Question

The salaries of physicians in a certain specialty are approximately normally distributed. If $25$percent of these

physicians earn less than $\backslash (180,000$and $25$percent earn more than $\backslash )320,000$, approximately what fraction earn

(a) less than $\backslash (200,000$?

(b) between $\backslash )280,000\text{and}\$320,000$?

Short answer

(a) The fraction that earn less than $\$200,000$is$0.3149$.

(b) The fraction that earn between $\$280,000$and $\$320,000$is 0.1348.

Step-by-step solution

Part (a) Step 1. Given information.

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$>\$320,000$

Part (a) Step 2. Find the value of μ and σ.

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

So,

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

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" $$\begin{gathered} z=-0.675 \\ \frac{180-\mu }{\sigma }=-0.675 \\ 180=\mu -0.675\sigma ...................\left(1\right) \end{gathered}$$

and

localid="1646736786547" $$\begin{gathered} P(X>320)=0.25 \\ P\left(\frac{X-\mu }{\sigma }>\frac{320-\mu }{\sigma }\right)=0.25 \\ 1-P(Z\le z)=0.25 \\ P(Z\le z)=0.75 \end{gathered}$$

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,

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

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

$$\begin{gathered} 2\mu =500 \\ \therefore \mu =250 \end{gathered}$$

Substituting the value of $\mu$in equation $1$, we get

$$\begin{gathered} 180=250-0.675\sigma \\ 0.675\sigma =70 \\ \therefore \sigma =103.704 \end{gathered}$$

Part (a) Step 3. Calculate the fraction of physicians that earn less than $200,000.

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

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

Part (b) Step 1. Calculate the fraction of physicians that earn between $280,000 and $320,000.

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

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