Question

The systolic blood pressure of adults in the USA is nearly normally distributed with a mean of 120 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? Give your answer rounded to two decimal places.

Answer 1

Using the normal distribution, it is found that around 0.62% of adults in the USA have stage 2 high blood pressure.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 120, \sigma = 16$.

The proportion with a stage 2 high blood pressure is one subtracted by the p-value of Z when X = 160, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{160 - 120}{16}$

Z = 2.5

Z = 2.5 has a p-value of 0.9938.

1 - 0.9938 = 0.0062.

0.0062 = 0.62% of adults in the USA have stage 2 high blood pressure.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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