Question

Math 132 final exam grade is normally distributed with mean 68 and standard deviation 23. Final exam score above 92 corresponds to an A Approximately what percent of the class gotan A?

Answer 1

Since the function is a normal distribution, we can use the z-score formula shown below

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

In our case,

$\mu=68,\sigma=23$

Then, set x=92 and solve for Z

$\begin{gathered} x=92 \\ \Rightarrow Z=\frac{92-68}{23}=\frac{24}{23} \\ \Rightarrow Z=\frac{24}{23}=1.04347\ldots \end{gathered}$

Using a z-score table, the cumulative probability of Z=24/23 is

$\Rightarrow P(X\le92)=0.8508$

Finally,

$\begin{gathered} \Rightarrow P(X>92)=1-P(X\le92)=1-0.8508=0.1492 \\ \Rightarrow P(X>92)=0.1492 \end{gathered}$

Thus, the answer is 0.1492, which is equivalent to 14.92%; rounded to the nearest percentage, the answer is 15%