JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 18)

Consider the following frequency distribution :

Class : 0-6 6-12 12-18 18-24 24-30
Frequency : $$a $$ $$b$$ 12 9 5

If mean = $${{309} \over {22}}$$ and median = 14, then the value (a $$-$$ b)2 is equal to _____________.
Answer
4

Explanation

Class Frequency $${x_i}$$ $${f_i}{x_i}$$
0-6 a 3 3a
6-12 b 9 9b
12-18 12 15 180
18-24 9 21 189
24-30 5 27 135
$$N = (26 + a + b)$$ $$(504 + 3a + 9b)$$


Mean = $${{3a + 9b + 180 + 189 + 135} \over {a + b + 26}} = {{309} \over {22}}$$

$$ \Rightarrow 66a + 198b + 11088 = 309a + 309b + 8034$$

$$ \Rightarrow 243a + 111b = 3054$$

$$ \Rightarrow 81a + 37b = 1018$$ $$\to$$ (1)

Now, Median $$ = 12 + {{{{a + b + c} \over 2} - (a + b)} \over {12}} \times 6 = 14$$

$$ \Rightarrow {{13} \over 2} - \left( {{{a + b} \over 4}} \right) = 2$$

$$ \Rightarrow {{a + b} \over 4} = {9 \over 2}$$

$$ \Rightarrow a + b = 18$$ $$\to$$ (2)

From equation (1) $ (2)

a = 8, b = 10

$$\therefore$$ $${(a - b)^2} = {(8 - 10)^2}$ = 4

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