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

The **Answer to the Question**

is below this banner.

**Here's the Solution to this Question**

$P(X>92)=1−P(X≤92)=$

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

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