Question

For males in a certain town, the systolic blood pressure is normally distributed with a mean of 125 and a standard deviation of 6. What percentage of males in the town that have a systolic blood pressure higher than 117, to the nearest tenth?

Answer 1

Let X ~ N(125, 6^2). We want P(X > 117).

Compute z = (117 − 125)/6 = −8/6 ≈ −1.3333.
P(X > 117) = P(Z > −1.3333) = Φ(1.3333) ≈ 0.9088.

So about 90.9% of males have systolic blood pressure higher than 117.