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Do we use normal approximation when discrete distributions are hard to solve?

For example, $P(X\ge 7000)$ where is $X\sim\operatorname{Binomial}(13000, 0.7)$.

Obviously, summing each case manually cannot be done and the summation is difficult to solve (I assume)?

In my textbook, sometimes when something is a dicrete distribution and I calculate it using discrete distribution, I flip to the answer and they used normal approximation to the discrete distribution.

Example:

enter image description here

"Why are you choosing to use normal approximation?!"

A_for_ Abacus
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  • Are you wondering about why an approximation in the first place or why the normal approximation? – user190080 Mar 08 '18 at 15:01
  • @user190080 Why an approximation in the first place? I know they used normal approximation because binomial distribution is approximately normal in the long-run. – A_for_ Abacus Mar 09 '18 at 02:43

1 Answers1

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As you readily point out, it is a matter of convenience. You can indeed evaluate the binomial distribution $n$ times and the add the results together, but that gets pretty boring really fast. Unless of course you have a computational tool at hand (calculator, computer, $\dots$)

Instead if $n$ is large enough you can use the Normal approximation and the results are strikingly good. The problem is always "what does it mean for $n$ to be large?". And to that there are a few rules you could follow (these are more suggestions that actual rules)

  • Both $np\gtrsim 5 $ and $n(1-p)\gtrsim 5$

  • The $|\gamma| < 1/3$ and $n > 5$, with $\gamma$ being the skewness

$$ \gamma = {\frac {1-2p}{\sqrt {np(1-p)}}} $$

I am pretty sure that there are more rules out there, but these have worked out for me in the past

caverac
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