Given a standard normal distribution with mean =136 and standard deviation = 39, find P(X>84)

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$X\sim N(136, 39^2)$

$P(X>84)=1-P(X\leq 84)$

$=1-P(Z\leq \dfrac{84-136}{39})=1-P(Z\leq-\dfrac{4}{3})$

$\approx1-P(Z\leq -1.3333)\approx0.9088$

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