Question

Almost all medical schools in the United States require applicants to take the Medical College Admission Test (MCAT). On one exam, the scores of all applicants on the biological sciences part of the MCAT were approximately Normal with mean 9.7 and standard deviation 2.1. For applicants who actually entered medical school, the mean score was 10.5 and the standard deviation was "1.6". (a) What percent of all applicants had scores higher than 13? ANSWER: % (b) What percent of those who entered medical school had scores between 9 and 11? ANSWER:

Answer 1

Answer:

a) 5.82%

b) 44.81%

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

(a) What percent of all applicants had scores higher than 13?

All applicants have $\mu = 9.7, \sigma = 2.1$

This probability is 1 subtracted by the pvalue of Z when X = 13. So

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

$Z = \frac{13 - 9.7}{2.1}$

$Z = 1.57$

$Z = 1.57$ has a pvalue of 0.9418

1 - 0.9418 = 0.0582

5.82% is the answer

(b) What percent of those who entered medical school had scores between 9 and 11?

Those who entered medical school have $\mu = 10.5, \sigma = 1.6$.

This probability is the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 9. So

X = 11

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

$Z = \frac{11 - 10.5}{1.6}$

$Z = 0.31$

$Z = 0.31$ has a pvalue of 0.6217

X = 9

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

$Z = \frac{9 - 10.5}{1.6}$

$Z = -0.94$

$Z = -0.94$ has a pvalue of 0.1736

0.6217 - 0.1736 = 0.4481 = 44.81%