Why normal approximation to binomial distribution uses np> 5 as a condition I was reading about normal approximation to binomial distribution and I dunno how it works for cases when you say for example p is equal to 0.3 where p is probability of success.
On most websites it is written that normal approximation to binomial distribution works well if average is greater than 5. I.e. np> 5
But I am unable to find where did this empirical formula came from?
If n is quite large and probability of success is equal to .5 then i agree that normal approximation to binomial distribution is going to be quite accurate. But what about other cases? How can one say np> 5 is the condition for doing normal approximation?

Here's how I'm thinking of these conditions.
Note that if a random variable is truncated near its mean (i.e. the absolute value of the z score of the truncated value isn't too large) then the random variable's distribution will be skewed away from the truncated value and toward its mean.
That being said, observe that a binomial random variable X~B(n,p) is truncated at 0 and n. The condition np>10 pushes the distribution away from the truncation at 0, while n(1-p)>10 pushes the distribution away from the truncation at n. This will assure us that the distribution of X won't be undesirably skewed in any direction.
Think of np and n(1-p) as the expected number of success and failures in a series of n trials, respectively.
Hope this helps.
