Chapter 14.3, Problem 25E

a.

To determine

To calculate: the arithmetic mean of the audience ratings.

Answer to Problem 25E

The arithmetic mean is $11.$

Explanation of Solution

Given information:

The frequency distribution shows the average audience rating for the top fifty network television shows for one season.

  

Formula used:

Formula of Arithmetic Mean

  $\overline{X}=\frac{{\displaystyle \sum f.X}}{f}$

Class Marks $=\frac{\text{Upper limit}+\text{Lower Limit}}{2}$

Calculation:

According to the question,

Class Limits	Class Marks $\left(X\right)$	$f$	$f.X$
$8-10$	$=\frac{10+8}{2}=9$	26	$26\times 9=234$
$10-12$	$=\frac{12+10}{2}=11$	12	$12\times 11=132$
$12-14$	$=\frac{14+12}{2}=13$	6	$6\times 13=78$
$14-16$	$=\frac{16+14}{2}=15$	2	$2\times 15=30$
$16-18$	$=\frac{18+16}{2}=17$	2	$2\times 17=34$
$18-20$	$=\frac{20+18}{2}=19$	0	$0\times 19=0$
$20-22$	$=\frac{22+20}{2}=21$	2	$2\times 21=42$
		$50$	$550$

Therefore, we have

  ${\displaystyle \sum f=50}$

  ${\displaystyle \sum f.X=}550$

The arithmetic mean −

  $\begin{array}{l}=\overline{X}=\frac{550}{50}\\ =\overline{X}=11\end{array}$

Hence, the arithmetic mean is $11.$

b.

To determine

To calculate: The standard deviation of the audience ratings.

Answer to Problem 25E

The standard deviation is approximately $2.9393$ .

Explanation of Solution

Given information:

The frequency distribution shows the average audience rating for the top fifty network television shows for one season.

  

Formula used:

Standard Deviation of the Data in a Frequency Distribution − If $X_1,X_2,\mathrm{....},X_k$ are the class marks in a frequency distribution with k classes , and $f_1,f_2,\mathrm{.....},f_k$ are the corresponding frequencies, then the standard deviation $\sigma$ of the data in the frequency distribution is found as follows:

  $\sigma =\sqrt{\frac{{\displaystyle \sum _{j=1}^k{\left(X_j-\overline{X}\right)}^2.f_i}}{{\displaystyle \sum _{i=1}^k{f}_i}}}$

Calculation:

According to the question,

Class Limits	Class Marks $\left(X\right)$	$f$	$f.X$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$	${\left(X_i-\overline{X}\right)}^2.f$
$8-10$	$=\frac{10+8}{2}=9$	26	$26\times 9=234$	$\begin{array}{l}=\left(9-11\right)\\ =-2\end{array}$	$\begin{array}{l}={\left(-2\right)}^2\\ =4\end{array}$	$\begin{array}{l}=4\times 26\\ =104\end{array}$
$10-12$	$=\frac{12+10}{2}=11$	12	$12\times 11=132$	$\begin{array}{l}=\left(11-11\right)\\ =0\end{array}$	$\begin{array}{l}={\left(0\right)}^2\\ =0\end{array}$	$\begin{array}{l}=0\times 12\\ =0\end{array}$
$12-14$	$=\frac{14+12}{2}=13$	6	$6\times 13=78$	$\begin{array}{l}=\left(13-11\right)\\ =2\end{array}$	$\begin{array}{l}={\left(2\right)}^2\\ =4\end{array}$	$\begin{array}{l}=4\times 6\\ =24\end{array}$
$14-16$	$=\frac{16+14}{2}=15$	2	$2\times 15=30$	$\begin{array}{l}=\left(15-11\right)\\ =4\end{array}$	$\begin{array}{l}={\left(4\right)}^2\\ =16\end{array}$	$\begin{array}{l}=16\times 2\\ =32\end{array}$
$16-18$	$=\frac{18+16}{2}=17$	2	$2\times 17=34$	$\begin{array}{l}=\left(17-11\right)\\ =6\end{array}$	$\begin{array}{l}={\left(6\right)}^2\\ =36\end{array}$	$\begin{array}{l}=36\times 2\\ =72\end{array}$
$18-20$	$=\frac{20+18}{2}=19$	0	$0\times 19=0$	$\begin{array}{l}=\left(19-11\right)\\ =8\end{array}$	$\begin{array}{l}={\left(8\right)}^2\\ =64\end{array}$	$\begin{array}{l}=64\times 0\\ =0\end{array}$
$20-22$	$=\frac{22+20}{2}=21$	2	$2\times 21=42$	$\begin{array}{l}=\left(21-11\right)\\ =10\end{array}$	$\begin{array}{l}={\left(10\right)}^2\\ =100\end{array}$	$\begin{array}{l}=100\times 2\\ =200\end{array}$
		$50$	$550$			$432$

Therefore, we have

  ${\displaystyle \sum f=50}$

  ${\displaystyle \sum f.X=}550$

The arithmetic mean $11.$

Also,

  ${\displaystyle \sum {\left(X_i-\overline{X}\right)}^2.f}=432$

The standard deviation −

  $\begin{array}{l}\sigma =\sqrt{\frac{432}{50}}\\ \sigma =\sqrt{8.64}\\ \sigma =2.9393\end{array}$

Hence , the standard deviation is approximately $2.9393$