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 \({\bf{75\;mmHg}}\).

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,

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

Let\(\bar X\) be the sample mean blood pressure for 16 randomly selected women.

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

\(\bar X \sim N\left( {{\mu _{\bar X}},{\sigma _{\bar X}}^2} \right)\)

Where,

\(\begin{array}{c}{\mu _{\bar X}} = 70.2\\{\sigma _{\bar X}} = \frac{{{\sigma _X}}}{{\sqrt n }}\\ = \frac{{11.2}}{{\sqrt {16} }}\\ = 2.8\end{array}\)

Compute the probability

The z score associated with 75 mm Hg on the distribution of\(\bar X\)is

\(\begin{array}{c}z = \frac{{\bar x - {\mu _{\bar X}}}}{{{\sigma _{\bar X}}}}\\z = \frac{{75 - 70.2}}{{2.8}}\\ = 1.714\end{array}\)

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

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

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.