The mean and variance of seven observations are 8 and 16 respectively. If five of the observations are $2,4…
The mean and variance of seven observations are 8 and 16 respectively. If five of the observations are $2,4,10,12,14$, then the product of remaining two observations is
45
44
48
40
Solution
Let the unknown numbers be $x$ and $y$.
$\begin{aligned}
& \text { Mean }=8 \\
& \Rightarrow \frac{2+4+10+12+14+x+y}{7}=8 \\
& \Rightarrow x+y=14 ...(i)\\
& \text { Variance }=16 \\
& \Rightarrow \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7}
\end{aligned}$
$-(\text { mean })^2=16$
$\begin{aligned}
& \Rightarrow 460+x^2+y^2=7\left[16+(8)^2\right] \\
& \Rightarrow 460+x^2+y^2=560 \\
& \Rightarrow x^2+y^2=100...(ii)
\end{aligned}$ Solving (i) and (ii), we get
$x=6, y=8 \text { or } x=8, y=6$
$\therefore \quad$ Product $=48$