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.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{z-score}[/tex],
[tex]x=\text{Random sample score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex].
Let us find z-score corresponding to sample score of 72%.
[tex]z=\frac{72\%-72\%}{5\%}[/tex]
[tex]z=\frac{0\%}{5\%}[/tex]
[tex]z=0[/tex]
Let us find z-score for sample score 87%.
[tex]z=\frac{87\%-72\%}{5\%}[/tex]
[tex]z=\frac{15\%}{5\%}[/tex]
[tex]z=3[/tex]
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.
[tex]P(a<z<b)=P(z<b)-P(z<a)[/tex]
Upon substituting our given values in above formula we will get,
[tex]P(0<z<3)=P(z<3)-P(z<0)[/tex]
Using normal distribution table we will get,
[tex]P(0<z<3)=0.99865-0.50000[/tex]
[tex]P(0<z<3)=0.49865[/tex]
Let us convert our given probability to percentage by multiplying 0.49865 by 100.
[tex]0.49865*100=49.865\%\approx 50\%[/tex]
Therefore, approximately 50% of students will earn a score between 72% and 87%.
