The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is [2023]

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Solution

Q.
The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is [2023]

1 1216

2 1456

3 1072

4 1792

Ans.
(3)

Let the missing observations be $x$ and $y$ .

$∴$ $Mean = \frac{1 + 3 + 5 + x + y}{5}$

$\Rightarrow 5 = \frac{9 + x + y}{5} \Rightarrow x + y = 16$ ...(i)

and variance $= \frac{x^{2} + y^{2} + 35}{5} - 25 \Rightarrow 8 = \frac{x^{2} + y^{2} + 35 - 125}{5}$

$\Rightarrow x^{2} + y^{2} = 40 + 90 = 130$ ..(ii)

Consider ${(x + y)}^{2} = x^{2} + y^{2} + 2 x y$

$256 = 130 + 2 x y \Rightarrow 2 x y = 126$

Now, consider ${(x - y)}^{2}$

$\Rightarrow x^{2} + y^{2} - 2 x y = 130 - 126 \Rightarrow {(x - y)}^{2} = 4$

$\Rightarrow x - y = 2$ ...(iii) or $x - y = - 2$ ...(iv)

Using (i) and (iii), we have $x = 9$ and $y = 7$

Using (i) and (iv), we have $x = 7$ and $y = 9$

$∴$ $Sum of cubes of x and y = 7^{3} + 9^{3} = 343 + 729 = 1072$

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