Question

The grades on the last math exam had a mean of 72%. Assume the population of grades on math exams is known to be distributed normally, with a standard deviation of 5%. Approximately what percent of students earn a score between 72% and 87%?

Answer 1

49.865%
round to the nearest tenth of a percent: 49.9% 
round to the nearest integer:  50%

Answer 2

Answer:

Approximately 50%.  

Step-by-step explanation:

We have been given that the grades on the last math exam are normally distributed and had a mean of 72% and a standard deviation of 5%.

To find the percent of students, who will earn a score between 72% and 87% we will use z-score formula.

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

$z=\text{z-score}$,

$x=\text{Random sample score}$,

$\mu=\text{Mean}$,

$\sigma=\text{Standard deviation}$.

Let us find z-score corresponding to sample score of 72%.

$z=\frac{72\%-72\%}{5\%}$

$z=\frac{0\%}{5\%}$

$z=0$

Let us find z-score for sample score 87%.

$z=\frac{87\%-72\%}{5\%}$

$z=\frac{15\%}{5\%}$

$z=3$

Let us find the probability of z-score between 0 and 3.

Since we now that probability between two z-scores can be found by subtracting the probability of smaller area from bigger area.

$P(a$

Upon substituting our given values in above formula we will get,

$P(0$

Using normal distribution table we will get,

$P(0$

$P(0$

Let us convert our given probability to percentage by multiplying 0.49865 by 100.

$0.49865*100=49.865\%\approx 50\%$

Therefore, approximately 50% of students will earn a score between 72% and 87%.