JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 18)
Consider the following frequency distribution :
If mean = $${{309} \over {22}}$$ and median = 14, then the value (a $$-$$ b)2 is equal to _____________.
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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