Question

An aptitude test has a mean score of 80 and a standard deviation of 5. The population of scores is normally distributed. What proportion of tests has scores over 90?

Answer 1

Answer:

2.28% of tests has scores over 90.

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:

$\mu = 80, \sigma = 5$

What proportion of tests has scores over 90?

This proportion is 1 subtracted by the pvalue of Z when X = 90. So

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

$Z = \frac{90 - 80}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

So 1-0.9772 = 0.0228 = 2.28% of tests has scores over 90.