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 ${\mathbf{P}}_{\mathbf{90}}$, 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,

$$\begin{gathered} X~N\left(\mu ,{\sigma }^2\right) \\ ~N\left(70.2,{11.2}^2\right) \end{gathered}$$

Compute the z score corresponding to the 90th percentile

Let $X=P_{90}$be the 90^th percentile score with the corresponding zscore of such that

Thus, as per the definition of percentile,

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

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

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

Thus,$z_{90}=1.28$ .

Compute the 90th percentile

Substituting the zscore in equation (1),

$$\begin{gathered} 1.28=\frac{P_{90}-70.2}{11.2} \\ {P}_{90}=84.536 \end{gathered}$$

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