Answer :
A z-table or calculator, we find that for a z-score of 0.5, about 16.8% of the scores are greater than 161.
The 68-95-99.7 rule states that for a normal distribution:
* 68% of the scores fall within 1 standard deviation (σ) of the mean
* 95% of the scores fall within 2 standard deviations (2σ) of the mean
* 99.7% of the scores fall within 3 standard deviations (3σ) of the mean
In this problem, we need to find the percentage of scores greater than 161. Since the mean is 101 and the standard deviation is 20, we can calculate:
161 - 101 = 60 (distance from the mean to the score)
Since this distance is greater than 2σ, we know that 95% of the scores are within 2σ of the mean. Therefore, only a small percentage of scores will be greater than 161.
To find the exact percentage, we can use a z-score calculation:
z = (x - μ) / σ = (161 - 101) / 20 = 0.5
Using a z-table or calculator, we find that for a z-score of 0.5, about 16.8% of the scores are greater than 161.
The percentage of scores greater than 161 is 32%.
The 68-95-99.7 rule states that:
* About 68% of the data points fall within 1 standard deviation (σ) of the mean (μ).
* About 95% of the data points fall within 2 standard deviations (σ) of the mean (μ).
* About 99.7% of the data points fall within 3 standard deviations (σ) of the mean (μ).
In this problem, we need to find the percentage of scores that are greater than 161. Since 161 is more than 1 standard deviation above the mean (101), we can use the 68-95-99.7 rule to estimate the percentage.
Since we're looking for scores greater than 161, we can assume that most of these scores are more than 1 standard deviation away from the mean. Let's use the 68% value from the rule, which corresponds to 1 standard deviation.
About 32% of the scores fall below the mean, and about 68% fall within 1 standard deviation of the mean. This means that about:
100% - 68% = 32% of the scores fall more than 1 standard deviation away from the mean, which includes scores greater than 161.
