The annual income of a head physician is normally distributed. If 25% of the head physician earn less than 180 000 and 25% of them earn more than 320 000, what is the percentage that earn:
(a) less than 200 000
(b) between 280 00 and 320 000?
Let X be a normal random variable that represents the annual income of a head physician (every unit correspons to a thousand). $$P(X< 180)=0,25$$ e $$P(X > 320)=0,25$$
$$P(X> 320)=1-P(x<320)=1-P((X-\mu )/\sigma \le (320-\mu )/\sigma )=0,25$$ $$P((X-\mu )/\sigma \le (320-\mu )/\sigma )=0,75 \approx\Phi (0,675)=P((X-\mu )/\sigma \le 0,675 )$$ Then $$ (320-\mu )/\sigma=0,675$$
$$\Phi (0,675)\approx 0,75\Rightarrow \Phi (-0,675)=1-\Phi (0,675)=1-0,75=0,25$$
$$P((X-\mu )/\sigma \le (180-\mu )/\sigma )=0,25 \approx\Phi (-0,675)=P((X-\mu )/\sigma \le -0,675 )\Rightarrow (180-\mu )/\sigma =-0,675$$
Then I calculate $$\mu=250$$ and $$\sigma=103,70$$
$$P(x<200)=P((X-\mu )/\sigma \le -0,4821 )=0,75=\Phi (-0,4821)=1-\Phi (0,4821)=0,3156$$
$$P(280<X<320)=P(x<320)-P(x<180)=2*\Phi (0,0,675)-1=0,5$$
But the solution for the second question should be $$0,136$$
why?
In your answer, you have a typo where you write $P(X < 180)$ where it should be $P(X < 280)$. Therefore, we get: $$P(280 < X < 320)=P(X < 320)-P(X < 280)=0.75-P\left(\frac{X-\mu}{\sigma} < 0.289\right)$$ Now, according to my calculator, the latter probability is $\approx 0.613$, so we get: $$0.75-0.613=0.136$$