Suppose that pulse rates among healthy adults are normally distributed with a mean of 80 beats/second and a standard deviation of 8 beats/second. What proportion of healthy adults have pulse rates that are more than 83 beats/sec? Round your answer to at least four decimal places.

We have been given that the pulse rates among healthy adults are normally distributed with a mean of 80 beats/second and a standard deviation of 8 beats/second. We are asked to find the proportion of healthy adults have pulse rates that are more than 83 beats/sec.

First of all, we will find z-score corresponding to sample score of 83 as:

[tex]z=\frac{x-\mu}{\sigma}[/tex], where,

z = Z-score,

x = Sample score,

[tex]\mu[/tex] = Mean,

[tex]\sigma[/tex] = Standard deviation.

Upon substituting our given values in z-score formula, we will get:

[tex]z=\frac{83-80}{8}=\frac{3}{8}=0.375\approx 0.38[/tex]

Now, we need to find the probability that a z-score is greater than 0.38.

Using formula [tex]P(z>a)=1-P(z<a)[/tex], we will get:

[tex]P(z>0.38)=1-P(z<0.38)[/tex]

Using normal distribution table, we will get:

[tex]P(z>0.38)=1-0.64803[/tex]

[tex]P(z>0.38)=0.35197[/tex]

[tex]P(z>0.38)\approx 0.3520[/tex]

Therefore, 0.3520 of healthy adults have pulse rates that are more than 83 beats/sec.