Hospital work-related injuries. According to an Occupational and Health Safety Administration (OHSA) 2014 report, a hospital is one of the most dangerous places to work. The major cause of injuries that resulted in missed work was overexertion. Almost half (48%) of the injuries that result in missed work were due to overexertion. Let x be the number of hospital-related injuries caused by overexertion.

a. Explain why x is approximately a binomial random variable.

b. Use the OHSA report to estimate p for the binomial random variable of part a.

c. Consider a random sample of 100 hospital workers who missed work due to an on-the-job injury. Use the p from part b to find the mean and standard deviation of, the proportion of the sampled workers who missed work due to overexertion.

d. Refer to part c. Find the probability that the sample proportion is less than .40.

The proportion of all hospital-related injuries resulting from missed work was due to overexertion$p=0.48$.

Let x be the number of hospital-related injuries caused by overexertion.

A random sample of $n=100$is selected.

a.

There are two outcomes, whether the injuries are due to overexertion. The proportion (probability) of injuries due to overexertion is constant for every patient. The number of patients surveyed is also fixed. These conditions of the binomial distribution are satisfied.

Therefore, x is approximately a binomial random variable.

b.

The probability estimate for a binomial random variable from the OHSA report is $p=0.48$.

c.

The mean of the sampling distribution of $\hat{p}$is:

$$\begin{gathered} E\left(\hat{p}\right)=p \\ =0.48 \end{gathered}$$.

Since$E\left(\hat{p}\right)=p$.

The standard deviation of the sampling distribution of $\hat{p}$is obtained as:

$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}} \\ =\sqrt{\frac{0.48\times 0.52}{100}} \\ =\sqrt{0.002496} \\ 0.04995998 \end{gathered}$$.

Therefore, ${\sigma }_{\hat{p}}=0.05$.

d.

The probability that the sample proportion is less than 0.40 is obtained as:

$\begin{array}{rcl}\left(\hat{p}<0.40\right)& =& P\left(\frac{\hat{p}-p}{{\sigma }_{\hat{p}}}<\frac{0.40-p}{{\sigma }_{\hat{p}}}\right)\\ & =& P\left(Z<\frac{0.40-0.45}{0.05}\right)\\ & =& P\left(Z<\frac{-0.05}{0.05}\right)\\ & =& P\left(Z<-1.00\right)\\ & =& 0.1587\end{array}$.

The probability is obtained using the z-table; in the z-table, the value at the intersection of -1.00 and 0.00 is the desired value.

Thus the required probability is 0.1587.