The median of the following data is 32.5.Find the value of x and y if the total frequency is 40.

It is given that the median of the following data is 32.5 and the total frequency is

40 .

From the given table, we have the frequency distribution table as,

We have,

N = \sum f_{i} = 40

\Rightarrow 31 + x + y = 40

\Rightarrow x + y = 9

It is given that the median is

32.5

. Clearly, it lies in the class

30 - 40 .

The general formula for median is given as,
Median

= l + \frac{\frac{N}{2} - C . F .}{f} \times h

Where,
l = lower limit of median class

= 30 .

C.F. = C.F. of previous class

= (x + 14) .

f = frequency of median class

= 12 .

h = width of median class = upper limit - lower limit

h = 40 - 30

h = 10

Substituting all the above values in given formula of median, we obtain,

\Rightarrow

32.5

= 30 + \frac{(\frac{40}{2} - (x + 14))}{12} \times 10

\Rightarrow

32.5

= 30 + \frac{(20 - x - 14)}{12} \times 10

\Rightarrow 195 = 180 + 100 - 5 x - 70

\Rightarrow 5 x = 15

\Rightarrow x = 3

Putting

x = 3

x + y = 9

, we get

y = 6 .

Therefore,

x = 3

and

y = 6 .

Hence, option (2) is correct.

The median of the following data is 32.5.Find the value of $x$ and $y$ if the total frequency is $40 .$