Question

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). A new version of the exam was introduced in spring 2015 and is intended to shift the focus from what applicants know to how well they can use what they know. One result of the change is that the scale on which the exam is graded has been modified, with the total score of the four sections on the test ranging from 472 to 528 . In spring 2015 the mean score was 500.0 with a standard deviation of 10.6 .What proportion of students taking the MCAT had a score over 512

Answer 1

Answer:

12.92%

Step-by-step explanation:

Mean of the scores= u = 500

Standard deviation = $\sigma$ = 10.6

We have to find what proportion of students scored more than 512 marks.

The distribution of scores in a test generally follows the Normal distribution. So we can assume that the distribution of MCAT scores is normally distributed about the mean.

Since, the distribution is normal, we can use the concept of z scores to find the proportion of students who scored above 512.

The formula for z scores is:

$z=\frac{x-u}{\sigma}$

So, z score for x = 512 will be:

$z=\frac{512-500}{10.6}=1.13$

Thus,

P(X > 512) is equivalent to P(z > 1.13)

So, the test scores of 512 is equivalent to a z score of 1.13. Using the z table we have to find the proportion of z scores being greater than 1.13, which comes out to be 0.1292

Since,

P(X > 512) = P(z > 1.13)

We can conclude that, the proportion of students taking the MCAT who had a score over 512 is 0.1292 or 12.92%

Answer 2

You can use the standard normal variate to get the needed proportion.

The proportion of students taking the MCAT having score over 512 is given approximately as 0.13 or 13/100 or 13%

How to convert a normal variate to standard normal variate?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

$X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$)

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

Using the above stated method to get the z scores and then the p value to get the needed proportion

Let the random variable X track the score obtained in the MCAT exam, then we have:

$\rm X \sim N(\mu = mean = 500, \sigma = std.\: deviation = 10.6)$

The probability that the score obtained will be bigger than 512 is given as P(X > 512)

Converting X to standard normal variate, we get the Z score for X = 512 as

$Z = \dfrac{X - \mu}{\sigma} = \dfrac{512 - 500}{10.6} = \dfrac{12}{10.6} \approx 1.13$

The p value from the z tables for Z = 1.13 is obtained as 0.8708

This value denotes $P(X \leq 512) = P(Z \leq 1.132) = 0.8708$

Thus, we have

$P(X > 512) = 1 - P(X \lew 512) = 1 - 0.87 = 0.13$

The probability is score out of 1.

This shows that the proportion of students taking MCAT getting score over 512 is 0.13 or 13/100

Thus,

The proportion of students taking the MCAT having score over 512 is given approximately as 0.13 or 13/100 or 13%

Learn more about standard normal distribution here:

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