The table below shows the daily expenditure on food of 30 households in a locality:
Daily expenditure (in Rs.) | Number of households |
100 - 150 | 6 |
150 - 200 | 7 |
200 - 250 | 12 |
250 - 300 | 3 |
300 - 350 | 2 |
Find the mean and median daily expenditure on food. [CBSE 2009C]
To find mean, we will solve by direct method:
DAILY EXPENDITURE (Rs.) | MID - POINT(xi) | NUMBER OF HOUSEHOLDS(fi) | fixi |
100 - 150 | 125 | 6 | 750 |
150 - 200 | 175 | 7 | 1225 |
200 - 250 | 225 | 12 | 2700 |
250 - 300 | 275 | 3 | 825 |
300 - 350 | 325 | 2 | 650 |
TOTAL | 30 | 6150 |
We have got
Σfi = 30 & Σfixi = 6150
∵ 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.
DAILY EXPENDITURE (Rs.) | NUMBER OF HOUSEHOLDS(fi) | Cf |
100 - 150 | 6 | 6 |
150 - 200 | 7 | 6 + 7 = 13 |
200 - 250 | 12 | 13 + 12 = 25 |
250 - 300 | 3 | 25 + 3 = 28 |
300 - 350 | 2 | 28 + 2 = 30 |
TOTAL | 30 |
So, N = 30
⇒ N/2 = 30/2 = 15
The cumulative frequency just greater than (N/2 = ) 15 is 25, so the corresponding median class is 200 - 250 and accordingly we get Cf = 13(cumulative frequency before the median class).
Now, since median class is 200 - 250.
∴ l = 200, h = 50, f = 12, N/2 = 15 and Cf = 13
Median is given by,
⇒ 
= 200 + 8.33
= 208.33
Hence, mean is 205 and median is 208.33