91影视

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.