
Answer:
40.13%
Step-by-step explanation:
We have been given that scores on the quantitative portion of an exam have a mean of 563 and a standard deviation of 144. Assume the scores are normally distributed.
We are asked to find the percentage of students taking the quantitative exam, who scored above 599​.
First of all we will find z-score corresponding to 599 using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
z = z-score, Â Â
x = Random sample score,
[tex]\mu[/tex] Â = Mean,
[tex]\sigma[/tex] Â = Standard deviation. Â
[tex]z=\frac{599-563}{144}[/tex]
[tex]z=\frac{36}{144}[/tex]
[tex]z=0.25[/tex]
Now we will use normal distribution table to find the probability of z-score greater than 0.25 that is [tex]P(z>0.25)[/tex].
Using formula [tex]P(z>a)=1-P(z<a)[/tex], we will get: Â
[tex]P(z>0.25)=1-P(z<0.25)[/tex]
[tex]P(z>0.25)=1-0.59871[/tex]
[tex]P(z>0.25)=0.40129[/tex]
Let us convert 0.40129 into percentage as:
[tex]0.40129\times 100\%=40.129\%\approx 40.13\%[/tex]
Therefore, approximately 40.13% of students taking the quantitative exam will score above 599.