Question

The systolic blood pressure (BP) of adults in the USA is nearly normally distributed with a mean of 121 and standard deviation of 16 . Someone qualifies as having Stage 2 high blood pressure if their systolic blood pressure is 160 or higher. (a) Around what percentage of adults in the USA have stage 2 high blood pressure

Answer 1

Around 0.73% of adults in the USA have stage 2 high blood pressure

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 121 and standard deviation of 16.

This means that
`\mu = 121, \sigma = 16`

Around what percentage of adults in the USA have stage 2 high blood pressure

The proportion is 1 subtracted by the p-value of Z when X = 160. So

`Z = (X - \mu)/(\sigma)`

`Z = (160 - 121)/(16)`

`Z = 2.44`

`Z = 2.44` has a p-value of 0.9927.

1 - 0.9927 = 0.0073

0.0073*100% = 0.73%

Around 0.73% of adults in the USA have stage 2 high blood pressure