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

Let X be the blood pressure measurement of women.

Then,

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

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

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

$\stackrel{炉}{X}~N\left({渭}_{\stackrel{炉}{X}},{{蟽}_{\stackrel{炉}{X}}}^2\right)$

Where,

$$\begin{gathered} {渭}_{\stackrel{炉}{X}}=70.2 \\ {蟽}_{\stackrel{炉}{X}}=\frac{{蟽}_X}{\sqrt{n}} \\ =\frac{11.2}{\sqrt{16}} \\ =2.8 \end{gathered}$$

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

$$\begin{gathered} z=\frac{\stackrel{炉}{x}-{渭}_{\stackrel{炉}{X}}}{{蟽}_{\stackrel{炉}{X}}} \\ z=\frac{75-70.2}{2.8} \\ =1.714 \end{gathered}$$

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

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

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.