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68%
0%
0/60
95%
17%
10/60
96.5%
97.5%
52%
31/60
99.7%
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This subject scored 2 standard deviations (SD) higher than the mean and thus scored higher than 97.5% of participants. The following represents what percent of the population is contained within a normal distribution: +/- 1 SD = 68%, +/- 2 SD = 95%, +/- 3 SD = 99.7% When calculating the area under the curve to the left of 2 SD (this represents the percent of the participants the student scored higher than): 50% + (95%/2) = 97.5%. 50% represents the first half of the population the student outscored, and 95%/2 represents the area under a normal distribution curve between 2 SD and 0 SD. Incorrect Answers: Answer 1: 68% represents the percent of respondents who scored within +/- 1SD. Answer 2: 95% represents the percent of respondents who scored within +/- 2SD, but does not represent the % of respondents that the student outscored. This percent leaves out the area of the curve to the left of -2 SD. Answer 3: 96.5% is not a reasonable number that could be calculated without a standard normal table. Answer 5: 99.7% represents the area under the curve that lies within +/- 3 SD. Bullet Summary: In a normal distribution, the area under the curve for +/- 1 SD = 68%, +/- 2 SD = 95%, and +/- 3 SD = 99.7%.
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