91Ó°ÊÓ

Here, it is given that the salaries of physicians in a certain specialty are approximately normally distributed.

$25\%$of these physicians earn $<\$180,000$and

$25\%$of these physicians earn$>\$320,000$

Let $X$be the random variable that represents the salary in thousand dollars.

So,

localid="1646737232885" $$\begin{gathered} P(X<180)=0.25 \\ P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}<\frac{180-\mathrm{μ}}{\mathrm{σ}}\right)=0.25 \\ P(Z<z)=0.25 \end{gathered}$$

From standard normal table values, the critical value of $z$for the one tail test and the corresponding cumulative area of $0.25$is $-0.675$.

So,

localid="1646736797891" $$\begin{gathered} z=-0.675 \\ \frac{180-\mathrm{μ}}{\mathrm{σ}}=-0.675 \\ 180=\mathrm{μ}-0.675\mathrm{σ}...................\left(1\right) \end{gathered}$$

and

localid="1646736786547" $$\begin{gathered} P(X>320)=0.25 \\ P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}>\frac{320-\mathrm{μ}}{\mathrm{σ}}\right)=0.25 \\ 1-P(Z≤z)=0.25 \\ P(Z≤z)=0.75 \end{gathered}$$

From standard normal table values, the critical value of $z$for the one tail test and the corresponding cumulative area of $0.75$is $0.675$.

So,

$$\begin{gathered} z=0.675 \\ \frac{320-\mathrm{μ}}{\mathrm{σ}}=-0.675 \\ 320=\mathrm{μ}+0.675\mathrm{σ}...................\left(2\right) \end{gathered}$$

From equation $\left(1\right)\text{and}\left(2\right)$, we get

$$\begin{gathered} 2\mathrm{μ}=500 \\ ∴\mathrm{μ}=250 \end{gathered}$$

Substituting the value of $\mathrm{μ}$in equation $1$, we get

$$\begin{gathered} 180=250-0.675\mathrm{σ} \\ 0.675\mathrm{σ}=70 \\ ∴\mathrm{σ}=103.704 \end{gathered}$$

$$\begin{gathered} P(X<200)=P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}<\frac{200-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P\left(z<\frac{200-250}{103.704}\right) \\ =P\left(z<-0.48\right) \\ =0.3149 \end{gathered}$$

Therefore, the fraction of physicians that earn less than$\$200,000$is$0.3149$.

$$\begin{gathered} P(280<X<320)=P\left(\frac{280-\mathrm{μ}}{\mathrm{σ}}<\frac{X-\mathrm{μ}}{\mathrm{σ}}<\frac{320-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P\left(\frac{280-250}{103.704}<z<\frac{200-250}{103.704}\right) \\ =P\left(0.29<z<0.67\right) \\ =P\left(z<0.67\right)-P(z<0.29) \\ =0.7486-0.6138 \\ =0.1348 \\ =0.3149 \end{gathered}$$

Therefore, the fraction of physicians that earn between$\$280,000\text{and}\$320,000$is$0.3149$.