Use Chebyshev's Inequality. Chebyshev's inequality states that for any distribution (does not even have to be a normal one), the proportion of observations outside k standard deviations of the mean is at most
1/k2.
We can use a handy measure called a z-score to figure out how many standard deviations 110 is from the mean of 87. That calculation is
(110 - 87)/8 = 2.875.
OK! That looks pretty high. Let's see if the officer makes the upper 15th percentile.
1/k2 = 1/2.8752 = 0.121
In other words, at most 12.1% of the population of officer test scores is farther than 2.875 standard deviations away from the mean. So, at worst, 110 is at the 100% - 12.1% or 87.9% percentile. He makes the 85 percentile requirement.