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 diastolic blood pressure level between \({\bf{60\;mmHg}}\) and \({\bf{80\;mmHg}}\).

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.

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

Then,

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

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

\(\begin{array}{c}{z_1} = \frac{{{x_1} - \mu }}{\sigma }\\ = \frac{{60 - 70.2}}{{11.2}}\\ \approx - 0.91\end{array}\)

\(\begin{array}{c}{z_2} = \frac{{{x_2} - \mu }}{\sigma }\\ = \frac{{80 - 70.2}}{{11.2}}\\ \approx 0.88\end{array}\)

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

The probability to the left of -0.91 is the value in the cell interaction for the row with -0.9 and the column with 0.01 in the standard normal probability table, which is 0.1814.

Thus, the probability that women will have blood pressure between 60 mm Hg and 80 mm Hg is

\(\begin{array}{l}P(60 < X < 80) = P( - 0.91 < Z < 0.88)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {\rm{Area}}\;{\rm{left}}\;{\rm{of}}\;0.88 - {\rm{Area}}\;{\rm{left}}\;{\rm{of}} - 0.91\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = P\left( {Z < 0.88} \right) - P\left( {Z < - 0.91} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.8106 - 0.1814\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.6292\end{array}\).

Thus, the probability that women have blood pressure between 60 mm Hg and 80 mm Hg is 0.6292.