Question

On a normally distributed anxiety test with mean 48 and standard deviation 4, approximately what anxiety test score would put someone in the top 5 percent? Group of answer choices

Answer 1

Answer:

Anxiety score close to 54.58.

Step-by-step explanation:

When the distribution is normal, we use 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:

$\mu = 48, \sigma = 4$

Approximately what anxiety test score would put someone in the top 5 percent?

We have to find the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.

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

$1.645 = \frac{X - 48}{4}$

$X - 48 = 1.645*4$

$X = 54.58$

Anxiety score close to 54.58.