Statistics
Mean-Variance to Find Values — Probability
nta_pyq_2026_jan
Grade None
Question:
Let the mean and variance of 7 observations $2,4,10,x,12,14,y$, $x>y$, be 8 and 16 respectively. Two numbers are chosen from $\{1,2,3,x-4,y,5\}$ one after another without replacement, then the probability that the smaller number among the two chosen numbers is less than 4, is:
$\dfrac{3}{5}$
$\dfrac{1}{3}$
$\dfrac{2}{5}$
$\dfrac{4}{5}$
Step-by-Step Solution
Key Concept: Mean $=8$: $x+y=14$. Variance $=16$: $(x-8)^2+(y-8)^2=4$. Solving with $x>y$: $x=8+\sqrt{2}$? ... Actually $(x-8)^2+(y-8)^2=4$ and $x+y=14\Rightarrow(x-8)+(y-8)=-2$. Let $u=x-8,v=y-8$: $u+v=-2$, $u^2+v^2=4\Rightarrow uv=-1$... wait: $u^2+v^2=(u+v)^2-2uv=4-2uv=4$, so $uv=0$. Thus $x=8,y=6$.
$x=8$, $y=6$. $P=\tfrac{4}{5}$.
Correct Answer: 4