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It is well known that

$$\sum_{k=0}^{n-1} a \,r^{k} = a\frac{1-r^n}{1-r}$$

Is there a similar formula for summing over

$$\sum_{x=0}^{m} C^{n}_{x}p^{x}(1-p)^{n-x}$$

The reason I ask this question is because I am trying to find the median for $n=100$ and $p=0.15$, to this a I have to find the value of $m$ such that

$$\sum_{x=0}^{m} C^{100}_{x}0.15^{x}0.85^{100-x}=\frac{1}{2}$$

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  • $\begingroup$ Use a table for binomial distribution $\endgroup$ Jan 2, 2017 at 23:07
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    $\begingroup$ Use normal approximation $\endgroup$ Jan 2, 2017 at 23:08

1 Answer 1

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Since I cannot flag or comment, I'll write a short answer.

First things first (and to answer your question), there is no so-called closed form. Your question is very similar to this one. Now what options do you have?

Sum it with a calculator

Of course you can just calculate it with a calculator, wouldn't take much time, maybe half an hour. Would be boring, though.

Use an approximation

Here are some guidelines for an approximation. If you use something like "approximate binomial distribution" as key words, you can probably even find a formula to measure your error and so quickly find out which $m$ would be the right one.

Use a program

If you already know a programming language, it shouldn't take much time. If you don't, then this is a good problem to get started! In fact, a similar problem was the reason I started programming because sometimes calculators are just not enough. I'm trying to say, learning to program is an investment for future problems to arise. Python would be a good start in my opinion, their homepage even shows an implementation of the Fibonacci numbers to start with.

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  • $\begingroup$ Ordinarily, the flagging and comment rules are to prevent people from abusing them, and encouraging them to write good answers. Most of this would be proper as comments, but you thought it out so well that I actually like it. Well done, just don't make a habit of it. :) $\endgroup$
    – The Count
    Jan 2, 2017 at 23:51
  • $\begingroup$ @TheCount Thanks for the encouragement :-) I'll be able to comment soon anyway, but I have to start getting points for it somewhere ;) $\endgroup$
    – Ayutac
    Jan 3, 2017 at 0:03
  • $\begingroup$ I forgot what the cutoff if, but you should be darn close at this point. $\endgroup$
    – The Count
    Jan 3, 2017 at 0:05

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