Find the mean, median and mode of the following data:
Class | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 | 120 - 140 |
Frequency | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
To find mean, we will solve by direct method:
CLASS | MID - POINT(xi) | FREQUENCY(fi) | fixi |
0 - 20 | 10 | 6 | 60 |
20 - 40 | 30 | 8 | 240 |
40 - 60 | 50 | 10 | 500 |
60 - 80 | 70 | 12 | 840 |
80 - 100 | 90 | 6 | 540 |
100 - 120 | 110 | 5 | 550 |
120 - 140 | 130 | 3 | 390 |
TOTAL | 50 | 3120 |
We have got
Σfi = 50 & Σfixi = 3120
∵ 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 - 20 | 6 | 6 |
20 - 40 | 8 | 6 + 8 = 14 |
40 - 60 | 10 | 14 + 10 = 24 |
60 - 80 | 12 | 24 + 12 = 36 |
80 - 100 | 6 | 36 + 6 = 42 |
100 - 120 | 5 | 42 + 5 = 47 |
120 - 140 | 3 | 47 + 3 = 50 |
TOTAL | 50 |
So, N = 50
⇒ N/2 = 50/2 = 25
The cumulative frequency just greater than (N/2 = ) 25 is 36, so the corresponding median class is 60 - 80 and accordingly we get Cf = 24(cumulative frequency before the median class).
Now, since median class is 60 - 80.
∴ l = 60, h = 20, f = 12, N/2 = 25 and Cf = 24
Median is given by,
⇒
= 60 + 1.67
= 61.67
And we know that,
Mode = 3(Median) – 2(Mean)
= 3(61.67) – 2(62.4)
= 185.01 – 124.8
= 60.21
Hence, mean is 62.4, median is 61.67 and mode is 60.21.