Using the 68-95-99.7 rule:
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:
Suggest you make a drawing and label first…
a. (0.1 point) Percentage of scores less than 100
b. (0.1 point) Relative frequency of scores less than 120
c. (0.1 point) Percentage of scores less than 140
d. (0.1 point) Percentage of scores less than 80
e. (0.1 point) Relative frequency of scores less than 60
f. (0.1 point) Percentage of scores greater than 120
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To use the 68-95-99.7 rule, also known as the empirical rule, in this scenario, we can make a normal distribution curve to visualize the data.
First, let's calculate the values for the different standard deviations from the mean:
a. The mean is 100, so the percentage of scores less than 100 is 50% since it is the exact middle of the distribution.
b. To find the relative frequency of scores less than 120, we'll use the fact that approximately 68% of the data lies within one standard deviation from the mean. Since one standard deviation is 20, we can find the z-score for 120:
z = (120 - 100) / 20 = 1
Looking up the z-score in a standard normal distribution table, we find that 1 corresponds to a cumulative percentage of 84.13%. Therefore, the relative frequency of scores less than 120 is approximately 84.13%.
c. Similar to the previous calculation, we can find the z-score for 140:
z = (140 - 100) / 20 = 2
The cumulative percentage corresponding to a z-score of 2 is 97.72%. Hence, the percentage of scores less than 140 is approximately 97.72%.
d. To find the percentage of scores less than 80, we can find the z-score for 80:
z = (80 - 100) / 20 = -1
The cumulative percentage corresponding to a z-score of -1 is 15.87%. Therefore, the percentage of scores less than 80 is approximately 15.87%.
e. Using the same process, we can find the z-score for 60:
z = (60 - 100) / 20 = -2
The cumulative percentage corresponding to a z-score of -2 is 2.28%. Thus, the relative frequency of scores less than 60 is approximately 2.28%.
f. To find the percentage of scores greater than 120, we need to subtract the cumulative percentage of scores less than 120 from 100%. Using the z-score of 1 from part b:
Percentage of scores greater than 120 = 100% - 84.13% = 15.87%.
Therefore, the percentage of scores greater than 120 is approximately 15.87%.