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:
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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
To find the mentioned quantities using the 68-95-99.7 rule, we need to understand the concept of z-scores.
The z-score measures the number of standard deviations a particular value is from the mean. It is calculated using the formula:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
Now, let's solve the problems step by step:
a. Percentage of scores less than 100:
Since the mean is 100 and the scores are less than 100, we look at the area to the left of the mean on a standard normal distribution. According to the 68-95-99.7 rule, approximately 50% of the scores lie to the left of the mean. Therefore, the percentage of scores less than 100 is approximately 50%.
b. Relative frequency of scores less than 120:
To find the relative frequency less than 120, we need to find the z-score for 120 and then find the corresponding area under the curve.
z = (120 - 100) / 20 = 1
Looking at the z-table, we find that the area corresponding to a z-score of 1 is approximately 0.8413. Therefore, the relative frequency of scores less than 120 is approximately 0.8413.
c. Percentage of scores less than 140:
To find the percentage of scores less than 140, we need to find the z-score for 140 and then find the corresponding area under the curve.
z = (140 - 100) / 20 = 2
Using the z-table, we find that the area corresponding to a z-score of 2 is approximately 0.9772. Therefore, the percentage of scores less than 140 is approximately 97.72%.
d. Percentage of scores less than 80:
To find the percentage of scores less than 80, we again need to find the z-score for 80 and the corresponding area under the curve.
z = (80 - 100) / 20 = -1
Using the z-table, we find the area corresponding to a z-score of -1 is approximately 0.1587. Therefore, the percentage of scores less than 80 is approximately 15.87%.
e. Relative frequency of scores less than 60:
To find the relative frequency less than 60, we will calculate the z-score for 60 and then find the corresponding area under the curve.
z = (60 - 100) / 20 = -2
Using the z-table, we find the area corresponding to a z-score of -2 is approximately 0.0228. Therefore, the relative frequency of scores less than 60 is approximately 0.0228.
f. Percentage of scores greater than 120:
To find the percentage of scores greater than 120, we subtract the percentage of scores less than 120 (found in part b) from 1.
Percentage = 1 - 0.8413 = 0.1587
Therefore, the percentage of scores greater than 120 is approximately 15.87%.
Z = ± 1 = ± 34% = 68%
Z = ± 1.96 (about 2) = ± 47.5% = 95%
Z = ± 2.96 (about 3) = ± 49.87% = 99.7%
This information should help you to find your own answers. Although it might require more time and effort, you will learn more if you do your own work. Isn't that why you go to school?