Let the mean and variance of 8 numbers be and , respectively. Then the mean of 4 numbers is: [2026]

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Solution

Q.
Let the mean and variance of 8 numbers $- 10, - 7, - 1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$ , respectively. Then the mean of 4 numbers $x, y, x + y + 1,$ $| x - y |$ is: [2026]

1 9

2 12

3 10

4 11

Ans.
(4)

$Mean = \frac{- 18 + x + y + 2 + 9 + 16}{8} = \frac{7}{2}$

$\Rightarrow \frac{x + y + 9}{8} = \frac{7}{2} \Rightarrow x + y + 9 = 28 . . . . . (1)$

$Variance = \frac{\sum z_{i}^{2}}{8} - μ^{2} = \frac{293}{4}$

$\Rightarrow \frac{10^{2} + 7^{2} + 1^{2} + x^{2} + y^{2} + 2^{2} + 9^{2} + 16^{2}}{8} - {(\frac{7}{2})}^{2} = \frac{293}{4} . . . . (2)$

$Solving (1) & (2) \Rightarrow x = 12, y = 7$

$Mean of (1 + x + y), x, y,$ $| y - x |$ $is$

$\Rightarrow \frac{20 + 12 + 7 + 5}{4} = \frac{44}{4} = 11$

$Option (4)$

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