A Binomial(*p, n*) has a mean and standard deviation given by:

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\mu=np |

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\sigma =\sqrt{np(1-p)} |

From Central Limit Theorem, as *n* gets large the number of observed successes *s* will tend to:

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s\approx Normal(np,\sqrt{np(1-p)}) |

Equation 2 for the binomial method can then be rewritten and p can be approximated by a Normal when n is large, as follows:

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p\approx \frac{Normal(n\frac{s}{n},\sqrt{n\frac{s}{n}(1-\frac{s}{n})}}{n} |

(1)

which can be rearranged to:

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p\approx Normal(\frac{s}{n},\sqrt{s\bigg(\frac{1}{n^{2}}-\frac{s}{n^{3}}\bigg)} |

(2)

and which results in the following equation:

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p\approx Normal \Big(\frac{s}{n},\sqrt{\frac{s(n-s)}{n^{3}}}\Big) |

(3)

**Figure 1**: Example of Equation 3 estimate of p where *s* = 5, *n* = 10

**Figure 2**: Example of Equation 3 estimate of p where *s* = 1, *n* = 10

Equation 3 works nicely in the plot above for small n (10) because the number of successes was half of n, and so the uncertainty distribution is symmetric about 0.5, which nicely matches the properties of a Normal distribution. However, if one had observed just 1 success from 10 trials, it would look quite different, as shown in Figure 2: now the Normal approximation of Equation 3 is completely inaccurate, assigning considerable confidence to negative values, and fails to reflect the asymmetric nature of the uncertainty distribution.