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

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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


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