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MGT 235
4: The Normal Distribution
4.4: Estimating the Binomial with the Normal Distribution

4.4: Estimating the Binomial with the Normal Distribution

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We found earlier that various probability density functions are the limiting distributions of others; thus, we can estimate one with another under certain circumstances. We will find here that the normal distribution can be used to estimate a binomial process. The Poisson was used to estimate the binomial previously, and the binomial was used to estimate the hypergeometric distribution.

In the case of the relationship between the hypergeometric distribution and the binomial, we had to recognize that a binomial process assumes that the probability of a success remains constant from trial to trial: a head on the last flip cannot have an effect on the probability of a head on the next flip. In the hypergeometric distribution this is the essence of the question because the experiment assumes that any "draw" is without replacement. If one draws without replacement, then all subsequent "draws" are conditional probabilities. We found that if the hypergeometric experiment draws only a small percentage of the total objects, then we can ignore the impact on the probability from draw to draw.

Imagine that there are 312 cards in a deck comprised of 6 normal decks. If the experiment called for drawing only 10 cards, less than 5% of the total, than we will accept the binomial estimate of the probability, even though this is actually a hypergeometric distribution because the cards are presumably drawn without replacement.

The Poisson likewise was considered an appropriate estimate of the binomial under certain circumstances. In Figure \(\PageIndex{11}\) shows a symmetrical normal distribution transposed on a graph of a binomial distribution where \(p = 0.2\) and \(n = 5\). The discrepancy between the estimated probability using a normal distribution and the probability of the original binomial distribution is apparent. The criteria for using a normal distribution to estimate a binomial thus addresses this problem by requiring BOTH \(np\) AND \(n(1 − p)\) are greater than five. Again, this is a rule of thumb, but is effective and results in acceptable estimates of the binomial probability.

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Figure \(\PageIndex{11}\)

\(1-[p(X=0)+p(X=1)+p(X=2)+\ldots+p(X=16)]=p(X>16)=p(Z>2)=0.0228\)