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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:

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

Therefore, ${蟽}_{\hat{p}}=0.05$.

d.

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

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

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.