The following table gives the marks obtained by 50 students in a class test:
Marks | 11 - 15 | 16 - 20 | 21 - 25 | 26 - 30 | 31 - 35 | 36 - 40 | 41 - 45 | 46 - 50 |
Number of students | 2 | 3 | 6 | 4 | 14 | 12 | 4 | 2 |
Calculate the mean, median and mode for the above data.
For equal class intervals, we will solve by finding mid points of these classes using direct method.
MARKS | MID - POINT(xi) | NUMBER OF STUDENTS(fi) | fixi |
10.5 - 15.5 | 13 | 2 | 26 |
15.5 - 20.5 | 18 | 3 | 54 |
20.5 - 25.5 | 23 | 6 | 138 |
25.5 - 30.5 | 28 | 4 | 112 |
30.5 - 35.5 | 33 | 14 | 462 |
35.5 - 40.5 | 38 | 12 | 456 |
40.5 - 45.5 | 43 | 4 | 172 |
45.5 - 50.5 | 48 | 2 | 96 |
TOTAL | 47 | 1516 |
We have got
Σfi = 47 and Σfixi = 1516
∵ mean is given by
⇒ 
⇒ 
Thus, mean is 32.26.
To find median, Assume
Σfi = N = Sum of frequencies,
h = length of median class,
l = lower boundary of the median class,
f = frequency of median class
and Cf = cumulative frequency
Lets form a table.
MARKS | NUMBER OF STUDENTS(fi) | NUMBER OF STUDENTS(fi) |
10.5 - 15.5 | 2 | 2 |
15.5 - 20.5 | 3 | 2 + 3 = 5 |
20.5 - 25.5 | 6 | 5 + 6 = 11 |
25.5 - 30.5 | 4 | 11 + 4 = 15 |
30.5 - 35.5 | 14 | 15 + 14 = 29 |
35.5 - 40.5 | 12 | 29 + 12 = 41 |
40.5 - 45.5 | 4 | 41 + 4 = 45 |
45.5 - 50.5 | 2 | 45 + 2 = 47 |
TOTAL | 47 |
We have got
So, N = 47
⇒ N/2 = 47/2 = 23.5
The cumulative frequency just greater than (N/2 = ) 23.5 is 29, so the corresponding median class is 30.5 - 35.5 and accordingly we get Cf = 15(cumulative frequency before the median class).
Now, since median class is 30.5 - 35.5.
∴ l = 30.5, h = 5, f = 14, N/2 = 23.5 and Cf = 15
Median is given by,
⇒ 
= 30 + 3.03
= 33.03
Thus, median is 33.03.
Since, we have got mean = 32.26 and median = 33.03
Applying the empirical formula,
Mode = 3(Median) – 2(Mean)
⇒ Mode = 3(33.03) – 2(32.26)
⇒ Mode = 99.09 – 64.52 = 34.57
∴ Mean = 32.26, Median = 33.03 and Mode = 34.57