Answer to Question #290414 in Statistics and Probability for lusientoo

Question

Given a standard normal distribution with mean =58 and standard deviation = 40, find P(X<96)

Answer 1

By condition,

$a = 58,\,\,\sigma = 40$.

Then

$P(X < 96) = P( - \infty < X < 96) = \Phi \left( {\frac{{\beta - a}}{\sigma }} \right) - \Phi \left( { - \infty } \right) = \Phi \left( {\frac{{96 - 58}}{{40}}} \right) + \Phi \left( \infty \right) = \Phi \left( {0,95} \right) + \Phi \left( \infty \right) \approx 0.3289 + 0.5 = 0.8289$

Answer: 0.8289