Solving Probability using standard Deviation and Mean

Question

Given a normal random variable with mean 90 and a standard deviation 3, find the probability that x is greater than or equal to 86.67 and less than or equal to 96.33

Answer 1

Hint: If $Y\text{~}N(\mu,\sigma^2)$, then $X=\frac{Y-\mu}{\sigma}\text{~}N(0,1)$.