Chapter 11.2, Problem 13HP

(a)

To determine

To find: The mean and the median for each of the distribution.

Answer to Problem 13HP

Themean of the graph is 14 and the median is 13.

Explanation of Solution

Given:

The given graph is shown in Figure 1

  

Figure 1

Calculation:

Consider the table from the graph shown in Figure 1

Table 1

Class interval	Frequency	Class width
0 to 4	3	$\frac{0+4}{2}=2$
5 to 9	6	7
10 to 14	5	12
15 to 19	4	17
20 to 24	3	22
25 to 29	2	27
30 to 34	1	32

Thus, the mean from the above data is obtained as,

  $\begin{array}{c}\overline{x}=\frac{2\cdot 3+7\cdot 6+12\cdot 5+17\cdot 4+22\cdot 3+27\cdot 2+32\cdot 1}{24}\\ =13.66\\ \approx 14\end{array}$

Consider the Table 2as,

Class interval	Frequency	Class boundary	Cumulative frequency
0 to 4	3	-0.5 to 4.5	3
5 to 9	6	4.5 to 9.5	3+6=9
10 to 14	5	9.5 to 14.5	14
15 to 19	4	14.5 to 19.5	19
20 to 24	3	19.5 to 24.5	22
25 to 29	2	24.5 to 29.5	24
30 to 34	1	29..5 to 34.5	25

Obtain the mean class as,

  $\begin{array}{c}\left(\frac{n+1}{2}\right)=\left(\frac{25+1}{2}\right)\\ =13\end{array}$

The value $13$ is less hen the cumulative frequency 14 for the class boundary of 9.5 to 14.5.

The formula for the median class is,

  $m=l+\frac{c}{f}\left(\frac{n}{2}-\text{sum}\text{of}\text{frequecies}\text{preceeding}\text{median}\text{class}\right)$

Then,

  $\begin{array}{c}m=9.5+\frac{5}{5}\left(\frac{25}{2}-9\right)\\ =13\end{array}$

Thus, the approximate median of the graph is $13$ .

(b)

To determine

To find: The mean and the median for each of the distribution.

Answer to Problem 13HP

The mean of the graph is $20.33$ and the median is $21.5$ .

Explanation of Solution

Given:

The given graph is shown in Figure 1

  

Figure 1

Calculation:

Consider the table from the graph shown in Figure 1

Table 1

Class interval	Frequency	Class width
0 to 4	1	$\frac{0+4}{2}=2$
5 to 9	2	7
10 to 14	3	12
15 to 19	4	17
20 to 24	5	22
25 to 29	6	27
30 to 34	3	32

Thus, the mean from the above data is obtained as,

  $\begin{array}{c}\overline{x}=\frac{2\cdot 1+7\cdot 2+12\cdot 3+17\cdot 4+22\cdot 5+27\cdot 6+32\cdot 3}{24}\\ =20.33\end{array}$

Consider the Table 2as,

Class interval	Frequency	Class boundary	Cumulative frequency
0 to 4	1	-0.5 to 4.5	1
5 to 9	2	4.5 to 9.5	3
10 to 14	3	9.5 to 14.5	6
15 to 19	4	14.5 to 19.5	10
20 to 24	5	19.5 to 24.5	15
25 to 29	6	24.5 to 29.5	21
30 to 34	3	29..5 to 34.5	24

Obtain the mean class as,

  $\begin{array}{c}\left(\frac{n+1}{2}\right)=\left(\frac{24+1}{2}\right)\\ =12.5\end{array}$

The value $13$ is less hen the cumulative frequency 14 for the class boundary of 9.5 to 14.5.

The formula for the median class is,

  $m=l+\frac{c}{f}\left(\frac{n}{2}-\text{sum}\text{of}\text{frequecies}\text{preceeding}\text{median}\text{class}\right)$

Then,

  $\begin{array}{c}m=19.5+\frac{5}{5}\left(\frac{24}{2}-10\right)\\ =21.5\end{array}$

Thus, the approximate median of the graph is $13$ .

(c)

To determine

To find: The mean and the median for each of the distribution.

Answer to Problem 13HP

The mean of the graph is $17$ and the median is $17.5$ .

Explanation of Solution

Given:

The given graph is shown in Figure 1

  

Figure 1

Calculation:

Consider the table from the graph shown in Figure 1

Table 1

Class interval	Frequency	Class width
0 to 4	1	$\frac{0+4}{2}=2$
5 to 9	3	7
10 to 14	5	12
15 to 19	6	17
20 to 24	5	22
25 to 29	3	27
30 to 34	1	32

Thus, the mean from the above data is obtained as,

  $\begin{array}{c}\overline{x}=\frac{2\cdot 1+7\cdot 3+12\cdot 5+17\cdot 6+22\cdot 5+27\cdot 3+32\cdot 1}{24}\\ =17\end{array}$

Consider the Table 2as,

Class interval	Frequency	Class boundary	Cumulative frequency
0 to 4	1	-0.5 to 4.5	1
5 to 9	3	4.5 to 9.5	4
10 to 14	5	9.5 to 14.5	9
15 to 19	6	14.5 to 19.5	15
20 to 24	5	19.5 to 24.5	20
25 to 29	3	24.5 to 29.5	23
30 to 34	1	29..5 to 34.5	24

Obtain the mean class as,

  $\begin{array}{c}\left(\frac{n+1}{2}\right)=\left(\frac{24+1}{2}\right)\\ =12.5\end{array}$

The value $13$ is less hen the cumulative frequency 14 for the class boundary of 9.5 to 14.5.

The formula for the median class is,

  $m=l+\frac{c}{f}\left(\frac{n}{2}-\text{sum}\text{of}\text{frequecies}\text{preceeding}\text{median}\text{class}\right)$

Then,

  $\begin{array}{c}m=14.5+\frac{5}{5}\left(\frac{24}{2}-9\right)\\ =17.5\end{array}$

Thus, the approximate median of the graph is $13$ .