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 $\square$.
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 83\), \(\mu=76.0\), and \(\sigma = 12.5\).

$$z=\frac{83 - 76}{12.5}=\frac{7}{12.5}=0.56$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt83)\), which is equivalent to \(P(Z\lt0.56)\) in the standard normal distribution (\(X\) is the pulse rate variable and \(Z\) is the standard normal variable).
Looking up the value of \(z = 0.56\) in the standard normal distribution table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: normalcdf(-1000,0.56)), we find that \(P(Z\lt0.56)=0.7123\)

Answer:

\(0.7123\)