Answer to Question #290413 in Statistics and Probability for lusientoo

Question

Given a standard normal distribution with mean =62 and standard deviation = 44, find P(X>92)

Answer 1

$$P(X>92)=1-P(X\leq92)=$$

$$=1-P(Z\leq\dfrac{92-\mu}{\sigma})=1-P(Z\leq\dfrac{92-62}{44})$$

$$\approx 1-P(Z\leq0.681818)\approx0.247677$$