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Let $n$ be any natural number, let $k\in\{0, \dots, n\}$, and let $p \in [0, 1]$.

If $\sum_{i=k}^n {n \choose i} p^{i}(1-p)^{n-i} \approx 0.05$, how can we find $k$ (in terms of $n$ and $p$)?

Rasputin
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  • See this link on wikipedia. – Alexey Burdin Jun 22 '20 at 18:53
  • Without more information on the magnitudes of $n$ and $p$, there is not much better that you can do than summing the terms explicitly, starting from $i=n$. –  Jun 22 '20 at 18:55
  • What is the context of your question? – callculus42 Jun 22 '20 at 19:00
  • You could approximate this binomial process by a normal, for which the equivalent problem is considerably easier. If more precise answers are needed, at least this gives you a good starting point. – lulu Jun 22 '20 at 19:02
  • While software could give you exact answers, binomial tables are also available if you are looking for an approximate answer. The solution may be even simpler depending on $n$ and $p$. – StubbornAtom Jun 22 '20 at 19:19

1 Answers1

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In some cases you can use a normal approximation with continuity correction and say $$\Phi\left(\frac{k-0.5 - np}{\sqrt{np(1-p)}}\right) \approx 1-0.05$$

Since $\Phi(1.644854) \approx 0.95$ you could then say in these cases $$k \approx np+0.5+ 1.644854\sqrt{np(1-p)}$$

As an example, suppose $n=40$ and $p=\frac14$. Then this suggests $k \approx 15$

As a check $\sum\limits_{i=15}^{40} {40 \choose i} \left(\frac14\right)^i\left(\frac34\right)^{40-i} \approx 0.054$ so this is not a bad approximation; $k=14$ would give about $0.103$ while $k=16$ would give about $0.026$

Henry
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