Question

The median of the following data is 525. Find the values of x and y, if the total frequency is 100.

$\begin{array}{cc}ClassInterval& Frequency\\ 0-100& 2\\ 100-200& 5\\ 200-300& x\\ 300-400& 12\\ 400-500& 17\\ 500-600& 20\\ 600-700& y\\ 700-800& 9\\ 800-900& 7\\ 900-1000& 4\end{array}$ [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

$\begin{array}{ccc}ClassInterval& Frequency& Cumulativefrequency\\ 0-100& 2& 2\\ 100-200& 5& 7\\ 200-300& x& 7+x\\ 300-400& 12& 19+x\\ 400-500& 17& 36+x\\ 500-600& 20& 56+x\\ 600-700& y& 56+x+y\\ 700-800& 9& 65+x+y\\ 800-900& 7& 72+x+y\\ 900-1000& 4& 76+x+y\end{array}$

It is given that n = 100

So, 76 + x + y = 100, i.e., x + y = 24---(1)

The median is 525, which lies in the class 500 - 600

So, l = 500, f = 20, $c_f$ = 36 + x, h = 100

Using the formula: Median $=l+\left(\frac{\frac{n}{2}-c_f}{f}\right)h$, we get

i.e., $525=500+\left(\frac{50-36-x}{20}\right)\times 100$

i.e., $525-500=(14-x)\times 5$

i.e., $25=70-5x$

i.e., $5x=70-25=45$

i.e., $x=9$

From (1), we get $9+y=24$
y = 24 - 9= 15

The values of x,y are 9,15 respectively.