Question

In Exercises 6–10, assume that women have diastolic blood pressure measures that are normally distributed with a mean of 70.2 mm Hg and a standard deviation of 11.2 mm Hg (based on Data Set 1 “Body Data” in Appendix B).

Diastolic Blood Pressure Find \({{\bf{P}}_{{\bf{90}}}}\), the 90th percentile for the diastolic blood pressure levels of women.

Short answer

The 90^th percentile for the blood pressure of women is 84.5 mm Hg.

Step-by-step solution

Given information

Diastolic blood pressure measures for women are normally distributed with a mean of 70.2 mm Hg and a standard deviation of 11.2 mm Hg.

Define the random variable

Let X be the random variable for the blood pressure of women.

Then,

\(\begin{array}{c}X \sim N\left( {\mu ,{\sigma ^2}} \right)\\ \sim N\left( {70.2,{{11.2}^2}} \right)\end{array}\)

Compute the z score corresponding to the 90th percentile

Let \(X = {P_{90}}\) be the 90^th percentile score with the corresponding zscore of \({z_{90}}\)such that

\({z_{90}} = \frac{{{P_{90}} - \mu }}{\sigma }\;\;\;\;\;\;\;\;\;\;...\left( 1 \right)\)

Thus, as per the definition of percentile,

\(\begin{array}{c}P\left( {X < {P_{90}}} \right) = P\left( {Z < {z_{90}}} \right)\\ = 0.90\end{array}\)

In the standard normal table, the zscore 1.28 has the closest value of 0.8997 for the cumulative probability of 0.90.

Thus, \({z_{90}} = 1.28\) .

Compute the 90th percentile

Substituting the zscore in equation (1),

\(\begin{array}{c}1.28 = \frac{{{P_{90}} - 70.2}}{{11.2}}\\{P_{90}} = 84.536\end{array}\)

Thus, the 90^th percentile is 84.5 mm Hg.