Reports on a student’s ACT, SAT, or MCAT usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than this one. In 2012, the total MCAT scores were close to Normal with mean 25.3 and standard deviation 6.5. William scored 31.A) What was his percentile? (Use table A; z table) He scored better than about (± ± 0.01%) % of all MCAT takers.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(25.3,6.5)[/tex]

Where [tex]\mu=25.3[/tex] and [tex]\sigma=6.5[/tex]

And for this case we have a score of 31

The percentile is calculated with the following probability:

[tex] P(X<31) [/tex]

And using the z score we got:

[tex] P(X<31)=P(Z < \frac{31-25.3}{6.5}) = P(Z<0.877) =0.8098 [/tex]

So then we can conclude that 31 represent approximately the percentile 81 on the distribution given

He scored better than about 80.98 % of all MCAT takers.