Chapter 2.CR, Problem 2CR

To determine

To create:

The frequency distribution with class boundaries, midpoint, relative frequency and cumulative frequency.

Answer to Problem 2CR

Solution:

The frequency distribution with class boundaries, midpoint, relative frequency and cumulative frequency is shown in Table 5.

Explanation of Solution

Given:

The data for number weights in pounds is given as,

$\begin{array}{l}195210225267231229301199\\ 178281245256278205217223\\ 279196235248262291302189\end{array}$

Calculation:

The number of classes with the class width of 25 will be 6. Calculate the number of values that fall within the class interval and complete the frequency table. The values lying between $175-199$ are $178,189,195,196,199$. Since, there is 5 values in the class interval $175-199$, so the frequency will be 5.

Similarly, completing all the frequencies as shown in table below,

Number of absent
Class	Frequency
$175-199$ $200-224$ $225-249$ $250-274$ $275-299$ $300-324$	5 4 5 4 4 2

Table 1

The formula to calculate the class boundary is,

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

For the first class,

Substitute $199$ for upper limit and $200$ for lower limit in the formula,

$\begin{array}{rll}\text{Class}\text{boundary}& =& \frac{200+199}{2}\\ & =& \frac{399}{2}\\ & =& 199.5\end{array}$

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

Class	Frequency	Class boundaries
$175-199$	5	$174.5-199.5$
$200-224$	4	$199.5-224.5$
$225-249$	5	$224.5-249.5$
$250-274$	4	$249.5-274.5$
$275-299$	4	$274.5-299.5$
$300-324$	2	$299.5-324.5$

Table 2

The formula to calculate the midpoint of any class is,

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

For the first class,

Substitute $199$ for upper limit and $174$ for lower limit in the formula,

$\begin{array}{rll}\text{Midpoint}& =& \frac{175+199}{2}\\ & =& \frac{374}{2}\\ & =& 187\end{array}$

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

Class	Frequency	Class boundaries	Midpoint
$175-199$	5	$174.5-199.5$	$187$
$200-224$	4	$199.5-224.5$	$212$
$225-249$	5	$224.5-249.5$	$237$
$250-274$	4	$249.5-274.5$	$262$
$275-299$	4	$274.5-299.5$	$287$
$300-324$	2	$299.5-324.5$	$312$

Table 3

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.

The sum of the frequencies is,

$\begin{array}{rll}N& =& 5+4+5+4+4+2\\ & =& 24\end{array}$

For the first class,

Substitute 5 for $f$ and 24 for $N$ in the formula,

$\begin{array}{rll}\text{Relative frequency}& =& \frac{5}{24}\\ & \approx & 0.21\\ & =& 21\%\end{array}$

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

Class	Frequency	Class boundaries	Midpoint	Relative frequency
$175-199$	5	$174.5-199.5$	$187$	$\frac{5}{24}=20.8\%$
$200-224$	4	$199.5-224.5$	$212$	$\frac{4}{24}=16.7\%$
$225-249$	5	$224.5-249.5$	$237$	$\frac{5}{24}=20.8\%$
$250-274$	4	$249.5-274.5$	$262$	$\frac{4}{24}=16.7\%$
$275-299$	4	$274.5-299.5$	$287$	$\frac{4}{24}=16.7\%$
$300-324$	2	$299.5-324.5$	$312$	$\frac{2}{24}=8.3\%$

Table 4

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

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

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
$175-199$	5	$174.5-199.5$	$187$	$\frac{5}{24}=20.8\%$	5
$200-224$	4	$199.5-224.5$	$212$	$\frac{4}{24}=16.7\%$	$5+4=9$
$225-249$	5	$224.5-249.5$	$237$	$\frac{5}{24}=20.8\%$	$9+5=14$
$250-274$	4	$249.5-274.5$	$262$	$\frac{4}{24}=16.7\%$	$14+4=18$
$275-299$	4	$274.5-299.5$	$287$	$\frac{4}{24}=16.7\%$	$18+4=22$
$300-324$	2	$299.5-324.5$	$312$	$\frac{2}{24}=8.3\%$	$22+2=24$

Table 5