Q82 / 150 · MHT CET 2024 (10 May Shift 2) · Mathematics

The mean and variance of 7 observations are 8 and 16 respectively. If first five observations are $2,4,10,12,14$, then absolute difference of remaining two observations is

  • A
    1
  • Correct answer
    B
    2
  • C
    3
  • D
    4

Solution

Let the unknown numbers be $x$ and $y$. Mean $=8$ $\begin{aligned} & \Rightarrow \frac{2+4+10+12+14+x+y}{7}=8 \\ & \Rightarrow x+y=14 ...(i)\\ & \text { Variance }=16 \\ & \Rightarrow \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} \\ & \Rightarrow 460+x^2+y^2=7\left[16+(8)^2\right] \\ & \Rightarrow 460+x^2+y^2=560 \\ & \Rightarrow x^2+y^2=100...(ii) \end{aligned}$
Solving (i) and (ii), we get $x=6, y=8 \text { or } x=8, y=6$ $\therefore \quad$ Required absolute difference $=|x-y|=2$

Hence, the correct answer is option: 'B'.

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