On an exam whose scores followed a normal distribution, a student's z-score was
+2.
What percentage of students scored better than he did? (Use the 68-95-99.7 Rule.)
Since you are only interested in number above Z = +2, you cannot apply the rule.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion for that Z score. Multiply by 100.
To determine the percentage of students who scored better than the student with a z-score of +2, we can use the 68-95-99.7 Rule, which states that in a normal distribution:
- Approximately 68% of the data falls within 1 standard deviation of the mean.
- Approximately 95% of the data falls within 2 standard deviations of the mean.
- Approximately 99.7% of the data falls within 3 standard deviations of the mean.
Since the student's z-score is +2, it means that the student's score is 2 standard deviations above the mean. According to the 68-95-99.7 Rule, approximately 95% of the data falls within 2 standard deviations of the mean.
Therefore, we can conclude that approximately 95% of the students scored worse than the student with a z-score of +2. As a result, the percentage of students who scored better than the student is 100% - 95% = 5%.
To find the percentage of students who scored better than the student with a z-score of +2, you can use the 68-95-99.7 rule, also known as the empirical rule or the three-sigma rule.
The 68-95-99.7 rule states that in a normal distribution:
- 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.
Since the student has a z-score of +2, it means they scored two standard deviations above the mean. According to the 68-95-99.7 rule, about 95% of the data falls within two standard deviations of the mean. Therefore, the percentage of students who scored better than the student is approximately 95%.
To explain how to get this answer, you need to know the formula to convert a z-score to a percentage using a z-table or a statistical software.
1. Look up the z-score of +2 in a standard normal distribution table (also known as a z-table). A z-table provides the percentage of data that falls below a given z-score.
2. In the z-table, find the row that corresponds to the digit before the decimal point (2) and the column that corresponds to the digit after the decimal point (0). The value at the intersection of the row and column represents the percentage of data below the z-score.
3. The value you find in the z-table represents the percentage of students who scored better than the student with a z-score of +2. In this case, since the z-score of +2 is quite high, the percentage is likely to be close to the maximum value provided by the z-table, which is approximately 0.9772 or 97.72%.
Therefore, approximately 97.72% of students scored better than the student with a z-score of +2.