Answer:
15.86%
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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.
Percent of area between the mean and 0.20 standard deviations from the mean:
pvalue of Z = 0.2 subtracted by the pvalue of Z = -0.2
Z = 0.2 has a pvalue of 0.5793
Z = -0.2 has a pvalue of 0.4207
0.5793 - 0.4207 = 0.1586
So this percentage is 15.86%