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Dear Math Helper,

How's it going? I'm in the business of printing T-shirts and just got an interesting order.

My sister is pre-med student and wants to show off how well she performed on the Medical College Admission Test (MCAT). I told her I’d make her a T-shirt, but she’s not making it easy for me.

She wants the back of the shirt to read, "I scored at the nth percentile on the MCAT!" But instead of giving me the value of n, she told me her individual score was 38 and that all the test scores are normally distributed with a mean of 25 and a standard deviation of 6.4.

I don’t want to let her think she’s smarter than me, even if she is going to be a doctor! Can you explain how I should use these numbers to find the value of n for her shirt? If so, maybe there's a free T-shirt in your future!

Many thanks,

T-shirt Guy

User Tyiesha
by
5.8k points

1 Answer

2 votes

Answer:

n = 98, that is, she scored at the 98th percentile.

Explanation:

When the distribution is normal, we use the z-score formula.

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


Z = (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.

In this question, we have that:

She scored 38, so
X = 38

Test scores are normally distributed with a mean of 25 and a standard deviation of 6.4.

This means that
\mu = 25, \sigma = 6.4

Find the percentile:

We have to find the pvalue of Z. So


Z = (X - \mu)/(\sigma)


Z = (38 - 25)/(6.4)


Z = 2.03


Z = 2.03 has a pvalue of 0.98(rounding to two decimal places).

So n = 98, that is, she scored at the 98th percentile.

User Ivan Zlatev
by
6.2k points