Question

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 batsman	5	x	y	12	6	2

Answer 1

We prepare the cumulative frequency table, as shown below:

Runs scored	Number of batsman `bb((f_i))`	Cumulative Frequency (cf)
2500 – 3500	5	5
3500 – 4500	x	5 + x
4500 – 5500	y	5 + x + y
5500 – 6500	12	17 + x + y
6500 – 7500	6	23 + x + y
7500 – 8500	2	25 + x + y
Total	N = Σ𝑓𝑖 = 60

Let x and y be the missing frequencies of class intervals 3500 – 4500 respectively. Then,

25 + x + y = 60 ⇒ x + y = 35            ……(1)

Median is 5000, which lies in 4500 – 5500.

So, the median class is 4500 – 5500.

∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x

Now,

Median, `"M" = "i" + (("N"/2−"cf")/"f") × "h"`

`⇒ 5000 = 4500 + ((60/2 −(5+x))/ "y")× 1000`

 `⇒ 5000 - 4500 = ((30−5−x)/"y") × 1000`

`⇒ 500 = ((25−x)/"y") × 1000`

⇒ y = 50 – 2x

⇒ 35 – x = 50 – 2x         [From (1)]

⇒ 2x – x = 50 – 35

⇒ x = 15

∴ y = 35 – x      

⇒ y = 35 – 15

⇒ y = 20

Hence, x = 15 and y = 20.