Given a standard normal distribution with mean =101 and standard deviation = 43, find P(X>145)

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$X\sim N(101, 43^2)$

$P(X>145)=1-P(X\leq 145)$ $=1-P(Z\leq \dfrac{145-101}{43})=1-P(Z\leq1.0232)=1-0.84689=0.1531$

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