Question #290414

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


1
Expert's answer
2022-01-26T18:47:05-0500

By condition,

a=58,σ=40a = 58,\,\,\sigma = 40.

Then

P(X<96)=P(<X<96)=Φ(βaσ)Φ()=Φ(965840)+Φ()=Φ(0,95)+Φ()0.3289+0.5=0.8289P(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


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