In the following data the median of the runs scored by 60 top batsmen of the world in one - day international cricket matches is 5000. Find the missing frequencies x and y.
Runs scored | 2500 - 3500 | 3500 - 4500 | 4500 - 5500 | 5500 - 6500 | 6500 - 7500 | 7500 - 8500 |
Number of batsmen | 5 | X | y | 12 | 6 | 2 |
Given: Median = 5000 & N = 60
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, where x is the unknown frequency.
RUNS SCORED | NUMBER OF BATSMEN(fi) | Cf |
2500 - 3500 | 5 | 5 |
3500 - 4500 | x | 5 + x |
4500 - 5500 | y | 5 + x + y |
5500 - 6500 | 12 | 5 + x + y + 12 = 17 + x + y |
6500 - 7500 | 6 | 17 + x + y + 6 = 23 + x + y |
7500 - 8500 | 2 | 23 + x + y + 2 = 25 + x + y |
TOTAL | 25 + x + y |
Median = 5000 (as already mentioned in the question)
5000 lies between 4500 - 5500 ⇒ Median class = 4500 - 5500
∴ l = 4500, h = 1000, f = y, N/2 = (25 + x + y)/2 and Cf = 5 + x
Median is given by,
⇒ 
⇒ 
⇒ 5000 – 4500 = (25000 – 1000x)/y
⇒ 500y = 25000 – 1000x
⇒ 2x + y = 50 …(i)
And given that N = 60
⇒ 25 + x + y = 60
⇒ x + y = 35 …(ii)
Solving equations (i) & (ii), we get
(2x + y) – (x + y) = 50 – 35
⇒ x = 15
Substituting x = 15 in eq.(ii),
15 + y = 35
⇒ y = 20
Thus, the unknown frequencies are x = 15 and y = 20.