Solved – Bernoulli and normal distribution

I know that a sum of RVs bernoulli distributed with the same parameter $p$ may be approximated with a normal distribution.
My question is whether a single bernoulli RV may be approximated with a normal distribution.

In particular I have a vector of zero and one values. I can compute the mean $mu=p$ and $sigma=sqrt{(p*(1-p))}$, i.e. $p(X=1) = p$ and $p(X=0)=1-p$.

I am going to approximate X with a normal distribution with parameters $mu$ and $sigma$. Is it correct? Please, note that with this approximate distribution I would like compute the probability of just $0$ and $1$.

Answered in comments: Short answer: no. Long answer. Also no but with a small caveat, and a "why would you do this? If you know $p$ you can compute the probability directly just as you show." – Glen_b

I'll repeat a comment I made to a similar question recently of course you can approximate a Bernoulli distribution with a Normal distribution. The question should focus on (a) why? and (b) how good can the approximation be? Since you're approximating a Bernoulli, you will be interested in only one nontrivial event: $Xle 0$, whose chance is $1−p$. From this you can easily calculate the chances of the events $X=0$ and $X=1$. The probability that $Xle 0$ can in turn be approximated with perfect accuracy by an infinite class of Normal distributions. – whuber

In short: You need to tell us why you want to do this!

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