Question

The scores on an exam are normally distributed, with a mean of 77 and a standard deviation of 10. What percentage of the scores are greater than 87? 34% 50% 84% 16%

Answer 1

Given

$\begin{gathered} \mu=77 \\ \sigma=10 \end{gathered}$

The z-score formula is

$z=\frac{x-\mu}{\sigma}$

In our case,

$\Rightarrow z=\frac{87-77}{10}=\frac{10}{10}=1$

Using a z-score table,

$\Rightarrow P(X<87)=0.8413$

Then,

$\begin{gathered} P(X>87)=1-P(X<87)=1-0.8413=0.1587=15.87\% \\ \Rightarrow P(X>87)\approx16\% \end{gathered}$

Thus, the answer is 16%