Question

Let $Y_1<\cdots<Y_n$ be order statistics from a continuous distribution symmetric about its mean $\mu$. Find the smallest $n$ (in integer) such that $P(Y_1<\mu<Y_n)\ge 0.99$.

Hint:

For symmetric continuous distributions, probabilities involving $\mu$ often reduce to powers of $\tfrac12$ since observations fall on either side independently with probability $\tfrac12$ each.

Answer 1

Correct Answer: 8

By symmetry and continuity, $P(X\le \mu)=P(X\ge \mu)=\tfrac12$ and ties have probability $0$. Thus \[ P(Y_1<\mu<Y_n)=1-P(\text{all }\le \mu)-P(\text{all }\ge \mu) =1-2\left(\tfrac12\right)^n=1-2^{1-n}. \] Require $1-2^{1-n}\ge 0.99 \;\Rightarrow\; 2^{1-n}\le 0.01 \;\Rightarrow\; 2^{n-1}\ge 100$. Hence $n-1\ge \log_2 100 \approx 6.6438$ so the smallest integer is $\boxed{8}$.