Let the mean and variance of 8 numbers x , y , 10 , 12 , 6 , 12 , 4 , 8 be 9 and 9.25 respectively. If x y , then 3 x ? 2 y is equal to _____. [2023]

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Solution

Q.
Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to _____. [2023]

Ans.
(25)

We have, $x, y, 10, 12, 6, 12, 4, 8$

Mean, $\bar{x} = \frac{x + y + 10 + 12 + 6 + 12 + 4 + 8}{8} = 9$

$\Rightarrow x + y = 20 \dots (i)$

Now, Variance $= \frac{x^{2} + y^{2} + {(10)}^{2} + {(12)}^{2} + {(6)}^{2} + {(12)}^{2} + {(4)}^{2} + {(8)}^{2}}{8} - 81$

$\Rightarrow 9.25 = \frac{x^{2} + y^{2} + 504}{8} - 81$ $\Rightarrow x^{2} + y^{2} = 218 \dots (i i)$

Now, ${(x + y)}^{2} = x^{2} + y^{2} + 2 x y$

$\Rightarrow {(20)}^{2} = 218 + 2 x y$ $\Rightarrow 2 x y = 182$

Now, ${(x - y)}^{2} = 218 - 182 = 36$ $\Rightarrow x - y = \pm 6 \dots (i i i)$

From (i) and (iii), we get $x = 13, y = 7$ $(∵ x > y)$

So, $3 x - 2 y = 39 - 14 = 25$

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