Find the mean, median and mode of the following data:
Class | 0 - 50 | 50 - 100 | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 |
Frequency | 2 | 3 | 5 | 6 | 5 | 3 | 1 |
To find mean, we will solve by direct method:
CLASS | MID - POINT(xi) | FREQUENCY(fi) | fixi |
0 - 50 | 25 | 2 | 50 |
50 - 100 | 75 | 3 | 225 |
100 - 150 | 125 | 5 | 625 |
150 - 200 | 175 | 6 | 1050 |
200 - 250 | 225 | 5 | 1125 |
250 - 300 | 275 | 3 | 825 |
300 - 350 | 325 | 1 | 325 |
TOTAL | 25 | 4225 |
We have got
Σfi = 25 & Σfixi = 4171
∵ 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 | FREQUENCY(fi) | Cf |
0 - 50 | 2 | 2 |
50 - 100 | 3 | 2 + 3 = 5 |
100 - 150 | 5 | 5 + 5 = 10 |
150 - 200 | 6 | 10 + 6 = 16 |
200 - 250 | 5 | 16 + 5 = 21 |
250 - 300 | 3 | 21 + 3 = 24 |
300 - 350 | 1 | 24 + 1 = 25 |
TOTAL | 25 |
So, N = 25
⇒ N/2 = 25/2 = 12.5
The cumulative frequency just greater than (N/2 = ) 12.5 is 16, so the corresponding median class is 150 - 200 and accordingly we get Cf = 10(cumulative frequency before the median class).
Now, since median class is 150 - 200.
∴ l = 150, h = 50, f = 6, N/2 = 12.5 and Cf = 10
Median is given by,
⇒ 
= 150 + 20.83
= 170.83
And we know that,
Mode = 3(Median) – 2(Mean)
= 3(170.83) – 2(169)
= 512.49 – 338
= 174.49
Hence, mean is 169, median is 170.83 and mode is 174.49.