Question #290413

Given a standard normal distribution with mean =62 and standard deviation = 44, find P(X>92)


1
Expert's answer
2022-01-26T14:46:26-0500
P(X>92)=1P(X92)=P(X>92)=1-P(X\leq92)=

=1P(Z92μσ)=1P(Z926244)=1-P(Z\leq\dfrac{92-\mu}{\sigma})=1-P(Z\leq\dfrac{92-62}{44})

1P(Z0.681818)0.247677\approx 1-P(Z\leq0.681818)\approx0.247677


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