Question

The scores for the Algebra 2 CFE are normally distributed with a mean score of 45 and a standard deviation of 5.6. If you score 52 on the test, what percentage of test takers scored lower than you?
87.49 %

89.44 %

90.32 %

91.15%

Answer 1

Answer:

89.44%

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 question:

$\mu = 45, \sigma = 5.6$

If you score 52 on the test, what percentage of test takers scored lower than you?

This is the pvalue of Z when X = 52.

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

$Z = \frac{52 - 45}{5.6}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944.

So 89.44% of test takers scored lower than you.