Assume that a set of test scores in an Introduction to Finance class is normally distributed with a mean of 72 and a standard deviation of 8. Use the 68-95-99.7 rule to
Find the percentage of scores between 64 and 88.
A) 95%
B) 68%
C) 99.7%
D) 81.5%
E) 49.85%

Question

Quizplus · Accepted Answer

By using the 68-95-99.7 rule, we know that approximately 68% of the scores fall within 1 standard deviation from the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations. Therefore, the percentage of scores between 64 and 88 is the same as the percentage of scores within 1.5 standard deviations from the mean. Using the formula z = (x - μ) / σ, where x is the score, μ is the mean, and σ is the standard deviation, we can calculate the z-scores for 64 and 88:

z1 = (64 - 72) / 8 = -1
z2 = (88 - 72) / 8 = 2

The area between these z-scores represents the percentage of scores between 64 and 88:

P(-1 < z < 2) = P(z < 2) - P(z < -1)
Using a standard normal distribution table, we can find that P(z < 2) is approximately 0.9772 and P(z < -1) is approximately 0.1587. Therefore,

P(-1 < z < 2) = 0.9772 - 0.1587 = 0.8185 = 81.85%

Rounded to the nearest tenth, this is 81.5%, which is answer choice D.

Assume That a Set of Test Scores in an Introduction