Chapter 10.3, Problem 29E

Solution

To calculate : The mean magnification power for the superzoom cameras from the given data and state the better average is mean or median for the given data.

The mean magnification power for the superzoom cameras is saye $5.28$ . Mean is the better average for the given data.

Given information :

Taking high-quality selfies require a high resolution "front camera". The table below summarizes the numbers of front camera megapixels offered by $25$ smartphone models.

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

Formula used :

The mean of a list of numbers $\left\{x_1,x_2,x_3,\mathrm{......},x_n\right\}$ with frequency $\left\{n_1,n_2,n_3,\mathrm{......},n_n\right\}$ is given by,

  $\overline{X}=\frac{x_1\cdot n_1+x_2\cdot n_2+x_3\cdot n_3+\mathrm{......}+x_n\cdot n_n}{n_1+n_2+n_3+\mathrm{....}+n_n}$

Calculation :

Consider the given data.

Make a new table with a column for product of frequency and megapixels.

Megapixels $\left(x\right)$	Frequency $\left(f\right)$	$fx$	Cumulative Frequency $\left(cf\right)$
$\begin{array}{l}1\\ 2\\ 3\\ 4\\ 5\\ 6\\ 7\\ 8\end{array}$	$\begin{array}{l}2\\ 2\\ 0\\ 1\\ 12\\ 0\\ 2\\ 6\end{array}$	$\begin{array}{l}2\\ 4\\ 0\\ 4\\ 60\\ 0\\ 14\\ 48\end{array}$	$\begin{array}{l}2\\ 4\\ 4\\ 5\\ 17\\ 17\\ 19\\ 25\end{array}$
	$\sum f=25$	$\sum fx=132$

Find the mean magnification power for the superzoom cameras.

  $\begin{array}{c}\overline{X}=\frac{132}{25}\\ \approx 5.28\end{array}$

  $\begin{array}{c}M=\frac{25+1}{2}\text{th term}\\ =13\text{th term}\\ =5\end{array}$

Therefore, the mean magnification power for the superzoom cameras is $5.28$ . Mean is the better average for the given data.