. 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:
(Hint: Make a drawing and label first)
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
a. In a normal distribution, the mean, median and mode have the same value. Does that help?
b. 120 = mean% + 34% (1/2 of 68%)
Use similar processes to solve the remainder of your questions.
if you have a body temperature of 99.00„aF, what is your percentile score?
a. According to the 68-95-99.7 rule, approximately 68% of the scores fall within one standard deviation of the mean. Since the mean is 100 and the standard deviation is 20, the range of scores within one standard deviation is from 100 - 20 = 80 to 100 + 20 = 120. However, we are looking for the percentage of scores less than 100, so only the scores within one standard deviation below the mean are considered. Therefore, the percentage of scores less than 100 is approximately 34%.
b. Continuing from the previous question, the range of scores within two standard deviations is from 100 - (2*20) = 60 to 100 + (2*20) = 140. We are interested in the relative frequency of scores less than 120, which is still within two standard deviations. Therefore, the relative frequency of scores less than 120 is approximately 95%.
c. From the 68-95-99.7 rule, we know that approximately 99.7% of the scores fall within three standard deviations of the mean. The range of scores within three standard deviations is from 100 - (3*20) = 40 to 100 + (3*20) = 160. We are interested in the percentage of scores less than 140, which is within three standard deviations. Therefore, the percentage of scores less than 140 is approximately 99.7%.
d. Similar to part a, the percentage of scores less than 80 can be determined by considering the range of scores within one standard deviation below the mean. Since the scores within one standard deviation below the mean range from 80 to 100, we can conclude that the percentage of scores less than 80 is also approximately 34%.
e. Following the same logic as in parts b and d, the range of scores within two standard deviations is from 100 - (2*20) = 60 to 100 + (2*20) = 140. We are interested in the relative frequency of scores less than 60, which is still within two standard deviations. Therefore, the relative frequency of scores less than 60 is approximately 2.5%.
f. The relative frequency of scores greater than 120 can be found by subtracting the relative frequency of scores less than 120 (as found in part b) from 100%. Therefore, the percentage of scores greater than 120 is approximately 5%.
To answer these questions, we can use the 68-95-99.7 rule, also known as the empirical rule or the three-sigma rule. This rule gives us the percentage or relative frequency of data points within a certain number of standard deviations from the mean in a normally distributed dataset.
Before we begin, let's make a drawing to visualize the normal distribution curve. We draw a horizontal axis with the mean (100) at the center and the standard deviation indicated with tick marks. The curve represents the distribution of the test scores, and its shape is symmetric around the mean.
Now let's find the quantities:
a. The percentage of scores less than 100: Since the mean is 100, this is simply 50%. Half of the scores are below the mean.
b. The relative frequency of scores less than 120: To find this, we need to calculate the z-score for 120 first. The formula for calculating the z-score is (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation. In this case, X = 120, μ = 100, and σ = 20. Plugging these values into the formula, we get (120 - 100) / 20 = 20 / 20 = 1.
Using the z-score table or a calculator, we can find that the area to the left of a z-score of 1 is approximately 0.8413. This means that about 84.13% of the scores are less than 120.
c. The percentage of scores less than 140: Similar to part b, we need to calculate the z-score for 140. Using the same formula, (140 - 100) / 20 = 40 / 20 = 2. The area to the left of a z-score of 2 is approximately 0.9772. Therefore, about 97.72% of the scores are less than 140.
d. The percentage of scores less than 80: To find this, we calculate the z-score for 80. Using the formula, (80 - 100) / 20 = -20 / 20 = -1. Since we want to find the area to the left of the z-score, we look up the absolute value of -1 in the z-score table, which gives us 0.1587. Therefore, about 15.87% of the scores are less than 80.
e. The relative frequency of scores less than 60: Again, we calculate the z-score for 60. Using the formula, (60 - 100) / 20 = -40 / 20 = -2. Looking up the absolute value of -2 in the z-score table, we get 0.0228. Hence, approximately 2.28% of the scores are less than 60.
f. The percentage of scores greater than 120: To find this, we can subtract the percentage of scores less than 120 from 100%. From part b, we know that about 84.13% of scores are less than 120. Therefore, the percentage of scores greater than 120 is 100% - 84.13% = 15.87%.
Using these calculations and the 68-95-99.7 rule, we can determine various quantities related to the given normal distribution of test scores.