91Ó°ÊÓ

The proportion of all dentists that uses nitrous oxide in their practice is p = 0.60.

A random sample of size n = 75 is selected.

Let $\hat{p}$represents the sample proportion that uses laughing gas in practice.

a

The mean of the sample proportion is obtained as:

$$\begin{gathered} E\left(\hat{p}\right)=p \\ =0.60 \end{gathered}$$.

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

Therefore, $E\left(\hat{p}\right)=0.60$.

b.

The standard deviation of the sample proportion is obtained as:

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}} \\ =\sqrt{\frac{0.60×0.40}{75}} \\ =\sqrt{\frac{32.24}{75}} \\ =\sqrt{0.4299} \\ =0.6557 \end{gathered}$$.

Therefore,${\mathrm{σ}}_{\hat{p}}=0.6557$.

c.

Here,

$$\begin{gathered} n\hat{p}=75×0.60 \\ =45 \\ >15 \end{gathered}$$,

and

$$\begin{gathered} n\left(1-\hat{p}\right)=75×0.40 \\ =30 \\ >15 \end{gathered}$$

The conditions to use normal approximation are satisfied. Therefore, the shape of the sampling distribution of the sample proportion is approximately symmetric (normal).

d.

To find the required probability normal approximation is used.

Therefore,

$$\begin{gathered} P\left(\hat{p}>0.70\right)=P\left(\frac{\hat{p}-p}{{\mathrm{σ}}_{\hat{p}}}>\frac{0.70-p}{{\mathrm{σ}}_{\hat{p}}}\right) \\ =P\left(Z>\frac{0.70-0.60}{0.6557}\right) \\ =P\left(Z>\frac{0.10}{0.6557}\right) \\ =P\left(Z>0.15\right) \\ =1-P\left(Zâ\copyright ½0.15\right) \\ =1-0.5596 \\ =0.4404 \end{gathered}$$

The z-table is used to find the probability; the value at the intersection of 0.10 and 0.05 is the probability of a z-score less than or equal to 0.15.

Hence, the required probability is 0.4404.