Chapter 2.1, Problem 2E

(a)

To determine

To find:

The class width for the given frequency distribution.

Answer to Problem 2E

Solution:

The class width of the given frequency distribution is $0.028$ which is approximately equal to $0.03$.

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,

Braking Times for Vehicles at 60 mph (in minutes)
Class	Frequency
$0.05-0.07$	12
$0.08-0.10$	15
$0.11-0.13$	14
$0.14-0.16$	15
$0.17-0.19$	14

The upper limit of the distribution is given as $0.19$ and the lower limit is given as $0.05$. 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 $0.19$ for upper limit and $0.05$ for lower limit and 5 for the number of classes in the formula,

$\begin{array}{rll}\text{Class}\text{width}& =& \frac{0.19-0.05}{5}\\ & =& \frac{0.14}{5}\\ & =& 0.028\\ & \approx & 0.03\end{array}$

(b)

To determine

To find:

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

Answer to Problem 2E

Solution:

The class boundary for each class of the given frequency distribution is shown in Table

Class	Frequency	Class boundaries
$0.05-0.07$	12	$0.045-0.075$
$0.08-0.10$	15	$0.075-0.105$
$0.11-0.13$	14	$0.105-0.135$
$0.14-0.16$	15	$0.135-0.165$
$0.17-0.19$	14	$0.165-0.195$

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,

Braking Times for Vehicles at 60 mph (in minutes)
Class	Frequency
$0.05-0.07$	12
$0.08-0.10$	15
$0.11-0.13$	14
$0.14-0.16$	15
$0.17-0.19$	14

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 0.07 and 0.08 is the smallest number in the next class, then, substitute $0.07$ for upper limit and $0.08$ for lower limit in the formula,

$\begin{array}{rll}\text{Class}\text{boundary}& =& \frac{0.07+0.08}{2}\\ & =& \frac{0.15}{2}\\ & =& 0.075\end{array}$

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
$0.05-0.07$	12	$0.045-0.075$
$0.08-0.10$	15	$0.075-0.105$
$0.11-0.13$	14	$0.105-0.135$
$0.14-0.16$	15	$0.135-0.165$
$0.17-0.19$	14	$0.165-0.195$

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

Answer to Problem 2E

Solution:

The midpoint of each class for the given frequency distribution is shown in Table

Class	Frequency	Class boundaries	Midpoint
$0.05-0.07$	12	$0.045-0.075$	$0.06$
$0.08-0.10$	15	$0.075-0.105$	$0.09$
$0.11-0.13$	14	$0.105-0.135$	$0.12$
$0.14-0.16$	15	$0.135-0.165$	$0.15$
$0.17-0.19$	14	$0.165-0.195$	$0.18$

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,

Braking Times for Vehicles at 60 mph (in minutes)
Class	Frequency
$0.05-0.07$	12
$0.08-0.10$	15
$0.11-0.13$	14
$0.14-0.16$	15
$0.17-0.19$	14

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 $0.07$ for upper limit and $0.05$ for lower limit in the formula,

$\begin{array}{rll}\text{Midpoint}& =& \frac{0.05+0.07}{2}\\ & =& \frac{0.12}{2}\\ & =& 0.06\end{array}$

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
$0.05-0.07$	12	$0.045-0.075$	$0.06$
$0.08-0.10$	15	$0.075-0.105$	$0.09$
$0.11-0.13$	14	$0.105-0.135$	$0.12$
$0.14-0.16$	15	$0.135-0.165$	$0.15$
$0.17-0.19$	14	$0.165-0.195$	$0.18$

(d)

To determine

To find:

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

Answer to Problem 2E

Solution:

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

Class	Frequency	Class boundaries	Midpoint	Relative frequency
$0.05-0.07$	12	$0.045-0.075$	$0.06$	$\frac{12}{70}=17\%$
$0.08-0.10$	15	$0.075-0.105$	$0.09$	$\frac{15}{70}=21\%$
$0.11-0.13$	14	$0.105-0.135$	$0.12$	$\frac{14}{70}=20\%$
$0.14-0.16$	15	$0.135-0.165$	$0.15$	$\frac{15}{70}=21\%$
$0.17-0.19$	14	$0.165-0.195$	$0.18$	$\frac{14}{70}=20\%$

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,

Braking Times for Vehicles at 60 mph (in minutes)
Class	Frequency
$0.05-0.07$	12
$0.08-0.10$	15
$0.11-0.13$	14
$0.14-0.16$	15
$0.17-0.19$	14

Calculation:

The sum of the frequencies is,

$\begin{array}{rll}N& =& 12+15+14+15+14\\ & =& 70\end{array}$

For the first class,

Substitute 12 for $f$ and 70 for $N$ in the formula,

$\begin{array}{rll}\text{Relative frequency}& =& \frac{12}{70}\\ & =& 0.171\\ & =& 17\%\end{array}$

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
$0.05-0.07$	12	$0.045-0.075$	$0.06$	$\frac{12}{70}=17\%$
$0.08-0.10$	15	$0.075-0.105$	$0.09$	$\frac{15}{70}=21\%$
$0.11-0.13$	14	$0.105-0.135$	$0.12$	$\frac{14}{70}=20\%$
$0.14-0.16$	15	$0.135-0.165$	$0.15$	$\frac{15}{70}=21\%$
$0.17-0.19$	14	$0.165-0.195$	$0.18$	$\frac{14}{70}=20\%$

(e)

To determine

To find:

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

Answer to Problem 2E

Solution:

The cumulative frequency of each class for the given frequency distribution is shown in Table

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
$0.05-0.07$	12	$0.045-0.075$	$0.06$	$\frac{12}{70}=17\%$	12
$0.08-0.10$	15	$0.075-0.105$	$0.09$	$\frac{15}{70}=21\%$	$12+15=27$
$0.11-0.13$	14	$0.105-0.135$	$0.12$	$\frac{14}{70}=20\%$	$27+14=41$
$0.14-0.16$	15	$0.135-0.165$	$0.15$	$\frac{15}{70}=21\%$	$41+15=56$
$0.17-0.19$	14	$0.165-0.195$	$0.18$	$\frac{14}{70}=20\%$	$56+14=70$

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,

Braking Times for Vehicles at 60 mph (in minutes)
Class	Frequency
$0.05-0.07$	12
$0.08-0.10$	15
$0.11-0.13$	14
$0.14-0.16$	15
$0.17-0.19$	14

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
$0.05-0.07$	12	$0.045-0.075$	$0.06$	$\frac{12}{70}=17\%$	12
$0.08-0.10$	15	$0.075-0.105$	$0.09$	$\frac{15}{70}=21\%$	$12+15=27$
$0.11-0.13$	14	$0.105-0.135$	$0.12$	$\frac{14}{70}=20\%$	$27+14=41$
$0.14-0.16$	15	$0.135-0.165$	$0.15$	$\frac{15}{70}=21\%$	$41+15=56$
$0.17-0.19$	14	$0.165-0.195$	$0.18$	$\frac{14}{70}=20\%$	$56+14=70$