JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Statistics

Consider the following frequency distribution :

\begin{tabular}{lccccc}
Class : &

0 - 6

6 - 12

12 - 18

18 - 24

24 - 30

\\
Frequency : & a & b & 12 & 9 & 5
\end{tabular}

If mean

= \frac{309}{22}

and median

= 14

, then the value

(a - b)^{2}

is equal to

Answer: 4

Solution

The class marks are

3, 9, 15, 21, 27

and the total frequency is

N = a + b + 26

. The mean condition gives

\frac{3 a + 9 b + 180 + 189 + 135}{a + b + 26} = \frac{309}{22} \Rightarrow 22 (3 a + 9 b + 504) = 309 (a + b + 26)

which reduces to

81 a + 37 b = 1018

. The median

14

falls in the class

12 - 18

, so with

L = 12

f = 12

h = 6

and cumulative frequency

a + b

before that class,

14 = 12 + \frac{\frac{a + b + 26}{2} - ( a + b )}{12} \times 6 \Rightarrow \frac{26 - a - b}{4} = 2 \Rightarrow a + b = 18

Substituting

b = 18 - a

into

81 a + 37 b = 1018

gives

44 a = 352

, so

a = 8

and

b = 10

. Hence

(a - b)^{2} = 4

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer