# Homework - Probability [duplicate]

Possible Duplicate:
Statistics Probability

I'm trying to figure out the steps needed to solve this problem:

A survey of adults ages $18-24$ found that $38\%$ get no news on an average day. You randomly select $250$ adults ages $18-24$ and ask them if they get news on an average day.

1. Find the probability that at least $113$ people say they get no news on an average day.

For this, I am finding $np$, $npq$, and then adding $.5$ to find the $z$-score. It never comes out right.

2. Find probability that fewer than $93$ people say they got new news on an average day.

Thank you.

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## marked as duplicate by Jonas Meyer, Raskolnikov, Asaf Karagila, J. M., t.b.Dec 22 '11 at 7:05

This question was marked as an exact duplicate of an existing question.

To what are you adding $0.5$ to find the $z$-score? and how is this question very different from this one that you asked just a short while ago? – Dilip Sarwate Dec 21 '11 at 20:24

I do not know why you would add $0.5$ to anything here. It seems to me that with $np$ (which I will denote as the mean) and $npq$ (when we really care about its square root, the standard deviation), you have all you need. $\dfrac{(\text{test number}) - \text{(mean})}{\text{standard deviation}} = \text{z-score}$, right?

Perhaps the problem is with the term 'at least.' Since you are looking for the probability that at least $113$ people do something, you might come at this from the other direction and find the probability that at most $112$ people get no news. This is complementary to what you want to find, as either 113 or more people get no news, or 112 or fewer people get no news. In particular, $P(N \geq 113) = 1 - P(N \leq 112)$.

EDIT (from Aaron's comment)

Aha! You should add $0.5$ as a continuity correction. And considering $P(N \leq 112.5)$ yields a still-better approximation.

I learn something every day.

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The adding 0.5 is a continuity correction for approximating binomial distributions with normal distributions (or more generally, approximating discrete distributions with continuous ones). Since $P(N\geq 113)=P(N>112)$ for a discrete variable but not for a continuous one, you can split the difference and view it as $P(N\geq 112.5)$, which will be slightly closer to the answer you want when you take a continuous approximation. – Aaron Dec 21 '11 at 21:33
@Aaron: I never knew that! At least, I don't think so. But a quick series of checks shows that it true. I think I've always operated under the idea that if $n > 30$ or so, then the approximation is sufficiently close. Cool! – mixedmath Dec 21 '11 at 21:44