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 If 16 women are randomly selected, find the probability that the mean of their diastolic blood pressure levels is less than$\mathbf{75}\mathbf{}\mathbf{mmHg}$.

Short answer

The probability that the mean of their diastolic blood pressure levels is less than 75 mm Hg is$0.9564.$

Step-by-step solution

Given information

The blood pressure is normally distributed for a mean of 70.2 mm Hg and a standard deviation of 11.2 mm Hg.

Define random variable

Let X be the blood pressure measurement of women.

Then,

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

Let $\overline{X}$ be the sample mean blood pressure for 16 randomly selected women.

As the population is normally distributed, the sample mean distribution will be normal.

$\overline{X}~N\left({\mu }_{\overline{X}},{{\sigma }_{\overline{X}}}^2\right)$

Where,

$$\begin{gathered} {\mu }_{\overline{X}}=70.2 \\ {\sigma }_{\overline{X}}=\frac{{\sigma }_X}{\sqrt{n}} \\ =\frac{11.2}{\sqrt{16}} \\ =2.8 \end{gathered}$$

Compute the probability 

The z score associated with 75 mm Hg on the distribution of is

$$\begin{gathered} z=\frac{\overline{x}-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}} \\ z=\frac{75-70.2}{2.8} \\ =1.714 \end{gathered}$$

The probability that the mean blood pressure is lesser than 1.71 is

$P\left(\overline{X}<75\right)=P\left(Z<1.71\right)$

From the standard normal table, the left-tailed area of 1.71 is obtained corresponding to rows 1.7 and 0.01, which is 0.9564.

Thus, the probability that the mean blood pressure is lesser than 1.71 is 0.9564.