What is meant by the condition for a binomial distribution that the number of trials must be fixed?
Does this mean the number of trials must be fixed before the experiment?
Solved – Fixed number of trials in binomial distribution
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Related Solutions
Look at the Negative binomial distribution. The usual way of stating the negative binomial is the number of failures before you see x successes, but if you add the number of failures and the number of success then you get the total n which you are asking about.
The negative binomial does assume that the last observed value (when you stopped) is a success, this may match with what you want. But if you only know the number of success and not if there were any more failures after the last success then you will need to expand this.
If the negative binomial does not answer the question then you may want to try a Baysian approach, choose a prior for n, hold the others constant (or give them priors as well) and apply Bayes theorem.
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To your question:
The standard error for the normal approximation is $\sqrt{\frac{pq}{n}}$ if you draw once from a $Bi(n,p)$ and average over these n draws, i.e. divide by $n$, or this is - as you say - equivalent to drawing $n$ times from $Bi(1,p)$ and considering their mean. So yes, if I understand you correctly, we apply the normal approximation to the $Bi(n,p)$ (if n sufficiently large), because it is the sum of $n$ Bernoulli trials and we use the normal approximation if we consider a large enough sum of independent Random Variables.
This is basically explained in the link that you mention: The Variance for a $Bi(25,p)$ is $25\cdot p(1-p)$ (as the variance of a Bernoulli trial is $p(1-p)$ and $Var(\sum_{i=1}^nX_i)= nVar(X_i)$, if $X_i$ are iid). In order to get the variance of the sum of $100$ of these Random Variables you obtain $100\cdot25\cdot p(1-p)$ (using the same rule as before). If you consider the mean and not the sum of these 100 experiments you divide by $100^2$, as ( $Var(X/n)=\frac{1}{n^2}Var(X)$). Taking the square root of the variance of the mean of these 100 experiments gets you $\sqrt{\frac{25\cdot p(1-p)}{100}}=\sqrt{\frac{p\cdot(1-p)}{4}}$, which is the standard error for binomial distribution, i.e. $[25\cdot \hat{p}-\sqrt{\frac{p(1-p)}{4}}, 25\cdot \hat{p}+\sqrt{\frac{p(1-p)}{4}}]$ contains the value $25\cdot p$ around 68% of the time, where $\hat{p}=\frac{1}{400}\sum_{i=1}^{2500}x_i$, i.e. the mean of all the 100 binomial trials (or 2500 Bernoulli trials).
Best Answer
It is hard to comment without further details, but your question can be given a general answer.
Binomial distribution describes the number of successes $k$ in $n$ trials, where the probability of success is $p$. So you have a fixed $n$ and $p$, i.e. they are parameters of your distribution and are not random. If you needed a distribution where the number of trials is also a random variable, you would have to define it in terms of a compound distribution rather then using simple binomial model. Another example of varying $n$ is a negative binomial distribution, where you stop your experiment after seeing $r$ failures -- so the sample size is not specified at start.