Question

5. Based on recent results, scores on the SAT test are normally distributed with a mean of 1511 and a standard deviation of 312. Scores on the ACT test are normally distributed with a mean of 21.1 and a standard deviation of 5.1. Assume that the two tests use different scales to measure the same aptitude. If someone gets an SAT score that is in the 95th percentile, find the actual SAT score and the equivalent ACT score.

Answer 1

Answer:

The actual SAT score is 2024.

The equivalent ACT score is 29.49.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

SAT score that is in the 95th percentile

Scores on the SAT test are normally distributed with a mean of 1511 and a standard deviation of 312, which means that $\mu = 1511, \sigma = 312$

95th percentile is X when Z has a pvalue of 0.95, so X when Z = 1.645. The score is:

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

$1.645 = \frac{X - 1511}{312}$

$X - 1511 = 1.645*312$

$X = 2024$

The actual SAT score is 2024.

Equivalent ACT score:

The equivalent ACT score is the 95th percentile of ACT scores.

Scores on the ACT test are normally distributed with a mean of 21.1 and a standard deviation of 5.1, which means that $\mu = 21.1, \sigma = 5.1$. So

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

$1.645 = \frac{X - 21.1}{5.1}$

$X - 21.1 = 1.645*5.1$

$X = 29.49$

The equivalent ACT score is 29.49.