The mean of the numbers $a, b, 8,5,10$ is 6 and the variance is $6 \cdot 8$. Then which of the following…

The mean of the numbers $a, b, 8,5,10$ is 6 and the variance is $6 \cdot 8$. Then which of the following gives possible values of $a$ and $b$ ?
  1. $\mathrm{a}=3, \mathrm{~b}=4$
  2. $\mathrm{a}=0, \mathrm{~b}=7$
  3. $\mathrm{a}=5, \mathrm{~b}=2$
  4. $\mathrm{a}=1, \mathrm{~b}=6$

Solution

$\begin{aligned} & \text { Mean }=6 \\ \therefore \quad & \frac{a+b+8+5+10}{5}=6 \\ \Rightarrow & a+b=7 \\ \Rightarrow & (a-6)=(1-b)...(i) \\ & 6.80=\frac{\sum\left(x_i-\bar{x}\right)^2}{n} \\ \Rightarrow & 6.80=\frac{(a-6)^2+(b-6)^2+4+1+16}{5} \end{aligned}$ $\begin{aligned} & \Rightarrow 34=(a-6)^2+(b-6)^2+21 \\ & \Rightarrow(a-6)^2+(b-6)^2=13 \\ & \Rightarrow(1-b)^2+(b-6)^2=13..[From (i)] \\ & \Rightarrow b^2-2 b+1+b^2-12 b+36=13 \\ & \Rightarrow 2 b^2-14 b+24=0 \\ & \Rightarrow b^2-7 b+12=0 \\ & \Rightarrow b=3,4 \\ & \therefore \quad b=3 \Rightarrow a=4 \text { and } \\ & \quad b=4 \Rightarrow a=3 \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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