Chapter 2.1, Problem 1E

(a)

To determine

To find:

The class width for the given frequency distribution.

Answer to Problem 1E

Solution:

The class width of the given frequency distribution is $4.8$ which is approximately equal to 5.

Explanation of Solution

Formula used:

The formula to calculate the class width is,

$\text{Class}\text{width}=\frac{\text{Upper}\text{limit}-\text{Lower}\text{limit}}{\text{Number}\text{of}\text{class}}$

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
$15-19$	7
$20-24$	8
$25-29$	10
$30-34$	2
$35-39$	3

The upper limit of the distribution is given as 39 and the lower limit is given as 15. The number of classes is 5.

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute 39 for upper limit and 15 for lower limit and 5 for the number of classes in the formula,

$\begin{array}{rll}\text{Class}\text{width}& =& \frac{39-15}{5}\\ & =& \frac{24}{5}\\ & =& 4.8\\ & \approx & 5\end{array}$

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries
$15-19$	7	$14.5-19.5$
$20-24$	8	$19.5-24.5$
$25-29$	10	$24.5-29.5$
$30-34$	2	$29.5-34.5$
$35-39$	3	$34.5-39.5$

Explanation of Solution

Formula used:

The formula to calculate the class boundary is,

$\text{Class}\text{boundary}=\frac{\text{Upper}\text{limit}\text{of}\text{one}\text{class}+\text{Lower}\text{limit}\text{of}\text{next}\text{class}}{\text{2}}$

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
$15-19$	7
$20-24$	8
$25-29$	10
$30-34$	2
$35-39$	3

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since the largest number is 19 and 20 is the smallest number in the next class, then, substitute 19 for upper limit and 20 for lower limit in the formula,

$\begin{array}{rll}\text{Class}\text{boundary}& =& \frac{19+20}{2}\\ & =& \frac{39}{2}\\ & =& 19.5\end{array}$

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
$15-19$	7	$14.5-19.5$
$20-24$	8	$19.5-24.5$
$25-29$	10	$24.5-29.5$
$30-34$	2	$29.5-34.5$
$35-39$	3	$34.5-39.5$

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint
$15-19$	7	$14.5-19.5$	17
$20-24$	8	$19.5-24.5$	22
$25-29$	10	$24.5-29.5$	27
$30-34$	2	$29.5-34.5$	32
$35-39$	3	$34.5-39.5$	37

Explanation of Solution

Formula used:

The formula to calculate the midpoint of any class is,

$\text{Midpoint}=\frac{\text{Upper}\text{limit}+\text{Lower}\text{limit}}{\text{2}}$

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
$15-19$	7
$20-24$	8
$25-29$	10
$30-34$	2
$35-39$	3

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute 19 for upper limit and 15 for lower limit in the formula,

$\begin{array}{rll}\text{Midpoint}& =& \frac{15+19}{2}\\ & =& \frac{34}{2}\\ & =& 17\end{array}$

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
$15-19$	7	$14.5-19.5$	17
$20-24$	8	$19.5-24.5$	22
$25-29$	10	$24.5-29.5$	27
$30-34$	2	$29.5-34.5$	32
$35-39$	3	$34.5-39.5$	37

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

Answer to Problem 1E

Solution:

The relative frequency of each class for the given frequency distribution is shown in Table 3.

Class	Frequency	Class boundaries	Midpoint	Relative frequency
$15-19$	7	$14.5-19.5$	17	$\frac{7}{30}=23\%$
$20-24$	8	$19.5-24.5$	22	$\frac{8}{30}=27\%$
$25-29$	10	$24.5-29.5$	27	$\frac{10}{30}=33\%$
$30-34$	2	$29.5-34.5$	32	$\frac{2}{30}=7\%$
$35-39$	3	$34.5-39.5$	37	$\frac{3}{30}=10\%$

Explanation of Solution

Formula used:

The formula to calculate the relative frequency of any class is,

$\text{Relative}\text{frequency}=\frac{f}{N}$

Here $N$ is the sum of the frequencies.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
$15-19$	7
$20-24$	8
$25-29$	10
$30-34$	2
$35-39$	3

Calculation:

The sum of the frequencies is,

$\begin{array}{rll}N& =& 7+8+10+2+3\\ & =& 30\end{array}$

For the first class,

Substitute 7 for $f$ and 30 for $N$ in the formula,

$\begin{array}{rll}\text{Relative frequency}& =& \frac{7}{30}\\ & =& 0.233\\ & =& 23\%\end{array}$

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
$15-19$	7	$14.5-19.5$	17	$\frac{7}{30}=23\%$
$20-24$	8	$19.5-24.5$	22	$\frac{8}{30}=27\%$
$25-29$	10	$24.5-29.5$	27	$\frac{10}{30}=33\%$
$30-34$	2	$29.5-34.5$	32	$\frac{2}{30}=7\%$
$35-39$	3	$34.5-39.5$	37	$\frac{3}{30}=10\%$

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
$15-19$	7	$14.5-19.5$	17	$\frac{7}{30}=23\%$	7
$20-24$	8	$19.5-24.5$	22	$\frac{8}{30}=27\%$	$7+8=15$
$25-29$	10	$24.5-29.5$	27	$\frac{10}{30}=33\%$	$15+10=25$
$30-34$	2	$29.5-34.5$	32	$\frac{2}{30}=7\%$	$25+2=27$
$35-39$	3	$34.5-39.5$	37	$\frac{3}{30}=10\%$	$27+3=30$

Explanation of Solution

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
$15-19$	7
$20-24$	8
$25-29$	10
$30-34$	2
$35-39$	3

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
$15-19$	7	$14.5-19.5$	17	$\frac{7}{30}=23\%$	7
$20-24$	8	$19.5-24.5$	22	$\frac{8}{30}=27\%$	$7+8=15$
$25-29$	10	$24.5-29.5$	27	$\frac{10}{30}=33\%$	$15+10=25$
$30-34$	2	$29.5-34.5$	32	$\frac{2}{30}=7\%$	$25+2=27$
$35-39$	3	$34.5-39.5$	37	$\frac{3}{30}=10\%$	$27+3=30$