How do you choose the counts to include in a binomial calculation if the range of counts is not discrete? For instance, given a binomial distribution of 110 counts and a probability of 0.301 of success for each, I'm trying to find the probability that the mean of 110 randomly chosen items will fall within 0.05 of the actual mean. For this, I'm doing the following:
$$X \sim B(110, 0.301)$$ $$\mu = np = 33.11$$ $$err = 0.05\cdot110 = 5.5$$ $$P(\mu-5.5 \le X \le \mu+5.5)$$ $$P(27.61 \le X \le 38.61)$$
At this point, calculating the probability is a matter of applying the binomial, and perhaps using a computer to do so; however, since the binomial is discrete, the values of 27.61 and 38.61 can't technically be applied (from the best of my knowledge you can't sum over a discrete probability mass function using fractions of a count), so what values do you use for this?
On one hand I could choose to start the count at 28 and go through 38, which gives 0.747 as the answer, or I could use 27 through 39, which gives 0.824 as the answer. Alternatively, blindly rounding each might be one approach, but does not seem to have much applicable logic behind it, but would have us include counts 28 through 39.
So I'm left with the question of which combination is best, and why? Gut feeling says despite rounding giving the same results, including 28 through 39 is the way to go, as 27.61 already covers most of the 28th count, and similarly 38.61 covers most of the 39th count, but I'm just not certain.
You can only get integer counts and cannot always get counts that will be exactly 5% above and 5% below the mean. So Because P(27.61 ≤ X ≤38.6) = P(27.61 ≤ X < 28) + P(28 ≤ X ≤ 38) + P(38 < X ≤ 38.61) = 0 + P(28 ≤ X ≤ 38) + 0 = P(28 ≤ X ≤ 38), you just add up the probabilities of getting 28 through 38. That is the probability of being within 5 percent of the mean.