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 quantiles: 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
40 60 80 100 120 140 160
This represents +/- 3 standard deviations.
68% falls within 80 and 120
95% falls within 60 and 140
99.7% falls within 40 and 160
Using this information and a diagram, I think you will be able to answer wthe questions.
a.percentage of scores less than 100
To find the quantiles using the 68-95-99.7 rule, you can use the standard deviation and mean of the distribution. Here are the answers to your specific questions:
a. The percentage of scores less than 100:
Since the mean of the distribution is 100, we know that 50% of the scores are below the mean. Therefore, the percentage of scores less than 100 is also 50%.
b. The relative frequency of scores less than 120:
Using the 68-95-99.7 rule, we know that approximately 68% of the scores fall within one standard deviation of the mean. In this case, one standard deviation is equal to 20. So, the relative frequency of scores less than 120 is approximately 68%.
c. The percentage of scores less than 140:
Using the same reasoning as before, approximately 95% of the scores fall within two standard deviations of the mean. Two standard deviations in this case is equal to 40. So, the percentage of scores less than 140 is approximately 95%.
d. The percentage of scores less than 80:
Since the mean is 100 and the standard deviation is 20, one standard deviation below the mean is 80. Using the 68-95-99.7 rule, we know that approximately 16% of the scores fall within one standard deviation below the mean. So, the percentage of scores less than 80 is approximately 16%.
e. The relative frequency of scores less than 60:
Similarly, two standard deviations below the mean is equal to 60. Using the 68-95-99.7 rule, we know that approximately 2.5% of the scores fall within two standard deviations below the mean. Hence, the relative frequency of scores less than 60 is approximately 2.5%.
f. The percentage of scores greater than 120:
Using the same rule, we know that approximately 32% of the scores fall within one standard deviation above the mean. Thus, the percentage of the scores greater than 120 is approximately 32%.
To find the quantiles using the 68-95-99.7 rule, you need to determine the area under the normal distribution curve.
a. The percentage of scores less than 100:
To find this, you need to calculate the area to the left of the score 100. Since the mean is 100 and it is normally distributed, the area to the left of the mean is 0.5 or 50%.
b. The relative frequency of scores less than 120:
To find this, you need to calculate the area to the left of the score 120. First, calculate the z-score using the formula:
z = (x - mean) / standard deviation
where x is the score and mean is the mean of the distribution. In this case:
z = (120 - 100) / 20 = 1.0
Now, find the area to the left of a z-score of 1.0 using a standard normal distribution table or a calculator. From the table, you will find that the area is approximately 0.8413.
c. The percentage of scores less than 140:
Similar to the previous calculation, calculate the z-score:
z = (140 - 100) / 20 = 2.0
Use the standard normal distribution table or a calculator to find the area to the left of a z-score of 2.0, which is approximately 0.9772.
d. The percentage of scores less than 80:
Again, calculate the z-score:
z = (80 - 100) / 20 = -1.0 (notice the negative sign represents the direction of the left of the mean)
Using the standard normal distribution table or a calculator, find the area to the left of a z-score of -1.0, which is approximately 0.1587.
e. The relative frequency of scores less than 60:
Like before, calculate the z-score:
z = (60 - 100) / 20 = -2.0
Use the standard normal distribution table or a calculator to find the area to the left of a z-score of -2.0, which is approximately 0.0228.
f. The percentage of scores greater than 120:
To find this, you need to calculate the area to the right of the score 120, which is the complement of the area to the left of 120. Using the value found in b, subtract it from 1 (since the total area of the distribution is 1) to get the area to the right of 120. Thus, it becomes approximately 1 - 0.8413 = 0.1587 or 15.87%.