Find the mean, median and mode of the following data:
Class | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 |
Frequency | 6 | 8 | 10 | 15 | 5 | 4 | 2 |
For equal class intervals, we will solve by finding mid points of these classes using direct method.
CLASS | MID - POINT(xi) | FREQUENCY(fi) | fixi |
0 - 10 | 5 | 6 | 30 |
10 - 20 | 15 | 8 | 120 |
20 - 30 | 25 | 10 | 250 |
30 - 40 | 35 | 15 | 525 |
40 - 50 | 45 | 5 | 225 |
50 - 60 | 55 | 4 | 220 |
60 - 70 | 65 | 2 | 130 |
TOTAL | 50 | 1500 |
We have got
Σfi = 50 and Σfixi = 1500
∵ mean is given by
⇒
⇒
Thus, mean is 30.
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 - 10 | 6 | 6 |
10 - 20 | 8 | 6 + 8 = 14 |
20 - 30 | 10 | 14 + 10 = 24 |
30 - 40 | 15 | 24 + 15 = 39 |
40 - 50 | 5 | 39 + 5 = 44 |
50 - 60 | 4 | 44 + 4 = 48 |
60 - 70 | 2 | 48 + 2 = 50 |
TOTAL | 50 |
We have got
So, N = 50
⇒ N/2 = 50/2 = 25
The cumulative frequency just greater than (N/2 = ) 25 is 39, so the corresponding median class is 30 - 40 and accordingly we get Cf = 24(cumulative frequency before the median class).
Now, since median class is 30 - 40.
∴ l = 30, h = 10, f = 15, N/2 = 25 and Cf = 24
Median is given by,
⇒
= 30 + 0.67
= 30.67
Thus, median is 30.67.
Since, we have got mean = 30 and median = 30.67
Applying the empirical formula,
Mode = 3(Median) – 2(Mean)
⇒ Mode = 3(30.67) – 2(30)
⇒ Mode = 92.01 – 60 = 32.01
∴ Mean = 30, Median = 30.67 and Mode = 32.01