Question

assume that females have pulse rates that are normally distributed with a mean of \\( \mu = 74.0 \\) beats per minute and a standard deviation of \\( \sigma = 12.5 \\) beats per minute. complete parts (a) through (c) below.

a. if 1 adult female is randomly selected, find the probability that her pulse rate is less than 80 beats per minute.
the probability is .6844 .
(round to four decimal places as needed.)
b. if 4 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 80 beats per minute.
the probability is .8315 .
(round to four decimal places as needed.)
c. why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
a. since the distribution is of sample means, not individuals, the distribution is a normal distribution for any sample size.
b. since the mean pulse rate exceeds 30, the distribution of sample means is a normal distribution for any sample size.
c. since the distribution is of individuals, not sample means, the distribution is a normal distribution for any sample size.
d. since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 80\), \(\mu=74.0\), and \(\sigma = 12.5\).

$$z=\frac{80 - 74}{12.5}=\frac{6}{12.5}=0.48$$

Step2: Find the probability using the z - score

Using the standard normal distribution table (or a calculator with a normal - distribution function), we find \(P(Z<0.48)\).
Looking up the value in the standard normal table, \(P(Z < 0.48)=0.6844\)

Step3: For part (b), calculate the standard error of the mean

The formula for the standard error of the mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(n = 4\), \(\sigma=12.5\). So \(\sigma_{\bar{x}}=\frac{12.5}{\sqrt{4}}=\frac{12.5}{2}=6.25\)

Step4: Calculate the z - score for the sample mean

The formula for the z - score of the sample mean is \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\), with \(\bar{x}=80\), \(\mu = 74.0\), and \(\sigma_{\bar{x}}=6.25\)

$$z=\frac{80 - 74}{6.25}=\frac{6}{6.25}=0.96$$

Step5: Find the probability using the z - score for the sample mean

Using the standard normal distribution table (or a calculator with a normal - distribution function), we find \(P(Z < 0.96)\)
Looking up the value in the standard normal table, \(P(Z<0.96)=0.8315\)

Step6: Explain the normal distribution of sample means

The Central Limit Theorem states that if the population is normally distributed, then the distribution of sample means \(\bar{X}\) is also normally distributed for any sample size \(n\). In part (b), since the population of female pulse rates is normally distributed (\(\mu = 74.0\), \(\sigma=12.5\)), the sampling distribution of sample means \(\bar{X}\) (where \(n = 4\)) is also normally distributed.

Answer:

a. The probability that a randomly - selected adult female has a pulse rate less than \(80\) is \(0.6844\).
b. The probability that \(4\) randomly - selected adult females have a mean pulse rate less than \(80\) is \(0.8315\).
c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.