$0$	$2$	$6$

Formula Used:

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

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 table and find the cumulative frequency.

Megapixels	Frequency	Cumulative Frequency
$\begin{array}{l}1\\ 2\\ 3\\ 4\\ 5\\ 6\\ 7\\ 8\end{array}$	$\begin{array}{l}2\\ 2\\ 0\\ 1\\ 12\\ 0\\ 2\\ 6\end{array}$	$\begin{array}{l}2\\ 4\\ 4\\ 5\\ 17\\ 17\\ 19\\ 25\end{array}$

Here, total of frequency is $25$ . So, the median with be,

  $\begin{array}{c}\text{Median}=\left(\frac{15+1}{2}\right)th\text{ number}\\ =8th\text{ number}\\ =5\end{array}$

  $\begin{array}{c}{Q}_1=\left(\frac{15+1}{4}\right)th\text{ number}\\ =4th\text{ number}\\ =2\end{array}$

  $\begin{array}{c}{Q}_3=3\left(\frac{15+1}{4}\right)th\text{ number}\\ =12th\text{ number}\\ =5\end{array}$

So,

  $\begin{array}{c}\text{Minimum value}=1\\ Q_1=2\\ \text{Median}=5\\ Q_3=5\\ \text{Maximum value}=8\end{array}$

Therefore, the five-number summary of a data set is $\left\{1,2,5,5,8\right\}$ .

The Inter-Quartile Range is,

  $\begin{array}{c}IQR=5-2\\ =3\end{array}$

Now, the upper fence is,

  $\begin{array}{c}{Q}_3+1.5IQR=5+1.5\left(3\right)\\ =5+4.5\\ =9.5\end{array}$

The lower fence is,

  $\begin{array}{c}{Q}_1-1.5IQR=2-1.5\left(3\right)\\ =2-4.5\\ =-2.5\end{array}$

Here, the data lies between $-2.5\text{ and }9.5$ .

Therefore, there is no outlier.