Question

Suppose X1, X2,..., X10 are independent normal random variables with mean O and variance 1. Let max{X1, X2, ..., X10} What is the largest value of t so that P(M <t) < 0.90? That is, find the 90th percentile of M.

Answer 1

The CDF for M here is computed as:

F(m) = P(M <= m) that is the probability that the maximum of all the X_i is less than m which is possible only when all the X_i's are less than m. Therefore, we get here:

F(m) = P(M <= m) = [P(X_i <= m)]¹⁰

P(M <= m) = [P(X_i <= m)]¹⁰

Now, we are to find t here such that:

P( M <= t) <= 0.9

Therefore,

[P(X_i <= m)]¹⁰ <= 0.9

$P(X_i \leq m) \leq 0.9^{1/10}$

$P(X_i \leq m) \leq 0.9895$

From standard normal tables, we have here:
P(Z < 2.309) = 0.9895

Therefore 2.309 is the required value of t here.