Answer to Question #292945 in Statistics and Probability for secret

Question

Given a standard normal distribution with mean =101 and standard deviation = 43, find P(X>145)

Answer 1

$X\sim N(101, 43^2)$

$$P(X>145)=1-P(X\leq 145)$$ $$=1-P(Z\leq \dfrac{145-101}{43})=1-P(Z\leq1.0232)=1-0.84689=0.1531$$