Minimum of an Order Statistic with probability Let $X_1,\ldots,X_n$ constitute a random sample of size $n$ from a normal distribution with $\mu = 0$ and var= 2. Find the smallest value of n such that $P(\min(X^2_1,\ldots,X^2_n)\leq .002) \geq .8$
Essentially we want the smallest value of n that would make the min of an order statistic less than .002 with 80% certainty.
 A: Let $p = \Pr(X_1^2 > 0.002) = 1 - \Pr(-\sqrt{0.002} < X_1 < \sqrt{0.002}\,)$.  Your software or a table in the back of the book will tell you what $p$ is numerically.
Then
\begin{align}
\Pr(\min > 0.002) & = \Pr(X_1^2>0.002\ \&\ \cdots\ \&\ X_n>0.002) \\[10pt]
& = \Pr(X_1^2 > 0.002)\cdots\Pr(X_n^2>0.002) \\[10pt]
& = p^n.
\end{align}
So you need $1-p^n\ge0.8$, or $p^n \le 0.2$, or $n\ge \dfrac{\log0.2}{\log p}$. ($\text{“}{\le}\text{''}$ changes to $\text{“}{\ge}\text{''}$ because $\log p$ is negative. Or if you use logarithms to a base that is less than $1$, so that the logarithms are positive, then the it changes because the logarithmic function itself is decreasing.)
A: I will assume that the $X_i$ are $iid$ $N(0,2)$.
Let
$$Y=\left(\frac{X_i}{\sqrt 2}\right)^2,$$ and notice that 
$X_i/\sqrt 2\sim N(0,1)$. Therefore, $Y\sim\text{Gamma}(1/2,1/2)$. Then $M = \min(2Y_1,\dotsc,2Y_n)$, and the problem becomes
\begin{align*}
P(M\leq .002) &= 1-P(M>.002)\\
&= 1-P(2Y_1>.002,\dotsc,2Y_n>.002)\\
&=1-\{P(Y_1>.001)\}^n\\
&\geq .8
\end{align*}
Using a calculator gives
$$1-(0.9747729)^n \geq .8$$
and this gives that $62.98978 \geq n$. So choose $n = 62$.
