Worker Fatigue. A study by M. Chen et al. titled "Heat Stress Evaluation and Worker Fatigue in a Steel Plant (American Industrial Hygiene Association, Vol. 64. Pp. 352-359) assessed fatigue in steelplant workers due to heat stress. If the mean post-work heart rate for casting workers equals the normal resting heart rate of $72$beats per minute (bpm), find the probability that a random sample of $29$ casting workers will have a mean post-work heart rate exceeding $78.3\mathrm{bpm}$ Assume that the population standard deviation of post-work heart rates for casting workers is $11.2$ bpm. State any assumptions that you are making in solving this problem.

The population standard variation of casting employees' post-work heart rates is $11.2$ beats per minute, whereas the mean post-work heart rate is $72$ beats per minute (bpm).

$\text{population mean and standard deviation:}{\mu }_{\overline{x}}=\mu \text{and}{\sigma }_{\hat{x}}=\sigma /\sqrt{n}\text{.}$

${\mu }_x=72\mathrm{bpm}$and ${\sigma }_x=11.2\mathrm{bpm}$are the values.

Let $X$represent the number of casting workers' post-work heart rates.

We need to calculate the likelihood that a random sample of $29$casting employees will have a mean post-work heart rate of more than $78.3\mathrm{bpm}$

A population variable $x$ has a normal distribution with a mean $\mu$ and standard deviation $\sigma$ The variable $\overline{x}$ is then normally distributed for samples of size $n$, with a mean $\mu$ and standard deviation $\sigma /\sqrt{n}$

Sample size $n=29$

Sampling distribution of the sample mean

$$\begin{gathered} {\mu }_{\overline{x}}=\mu x \\ =72 \end{gathered}$$

The sample standard deviation's sampling distribution

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{{\sigma }_x}{\sqrt{n}} \\ =\frac{11.2}{\sqrt{29}} \\ =2.08 \end{gathered}$$

The likelihood that a mean post-work heart rate surpasses $78.3\mathrm{bpm}$must be determined.

This equals $P(\overline{X}>78.3)$

$$\begin{gathered} P(\overline{X}>78.3)=P\left(\frac{\overline{X}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}>\frac{78.3-72}{2.08}\right) \\ =P(z>3.03) \\ =1-P(Z\le 3.03) \\ =1-0.9988 \\ =0.0012 \end{gathered}$$

The post-work heart rate of casting workers is assumed to be (roughly) regularly distributed in order to solve this problem.