Find the mean, mode and median of the following frequency distribution:
Class interval | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 |
Number of batsmen | 4 | 4 | 7 | 10 | 12 | 8 | 5 |
To find mean, we will solve by direct method:
CLASS INTERVAL | MID - POINT(xi) | NUMBER OF BATSMEN(fi) | fixi |
0 - 10 | 5 | 4 | 20 |
10 - 20 | 15 | 4 | 60 |
20 - 30 | 25 | 7 | 175 |
30 - 40 | 35 | 10 | 350 |
40 - 50 | 45 | 12 | 540 |
50 - 60 | 55 | 8 | 440 |
60 - 70 | 65 | 5 | 325 |
TOTAL | 50 | 1910 |
We have got
Σfi = 50 & Σfixi = 1910
∵ mean is given by
⇒ 
⇒ 
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.
CLASS INTERVAL | NUMBER OF WORKERS(fi) | Cf |
0 - 10 | 4 | 4 |
10 - 20 | 4 | 4 + 4 = 8 |
20 - 30 | 7 | 8 + 7 = 15 |
30 - 40 | 10 | 15 + 10 = 25 |
40 - 50 | 12 | 25 + 12 = 37 |
50 - 60 | 8 | 37 + 8 = 45 |
60 - 70 | 5 | 45 + 5 = 50 |
TOTAL | 50 |
So, N = 50
⇒ N/2 = 50/2 = 25
The cumulative frequency just greater than (N/2 = ) 25 is 37, so the corresponding median class is 40 - 50 and accordingly we get Cf = 25(cumulative frequency before the median class).
Now, since median class is 40 - 50.
∴ l = 40, h = 10, f = 37, N/2 = 25 and Cf = 25
Median is given by,
⇒ 
= 40 + 0
= 40
And we know that,
Mode = 3(Median) – 2(Mean)
= 3(40) – 2(38.2)
= 120 – 76.4
= 43.6
Hence, mean is 38.2, median is 40 and mode is 43.6.