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

a. Percentage of scores less than 100
b.Relative frequency of scores less than 120
c.Percentage of scores less than 140
d.Percentage of scores less than 80
e.Relative frequency of scores less than 60
f.Percentage of scores greater than 120

Z = (score-mean)/SD

Find the values and then use the rule, remembering that the percentages are on both sides of the mean.

Help me know if I am on the right track, I got:

a-50%, b-120/150 or 4/5, c- 93.3%, d-53.3%, e-60/150 or 2/5, f- 20%

To answer these questions using the 68-95-99.7 rule, we need to understand the concept of z-scores. A z-score is a measure of how many standard deviations an individual data point is from the mean of a distribution.

a. To find the percentage of scores less than 100, we need to find the area under the curve to the left of 100. Since the distribution is assumed to be normal, we can convert the raw score (100) to a z-score using the formula:

z = (x - μ) / σ

where x is the raw score, μ is the mean, and σ is the standard deviation.

Substituting the values, we have:

z = (100 - 100) / 20 = 0

According to the 68-95-99.7 rule, approximately 68% of the data falls within one standard deviation of the mean. Since 100 is the mean in this case and the z-score is 0, we can conclude that approximately 50% of the scores are less than 100.

b. To find the relative frequency of scores less than 120, we need to find the area under the curve to the left of 120. Again, we can convert the raw score to a z-score:

z = (120 - 100) / 20 = 1

According to the 68-95-99.7 rule, approximately 95% of the data falls within two standard deviations of the mean. Since 120 is one standard deviation above the mean, we can conclude that approximately 95% of the scores are less than 120.

c. To find the percentage of scores less than 140, we need to find the area under the curve to the left of 140. Convert the raw score to a z-score:

z = (140 - 100) / 20 = 2

According to the 68-95-99.7 rule, approximately 99.7% of the data falls within three standard deviations of the mean. Since 140 is two standard deviations above the mean, we can conclude that approximately 99.7% of the scores are less than 140.

d. To find the percentage of scores less than 80, we need to find the area under the curve to the left of 80, so convert the raw score to a z-score:

z = (80 - 100) / 20 = -1

According to the 68-95-99.7 rule, approximately 68% of the data falls within one standard deviation of the mean. Since 80 is one standard deviation below the mean, we can conclude that approximately 68% of the scores are less than 80.

e. To find the relative frequency of scores less than 60, we need to find the area under the curve to the left of 60, so convert the raw score to a z-score:

z = (60 - 100) / 20 = -2

According to the 68-95-99.7 rule, approximately 95% of the data falls within two standard deviations of the mean. Since 60 is two standard deviations below the mean, we can conclude that approximately 95% of the scores are less than 60.

f. To find the percentage of scores greater than 120, we need to find the area under the curve to the right of 120. First, find the area to the left of 120 (which we calculated in part b as approximately 95%), and subtract it from 1 (since we want the area to the right):

1 - 0.95 = 0.05

We can conclude that approximately 5% of the scores are greater than 120.