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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 observation 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=70 beats per minute.

The corresponding z-score is equal to:

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ =\frac{70-74.0}{12.5} \\ =-0.32 \end{gathered}$$

Thus, the required probability value is computed using the standard normal table as follows:

$$\begin{gathered} P\left(z>-0.32\right)=P\left(z<0.32\right) \\ =0.6255 \end{gathered}$$

Therefore, the probability that the pulse rate of the selected female is greater than 70 beats per minute is equal to 0.6255.

b.

Let the sample size be equal to n = 25.

The sample mean is equal to $\stackrel{炉}{x}=70\text{beats}\text{per}\text{minute}$.

The corresponding z-score is equal to:

$$\begin{gathered} z=\frac{\stackrel{炉}{x}-渭}{\frac{蟽}{\sqrt{n}}} \\ =\frac{70-74.0}{\frac{12.5}{\sqrt{25}}} \\ =-1.6 \end{gathered}$$

Thus, the required probability is computed using the standard normal table as follows:

$$\begin{gathered} P\left(z>-1.6\right)=P\left(z<1.6\right) \\ =0.9452 \end{gathered}$$

Therefore, the probability that for a sample of 25 females, their mean pulse rate is greater than 70 beats per minute is equal to 0.9452.

c.

Although the sample size (25) is less than 30, 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.