Chapter 3.1, Problem 22E

Enrollments for Selected Independent Religiously Controlled 4-Year Colleges Listed below are the enrollments for selected independent religiously controlled 4-year colleges that offer bachelor’s degrees only. Construct a grouped frequency distribution with six classes and find the mean and modal class.

To determine

The frequency distribution for given data; the mean and modal class of the given data.

Answer to Problem 22E

The frequency distribution for given data is tabulated below:

Class interval	Frequency
1000-1499	13
1500-1999	9
2000-2499	5
2500-2999	2
3000-3499	2
3500-3999	0
Total	31

The mean of the data is 1825.26. The modal class for data is 1000-1499.

Explanation of Solution

Given info:

The data shows the enrollments for selected independent religiously controlled 4-years colleges.

Use number of class 6 for given data.

Calculation:

Steps to obtain the frequency distribution of given data

Step 1: The lowest value in the data set is 1013. The highest value in the data set is 3008.

Step 2: Calculate the class width using the formula,

$\text{Class width}\approx \frac{\text{Maximum data value}-\text{Minimum data value}}{\text{number}\text{of}\text{classes}}$

Substitute 6 for number of classes, 1013 for minimum data value and 3008 for maximum data value in the above formula.

$\begin{array}{c}\text{Class}\text{width=}\frac{3008-1013}{6}\\ =\frac{2985}{6}\\ =497.5\\ \approx 500\left(\text{rounded up to a more convenient number}\right)\end{array}$

Step-3: Take a start value that is less than or equal to smallest value of data for this data take 1000 as start value.

Step 4: Add the class width of 500 to the starting value to obtain the second lower class limit is 1500. Continue to add the class width of 500 until six lower class limits are found. Thus, the lower class limits are 1000, 1500, 2000, 2500, 3000 and 3500.

Step 4: List the lower class limits vertically, to identify the corresponding upper class limits as shown in Table (1).

Class interval

1000-

1500-

2000-

2500-

3000-

3500-

Table (1)

Thus, the upper class limits are 1499, 1999, 2499, 2999, 3499-and 3999.

Step 5: Enter a tally mark for each data value in the appropriate class. Then add the tally marks to obtain the frequency for each class interval shown in Table (2).

Service times (sec)	Tally mark	Frequency
1000-1499	$\cancel{\left|\right|\left|\right|}\cancel{\left|\right|\left|\right|}\left|\right||$	13
1500-1999	$\cancel{\left|\right|\left|\right|}\left|\right|\left|\right|$	9
2000-2499	$\cancel{\left|\right|\left|\right|}|$	6
2500-2999	$\left|\right||$	3
3000-3499	$\left|\right|$	2
3500-3999		0
Total		33

Table (2)

From the Table (2), it can be observed that the total frequency is 33 and the number of classes is 6.

The total number of frequency is,

$\begin{array}{l}1013,1867,1268,2309,1231,3005,2895,2166,1136,1532,1461,1750,1069,1723\\ ,1827,1155,1714,2391,2155,1412,1688,2471,1759,3008,2511,2577,1082,1067\\ ,1062,1319,1037,2400\end{array}$

The formula to calculate the mean is,

$\begin{array}{c}\overline{X}=\frac{{\displaystyle \sum _1^{n}X}}{n}\\ =\frac{X_1+X_2+X_3+\mathrm{...}+X_n}{n}\end{array}$ (1)

Where,

$X$ is the number of enrollments in different colleges.

$n$ is the total number of colleges.

The sum of the total enrollments is give below:

$\begin{array}{c}{\displaystyle \sum _1^{33}X}=\left[\begin{array}{c}1013+1867+1268+2309+1231+3005+2895+2166+1136+1532+1461+\\ 1750+1069+1723+1827+1155+1714+2391+2155+1412+1688+2471+\\ 1759+3008+2511+2577+1082+1067+1062+1319+1037+2400\end{array}\right]\\ =60233.5\end{array}$

Substitute ${\displaystyle \sum _1^{33}X}=60233.5$ and $n=33$ in equation (1).

$\begin{array}{c}\text{Mean}\text{=}\frac{60233.5}{33}\\ =1825.26\end{array}$

The modal class is the class with largest frequency.

Since, the class 1000-1499 has the largest frequency. Hence, it is modal class for data. Sometimes midpoint of the class is used it is 1249.5 for this data.