Answer to Question #290412 in Statistics and Probability for lusientoo

Question

Given a standard normal distribution with mean =136 and standard deviation = 39, find P(X>84)

Answer 1

$X\sim N(136, 39^2)$

$$P(X>84)=1-P(X\leq 84)$$

$$=1-P(Z\leq \dfrac{84-136}{39})=1-P(Z\leq-\dfrac{4}{3})$$

$$\approx1-P(Z\leq -1.3333)\approx0.9088$$