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When we use normal approximation to a binomial distribution $Bin(n, p)$, do we assume it is a normal distribution and get $E(X)$ with $E(X)= \int_{-\infty}^{\infty}xf(x)dx$.

Or get $E(X)$ as a binomial distribution first with $E(X) = np$?

What about when you're normal approximating a distribution that you have no idea what distribution it is?

A_for_ Abacus
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    One approximation is to say $E(X)=np$ and $Var(X)=np(1-p)$ and use that in the normal distribution. If you do not know the distribution, you might estimate the mean and variance from a sample – Henry Mar 08 '18 at 08:44
  • Okay, so use E(X) and Var(X) based on the original distribution's definition of E(X) and Var(X). – A_for_ Abacus Mar 08 '18 at 08:56

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If you use the normal approximation of a binomial random variable $Y\sim \operatorname {Bin}(n,p)$ for sufficiently large $n$, then you should first ask yourself, what normal distribution is actually used for the approximation (without normalization):

It is $\mathcal {N}(np,\,np(1-p))$, hence you use while using the normal approximation the known parameter for the mean and variance of $Y\sim \operatorname {Bin}(n,p)$.

If you don't know the distribution but still want to use the normal approximation (which needs ofc some justification), then as Henry points out, you need to use sample estimates of mean and variance.

Some reference like this might be useful too, some examples are calculated and also the continuity correction (since $Y$ is discrete vs. $\mathcal N$ is continuous) is introduced.

user190080
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