Question

According to the College Board website, the scores on the math part of the SAT (SAT-M) in a recent year had a mean of 507 and standard deviation of 111. Assume that SAT-M scores have a normal probability distribution. One of the criteria for admission to a certain engineering school is an SAT-M score in the 98th percentile. This means the score is in the top 2% of scores. Find the actual SAT-M score marking the 98th percentile? Show your work. Interpret your result. Note: You will need the Inverse Normal Distribution Calculator below.

Answer 1

Answer:

The actual SAT-M score marking the 98th percentile is 735.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 507, \sigma = 111$

Find the actual SAT-M score marking the 98th percentile

This is X when Z has a pvalue of 0.98. So it is X when Z = 2.054. So

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

$2.054 = \frac{X - 507}{111}$

$X - 507 = 2.054*111$

$X = 735$