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%

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.

I think C is the best