Calculating probability.?

The "67-95-99.7" rule says there's an 0.15% chance of more that 6+3*2 = 12 errors on a page, but the same rule says that there's a 0.15% that the number of errors is less than 0--and that's clearly impossible.

So, there's not enough information about the distribution. The mean is only three standard deviations about 0 and it's impossible to have a negative number of errors on a page. Any assumptions about the distribution being approximately normal are invalid at a distance of 3 or more standard deviations from the mean.

Edit--re: Update

All of those response are correct. The strongest choice is "no more than 2%", but that's not all that good unless you know something more about the distribution than its mean and standard deviation.

PS: I tried to see if a binomial distribution makes sense, where each word on the page has an independed probability p of being misspelled, but your mean and standard deviation values can be used to solve for the page size and probability. Those values give 18 words per page with p=1/3 of a misspelling in any given word. Now the probability is exactly 0 for having 20 out of 18 word misspelled!

Refer to the z-table: http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf

P(x > 20)
= P(z > (20 - 6)/2)
= P(z > 7)
= P(z < -7)
= 0

(The zero probability means that the probability is extremely small and close to zero.)

Seven standard deviations is one in 400 billion.
Seven standard deviations for one tail is one in 800 billion.

You need stochastic process or other such to work at that margin. Something with a daily chance of happening of seven sigma would be expected to occur approximately once every two billion years. Mass extinction events are more common than that. This suggests the underlying model has a fat right tail, is skewed, is not normal after all, etc.