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The std of a discrete binomial distribution is $$\text{std}=\sqrt{nr(1-r)}$$ where $r$ is the probability of success on a given trial and $n$ is the total number of trials.

Based on the way I understand the formula, to reduce the std of a binomial distribution by half, the total number of trials must be reduced by fourth.

But I think I am probably wrong because:

  1. Reducing the number of trials modify the original distribution because the number of trials is one of the two parameters in the binomial distribution. What I am looking for is a change in the number of trials so that the same distribution has std reduced by half.

  2. Intuitively, the number of trials must be increased to lower std.

So my question is: What change must be made in the number of trials to reduce the std of the binomial distribution by half?

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  • $\begingroup$ The formula $\sqrt{nr(1-r)}$ is for the standard deviation of the sum of $n$ i.i.d. binomial variables. This gets larger as $n$ does. The formula for the standard deviation of the mean of $n$ i.i.d. binomial variables is $\sqrt{r(1-r)/n}$. This gets smaller as $n$ gets larger. $\endgroup$ – robjohn Nov 13 '17 at 5:23
  • $\begingroup$ Could you please explain a little more? What do you mean by std of the sum and of the mean? $\endgroup$ – A Slow Learner Nov 13 '17 at 5:42
  • $\begingroup$ Given $n$ random variables $x_1\ldots x_n$, the standard deviation of the sum is the standard deviation of $\sum\limits_{k=1}^nx_n$ and the standard deviation of the mean is the standard deviation of $\frac1n\sum\limits_{k=1}^nx_n$. $\endgroup$ – robjohn Nov 13 '17 at 5:50
  • $\begingroup$ I am not sure you are correct on that. I just checked my sources very carefully and the std of the mean that you provided is actually the std of the proportion of successes in n trials: $\sqrt{r(1-r)/n}=std(X/n)$ $\endgroup$ – A Slow Learner Nov 13 '17 at 7:38
  • $\begingroup$ And what is the difference between those? $\endgroup$ – robjohn Nov 13 '17 at 8:13

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