Question

15.34 runners. in a study of exercise, a large group of male runners walk on a treadmill for six minutes. after this exercise, their heart rates vary with mean 8.8 beats per five seconds and standard deviation 1.0 beats per five seconds. the researcher records the number of heartbeats per five seconds for each runner over a period of time. this distribution takes only whole - number values, so it is certainly not normal. a. let ( overline{x} ) be the mean number of beats per five seconds after measuring heart rate for 24 five - second intervals (two minutes). what is the approximate distribution of ( overline{x} ) according to the central limit theorem? b. what is the approximate probability that ( overline{x} ) is less than 8? c. what is the approximate probability that the heart rate of a runner is less than 100 beats per minute? (hint: restate this event in terms of ( overline{x} ).)

Explanation:

Step1: Determine the distribution of \(\bar{x}\)

According to the central limit theorem, if the sample size \(n = 24\), and the population mean \(\mu=8.8\) and population standard deviation \(\sigma = 1.0\), the mean of the sampling distribution of \(\bar{x}\) is \(\mu_{\bar{x}}=\mu = 8.8\), and the standard deviation of the sampling distribution of \(\bar{x}\) (standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{1.0}{\sqrt{24}}\approx0.204\). So \(\bar{x}\sim N(8.8,0.204^{2})\)

Step2: Calculate \(P(\bar{x}<8)\)

First, calculate the \(z\)-score: \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}=\frac{8 - 8.8}{0.204}\approx - 3.92\)
Then, using the standard normal distribution table \(P(Z < - 3.92)\approx0.00004\)

Step3: Restate the event for part c

Since there are \(12\) five - second intervals in a minute. If the heart rate is \(100\) beats per minute, then the number of beats per five - second interval is \(\frac{100}{12}\approx8.33\). So we want to find \(P(\bar{x}<8.33)\)
Calculate the \(z\)-score: \(z=\frac{8.33 - 8.8}{0.204}=\frac{-0.47}{0.204}\approx - 2.30\)
Using the standard normal distribution table \(P(Z < - 2.30)=0.0107\)

Answer:

a. \(\bar{x}\sim N(8.8,0.204^{2})\)
b. \(P(\bar{x}<8)\approx0.00004\)
c. \(P(\bar{x}<\frac{100}{12})\approx0.0107\)