Using the 68-95-99.7 Rule

Assume that a set of test scores is normally distributed with a meanof 100 and a standard deviationof 20. Use the 68-95-99.7 rule to find the following quantities:

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Percentage of scores less than 100
Relative frequency of scores less than 120
Percentage of scores less than 140
Percentage of scores less than 80
Relative frequency of scores less than 60
Percentage of scores greater than 120

mu = 100, sigma = 20

z = (x – mu)/sigma

(a) x = 100

z = (100 – 100)/20 = 0

Percentage of scores less than 100 = 100 * P(x < 100) = 100 * P(z < 0) = 100 * 0.5 = 50%

(b) x = 120

z = (120 – …

This solution provides the simple calculations to find the required quantities.