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 the probability that a randomly selected woman has a normal diastolic blood pressure level, which is below\({\bf{80\;mmHg}}\).

Short answer

The probability that a randomly selected woman has a normal diastolic blood pressure level is\(0.8106\).

Step-by-step solution

Given Information

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

Define the random variable

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

Then,

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

Compute the z-score for 80 mm Hg

The standardized score is the value \(x\) decreased by the mean and then divided by the standard deviation.

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{80 - 70.2}}{{11.2}}\\ = 0.875\\ \approx 0.88\end{aligned}\)

Compute the associated probability

The probability to the left of 0.88 is the value in the cell interaction for the row with 0.8 and in the column with 0.8 in the standard normal probability tablewhich is 0.8106.

Thus,

\(\begin{aligned}{l}P(X < 80) = P(Z < 0.88)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.8106\end{aligned}\)

Thus, the probability that women haveblood pressure below 80 mm Hg is 0.8106.