Given a standard normal distribution with mean =58 and standard deviation = 40, find P(X<96)
By condition,
a=58, σ=40a = 58,\,\,\sigma = 40a=58,σ=40.
Then
P(X<96)=P(−∞<X<96)=Φ(β−aσ)−Φ(−∞)=Φ(96−5840)+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 = 0.8289P(X<96)=P(−∞<X<96)=Φ(σβ−a)−Φ(−∞)=Φ(4096−58)+Φ(∞)=Φ(0,95)+Φ(∞)≈0.3289+0.5=0.8289
Answer: 0.8289
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