Question

What is the difference between class average and class median?

Answer 1

Hint: Here we will take an example of grouped data and find its mean and median. We will know certain terms like frequency (f), cumulative frequency (c.f.), lower class (L) and class interval (h). To find the median we will use the formula Median = $$L+\left({\displaystyle \frac{{\displaystyle \frac{N}{2}}-cf}{f}}\right)\times h$$ and to calculate mean we will use the formula Mean = ${\displaystyle \frac{\sum _{i=1}^n{x}_i{f}_i}{\sum _{i=1}^n{f}_i}}$ where $x_i$ is the average of upper limit and lower limit.

Complete step by step answer:
Here we have been asked to find the difference between class mean and class median. Let us take an example of grouped data and find the mean and median to understand the difference. Let us consider the following distribution: -

Marks	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	Total
No. of students	5	6	3	4	3	9	30

Using the above table let us form a frequency table, having three columns. Column 1 will have marks, column 2 will contain frequency or number of students, column 3 will contain cumulative frequency which is calculated by adding frequencies in each step. Therefore, the frequency table will look like: -

Marks	Frequency (f)	Cumulative frequency (c.f.)
0 – 10	5	5
10 – 20	6	5 + 6 = 11
20 – 30	3	11 + 3 = 14
30 – 40	4	14 + 4 = 18
40 – 50	3	18 + 3 = 21
50 – 60	9	21 + 9 = 30

(i) Let us calculate the class mean. Now, class mean is given by the formula Mean = ${\displaystyle \frac{\sum _{i=1}^n{x}_i{f}_i}{\sum _{i=1}^n{f}_i}}$ where $x_i$ is the average of upper limit and lower limit, mathematically given as $x_i={\displaystyle \frac{1}{2}}(L{L}_i+U{L}_i)$ where LL and UL denotes the lower class limit and upper class limit. So we get,
$\Rightarrow x_1={\displaystyle \frac{1}{2}}(0+10)=5,x_2={\displaystyle \frac{1}{2}}(10+20)=15,x_2={\displaystyle \frac{1}{2}}(20+30)=25$ and so on we can calculate the other values. Therefore substituting these values in the formula of mean we get,
$\Rightarrow$ Mean = ${\displaystyle \frac{(5\times 5)+(15\times 6)+(25\times 3)+(35\times 4)+(45\times 3)+(55\times 9)}{30}}$
$\Rightarrow$ Mean = 32
Therefore, the mean of the given distribution is 32.
(ii) Now let us find the median. We know that median = $$L+\left({\displaystyle \frac{{\displaystyle \frac{N}{2}}-cf}{f}}\right)\times h$$, where,
L = Lower class containing the median
N = Total number of students
f = frequency of the class containing median
c.f. = cumulative frequency before the median class
h = class interval = upper limit – lower limit
Here the median class is found by finding the c.f. value which is just greater than the value obtained by the relation ${\displaystyle \frac{N}{2}}$. We have ${\displaystyle \frac{N}{2}}=15$, that means the median class is 30 – 40. So we have,
$\Rightarrow$ L = 30, N = 30, f = 4, c.f. = 14 and h = 10 – 0 = 10.
Substituting these values in the formula for median we get,
$\Rightarrow$ Median = $$30+\left({\displaystyle \frac{{\displaystyle \frac{30}{2}}-14}{4}}\right)\times 10$$
$\Rightarrow$ Median = 32.5
Therefore, the median of the given distribution is 32.5.

Note: You can see that the value of the mean and the median is not so different. Mean is also known as the average value of the distribution while median is not exactly the average but its value is not very different from the mean. Here we have considered the example of the grouped data because we have been asked to consider the terms class average and class median so we took data in class intervals.