Question

Scores on a test have a mean of 78.2 and 8 percent of the scores are above 89. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth, if necessary.

Answer 1

Answer:

The standard deviation is 7.7

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 78.2$

8 percent of the scores are above 89

This means that X = 89 has a pvalue of 1 - 0.08 = 0.92. So when Z = 89, Z = 1.405.

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

$1.405 = \frac{89 - 78.2}{\sigma}$

$1.405\sigma = 10.8$

$\sigma = \frac{10.8}{1.405}$

$\sigma = 7.7$

The standard deviation is 7.7