Question

assume that females have pulse rates that are normally distributed with a mean of \\( \mu = 76.0 \\) 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 83 beats per minute.
the probability is 0.7123.
(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 83 beats per minute.
the probability is \\( \square \\).
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error for the sample mean

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 12.5$ and $n = 4$, we have $\sigma_{\bar{x}}=\frac{12.5}{\sqrt{4}}=\frac{12.5}{2}=6.25$.

Step2: Calculate the z - score

The z - score formula for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x} = 83$, $\mu=76.0$, and $\sigma_{\bar{x}} = 6.25$. So, $z=\frac{83 - 76}{6.25}=\frac{7}{6.25}=1.12$.

Step3: Find the probability using the standard normal distribution

We want to find $P(\bar{X}<83)$, which is equivalent to $P(Z < 1.12)$. Looking up the value of $P(Z < 1.12)$ in the standard - normal table (or using a calculator with a normal - distribution function, e.g., in Excel: =NORM.S.DIST(1.12,TRUE)), we get $P(Z < 1.12)=0.8686$.

Answer:

$0.8686$