Human proofreaders catch 70% of word errors. If a single essay contains 10 word errors, what is the distribution of the number of errors caught in this essay?

Aug 23, 2017

If $X$ is the number of errors caught, then the intended answer is most likely $X \text{ "~" ""Bin} \left(n = 10 , p = 0.7\right) .$

Explanation:

The wording of the question is a little confusing. There is a difference between "catching 70% of word errors" and "having a 70% chance of catching one error".

"Catching 70% of word errors" suggests a more accurate measure of the proofreaders' ability to catch errors, with minimal or no fluctuation in that 70% value. As such, it turns the question into a basic arithmetic problem, and the answer would be

70% xx 10 " errors" = 7 " errors",

with (nearly) 100% probability.

If, on the other hand, we interpret the question to mean "Human proofreaders have a 70% chance of catching a word error", then each individual error has a 70% chance of being spotted. This is more in line with a proper probability distribution, because it's like tossing a coin that's weighted to have a 70% chance of landing on heads. If we toss the coin 10 times, the most probable outcome is indeed 7 heads, but there is certainly a chance of tossing heads anywhere from 0 to 10 times.

If we let $X$ be the number of errors caught by our student proofreader, then the distribution for $X$ is Binomial, with each of the $n = 10$ word errors having a p=70% chance of being caught. This is a discrete probability distribution, written as

$X \text{ "~" Bin} \left(n = 10 , p = 0.7\right)$,

or often just

$X \text{ "~" Bin} \left(10 , 0.7\right)$.

Here is a graph of the distribution: