Using the empirical rule, also sometimes called the three-sigma or 68-95-99.7 rule
a. The percentage of scores less than 80 is 50 %.
b. The percentage of scores greater than 95 is 16 %
c. The percentage of scores between 50 and 95 is 34 %
a. The percentage of scores less than 80
Since the mean is µ = 80 and standard deviation σ = 15, to find the percentage of scores less than 80, we note that the mean is 80 and at the center of the distribution, so its z - score is 0. Thus, 50% of the values are less than the mean 80.
b. To find the percentage of scores greater than 95
Since we require the percentage of the scores greater than 95, we use the z-score z = (x - µ)/σ where x = 95
So, z = (x - µ)/σ
= (95 - 80)/15
= 15/15
= 1
Since the z - score is 1, this is one standard deviation away from the mean. Using the 68-95-99 rule, we have that 68 % of the scores are one standard deviation away from the mean. Now, there are 50 % of the quantities below the mean and 68 %/2 = 34 % above the mean up to the 95 score. So, the number of scores below 95 are 50 % + 34 % = 84 %.
Since the total score is 100 %, then the percentage above the 95 score is 100 % - 84 % = 16 %
c. To find the percentage of scores between 50 and 95, we note that
P(50 ≤ x ≤ 95) = P(0 ≤ z ≤ 1) = P(z = 1) - P(z = 0) = 84 % - 50 % = 34 %
Here is the complete question
Assume that a set of test scores is normally distributed with a mean of 80 and a standard deviation of 15
Use the 68-95-99.7 rule to find the following quantities.
a. The percentage of scores less than
80 is ___%.
(Round to one decimal place as needed.)
b. The percentage of scores greater than 95 is ___%
(Round to one decimal place as needed.)
c. The percentage of scores between 50 and 95 is ___%.
(Round to one decimal place as needed.)