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 batsmen	5	x	y	12	6	2

Answer 1

We prepare the cumulative frequency table, as shown below:

Runs scored	Number of batsmen (fi)	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 = ∑fi = 60

Let x and y be the missing frequencies of class intervals 3500−4500 and 4500−5500 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,

$$\begin{gathered} \mathrm{Median}=l+\left(\frac{{\displaystyle \frac{N}{2}}-cf}{f}\right)\times h \\ \Rightarrow 5000=4500+\left(\frac{{\displaystyle \frac{60}{2}}-\left(5+x\right)}{y}\right)\times 1000 \\ \Rightarrow 5000-4500=\left(\frac{30-5-x}{y}\right)\times 1000 \\ \Rightarrow 500=\left(\frac{25-x}{y}\right)\times 1000 \\ \Rightarrow y=50-2x \\ \Rightarrow 35-x=50-2x\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow 2x-x=50-35 \\ \Rightarrow x=15 \end{gathered}$$

∴ y = 35 − x [From (1)]
⇒ y = 35 − 15
⇒ y = 20

Hence, x = 15 and y = 20.