Let $X_1, X_2, X_3$ be three random variable following a normal distribution $N(6,4)$. What is the probability that the largest observation exceeds 8? Hint: $Y = \max(X_1, X_2, X_3)$.
Here is what I tried:
$P(Y\leq y) = P(X_i\leq y)$ for $i=1,2,3$. Then $$ P(Y\leq y) = P(X_1 \leq y)*P(X_2 \leq y) * P(X_3 \leq y). $$ To get $P(X_ i \leq y)$, I use the $P(X > 8) = 1 - P(X < 8) = 1 - 0.841 = 0.159$. Then I plug in $0.159$ to the previous equation: $P(X_1 \leq y) * P(X_2 \leq y) *P(X_3 \leq y)$, so that $(0.159)(0.159)(0.159) = 0.0040$.
However, my teacher said this is wrong.