To solve this problem using the 68-95-99.7 rule, we start by finding the z-score of 161. The z-score formula is given by:
z = (x - μ) / σ
Where x is the value we are interested in (161 in this case), μ is the mean (101), and σ is the standard deviation (20).
z = (161 - 101) / 20
z = 60 / 20
z = 3
Next, we use the z-score to find the percentage of scores greater than 161. According to the 68-95-99.7 rule, approximately 99.7% of scores fall within three standard deviations of the mean. Since we are interested in scores greater than 161, we subtract this percentage from 100%.
Percentage of scores greater than 161 = 100% - 99.7%
Percentage of scores greater than 161 = 0.3%
Therefore, approximately 0.3% of scores are greater than 161.
Use the 68-95-99.7 rule to solve the problem
Scores on a test are normally distributed with a mean of 101 and a standard deviation of 20. What percentage of scores are greater than 161?
1 answer
1 answer