On a particular test whose scores are distributed normally, the 2nd percentile is 1,720, while the 84th percentile is 1,990. What score, rounded to the nearest 10, most closely corresponds to the 16th percentile?
(A) 1,750
(B) 1,770
(C) 1,790
(D) 1,810
(E) 1,830
My answer is (1990-1720)/82 = 3.29, then 3.29*14 + 1720 = 1766.10, hence option B. But the answer is actually D.
From the given information you have two equations:
$$\Phi\left( \frac{1720-\mu}{\sigma} \right)=0.02 ,\Phi\left( \frac{1990-\mu}{\sigma} \right)=0.84$$
Take the inverse function of $\Phi(z)$
$$ \frac{1720-\mu}{\sigma} =\Phi^{-1}\left(0.02\right) ,\frac{1990-\mu}{\sigma} =\Phi^{-1}\left(0.84\right)$$
$$ \frac{1720-\mu}{\sigma} =-2.054 , \frac{1990-\mu}{\sigma} =0.994$$
Now you have a small equation system with two variables and two equations which can be solved.
I get $\mu=1901.95$ and $\sigma=88.5827$
Then evaluate the value of $\Phi^{-1}(0.16)$. Since the standard normal distribution is symmetric around 0 we have $\Phi^{-1}(0.16)=-\Phi^{-1}(0.84)$
Finally the following equation has to be solved
$$\frac{x-1901.95}{88.5827}=-0.994$$