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Scores on a university exam are normally distributed with a mean of 68 and a standard deviation of 9. Using the 68-95-99.7 rule, what percentage of students score above 77?

A. 2.5%
B. 5%
C. 16%
D. 32%

Answer :

I think C is the best

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Rewritten by : Barada

The rule is::

- 68% of data are in the range mean +/- 1*standard deviation
- 95% of data are in the range mean +/- 2*standard deviation
- 99.7% of data are in the range mean +/- 3*standard deviation

Here mean is 68 and standard deviation is 9.

Score 77 is 68 + 9, which is mean + 1 standard deviation.

Then, the difference 100% - 68% = 32% is the % of data out of the range. Of those, diven the symmetry of the normal distribution, half will be above the range, this is 16% of scores are above 77.