Q. 94.0( 29 Votes )
In the following
Answer :
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.

Given,
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 = 60/2=30
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.
Rate this question :





















