Assume that the random variable X normally distributed, with mean 90 and standard deviation 15. Compute...

Assume that the random variable X normally distributed, with mean 90 and standard deviation 15. Compute the probability P(X>102).

Expert Solution

P ( X > 102 ) = 1 - P ( X < 102 )
Standardizing the value

Z = ( 102 - 90 ) / 15
Z = 0.8

P ( Z > 0.8 )
P ( X > 102 ) = 1 - P ( Z < 0.8 )
P ( X > 102 ) = 1 - 0.7881
P ( X > 102 ) = 0.2119

orchestra answered 1 year ago

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