Assume that aset of test scores is normally distributed with a mean of 100 and a standard deviation of 20 use the 68-95-99?
Use it to find what?
To use the 68-95-99.7 rule in this scenario, we need to understand how it applies to a normal distribution. The 68-95-99.7 rule states that:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
Given that the mean is 100 and the standard deviation is 20, we can now calculate the ranges for each interval using the rule:
1. 68% interval: This falls within one standard deviation of the mean.
To find the interval, we take the mean ± one standard deviation.
Lower bound: 100 - 20 = 80
Upper bound: 100 + 20 = 120
Thus, approximately 68% of the data falls between 80 and 120.
2. 95% interval: This falls within two standard deviations of the mean.
To find the interval, we take the mean ± two standard deviations.
Lower bound: 100 - (2 * 20) = 60
Upper bound: 100 + (2 * 20) = 140
Thus, approximately 95% of the data falls between 60 and 140.
3. 99.7% interval: This falls within three standard deviations of the mean.
To find the interval, we take the mean ± three standard deviations.
Lower bound: 100 - (3 * 20) = 40
Upper bound: 100 + (3 * 20) = 160
Thus, approximately 99.7% of the data falls between 40 and 160.
Therefore, using the 68-95-99.7 rule, we can say that approximately 68% of the test scores range from 80 to 120, 95% of the scores range from 60 to 140, and 99.7% of the scores range from 40 to 160.