Question

For males in a certain town, the systolic blood pressure is normally distributed with a mean of 130 and a standard deviation of 9. What percentage of males in the town that have a systolic blood pressure higher than 111, to the nearest tenth?

Answer 1

z = (111 − 130)/9 = −19/9 ≈ −2.111.
P(X > 111) = P(Z > −2.111) = P(Z < 2.111) ≈ 0.9826 = 98.26%.

To the nearest tenth: 98.3%.