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What is a practical example where one would want to use a normal approximation of a binomial distribution over using properties of the binomial distribution itself?

I.e if I already know that the process can be modelled as a Bernoulli distribution, then in what circumstance would I want to use a normal approximation ? I already know the formula to find the probability of x successes in n trials, so why do I need a normal approximation ?

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    $\begingroup$ I use this approximation all the time when I'm not at a computer: I might be having a conversation with someone, for instance, and need a fast way to mentally estimate some binomial probabilities. When $n$ is much larger than $6$ or so, I'm just not that good at performing the sums--or, equivalently, evaluating the corresponding incomplete Beta integrals--needed to compute Binomial probabilities in my head. I don't know whether having this facility with mental approximations would constitute a "practical example" for you, though. $\endgroup$ – whuber Dec 21 '14 at 23:28
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    $\begingroup$ There have been occasions when I've needed to compute binomial probabilities for extremely large $n$ (tens of thousands, at least). In this situation, even with reasonably small $p$, the normal approximation can often be used with very little error. I also sometimes need to get an approximate calculation of binomial probabilities 'on the fly' for one reason or another. $\endgroup$ – Glen_b -Reinstate Monica Dec 22 '14 at 8:52
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If you had $X$ with a binomial distribution with parameters $n_x$ and $p_x$ and independently $Y$ with a binomial distribution with parameters $n_y$ and $p_y$, then you know that the distribution of $X+Y$ has mean $n_xp_x+n_yp_y$ and variance $n_xp_x(1-p_x)+n_yp_y(1-p_y)$, but you do not know the distribution in most cases.

The Normal distribution with continuity correction may provide a reasonable approximation for $X+Y$, at least away from the tails. Take the sum of several binomially distributed random variables, and it may be the only alternative to simulation.

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    $\begingroup$ Well, perhaps not the only alternative, since 'exact' numerical convolution is doable, but in some cases the normal approximation is a substantially more convenient approach and often quite sufficient. $\endgroup$ – Glen_b -Reinstate Monica Dec 22 '14 at 8:32
  • $\begingroup$ @Henry: Can you tell me how you got those beautifully formatted formulas. Quite a work of art! $\endgroup$ – Victor Dec 22 '14 at 15:18
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    $\begingroup$ @Victor - I used the MathJax version of TeX. See meta.math.stackexchange.com/questions/5020/… for some tips $\endgroup$ – Henry Dec 25 '14 at 17:10

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