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The population of female pulse rates is normally distributed with mean equal to 74.0 beats per minute and standard deviation equal to 12.5 beats per minute.

Let the population mean pulse rate be $渭=74.0\text{beats}\text{per}\text{minute}$.

Let the population standard deviation of beats per minute $蟽=12.5\text{beats}\text{per}\text{minute}$.

The z-score for a given sample value has the following expression:

$z=\frac{x-渭}{蟽}$

The z-score for the sample mean has the following expression:

$z=\frac{\stackrel{炉}{x}-渭}{\frac{蟽}{\sqrt{n}}}$

a.

The sample value given has a value equal to x=80 beats per minute.

The corresponding z-score is equal to:

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ =\frac{80-74.0}{12.5} \\ =0.48 \end{gathered}$$

Thus, the following probability needs to be determined:

$P\left(z<0.48\right)$

The corresponding left tailed probability value for the z-score equal to 0.48 can be observed from the table and has a value equal to 0.6844.

Therefore, the probability that the pulse rate of the selected female is less than 80 beats per minute is equal to 0.6844.

b.

Let the sample size be equal to n = 16.

The sample mean is equal to $\stackrel{炉}{x}=80\text{beats}\text{per}\text{minute}$.

The corresponding z-score is equal to:

$$\begin{gathered} z=\frac{\stackrel{炉}{x}-渭}{\frac{蟽}{\sqrt{n}}} \\ =\frac{80-74.0}{\frac{12.5}{\sqrt{16}}} \\ =1.92 \end{gathered}$$

Thus, the following probability needs to be determined:

$P\left(z<1.92\right)$

The corresponding left tailed probability value for the z-score equal to 1.92 can be observed from the table and has a value equal to 0.9726.

Therefore, the probability that for a sample of 16 females, their mean pulse rate is less than 80 beats per minute is equal to 0.9726.

c.

Here, the sample size (16) is less than 30 but it is given that the population of female pulse rates is normally distributed.

Hence, the sample mean female pulse rate can be assumed to follow the normal distribution.