Answer to Question #290411 in Statistics and Probability for lusientoo

Question

Given a standard normal distribution with mean =118 and standard deviation = 29, find P(76 < X < 144)

Answer 1

Given $\mu=118, \sigma=29$ then,

$p(76\lt x\lt 144)=p({76-118\over29}\lt Z\lt{144-118\over29})=p(-1.45\lt Z\lt 0.90)$

This is equivalent to,

$p(-1.45\lt Z\lt 0.90)=\phi(0.90)-\phi(-1.45)=0.8159-0.0735=0.7424$

Therefore,

$p(76\lt x\lt 144)=0.7424$