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?


The mean $\mu$ of a binomial = np. The standard deviation of a binomial = $\sqrt{np(1-p)}$

For a normal distribution, $\mu$ should be 3 standard deviations away from 0 and n.


$\mu$ - $3\sqrt{np(1-p)} > 0 \hspace{2cm}$ and $\hspace{2cm}\mu$ + $3\sqrt{np(1-p)}<n$

From that starting point, algebraically you can get to the inequalities:

$np>9(1-p)\hspace{2cm}$ and $\hspace{2cm}n(1-p)>9p$

To satisfy these inequalities, as n gets larger, p has a wider range. Or you could also say the closer p is to 0.5, the smaller n you can use.

Using n=10 (for example):


As n gets larger, p does not have to be so close to 0.5. For n = 100,


Remarkably, even with a p = 0.9, if n >100 then the mean will be 3 standard deviations away from 0 and n.

This relates to calculating np and n(1-p), as if both are greater than 5, usually these inequalities are satisfied. However something like n=15, p=0.65 does not work, so some textbooks say np>9.

This condition does not guarantee that the binomial will fit a normal dist. but just that the mean will not be skewed too far towards 0 or n.

New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

The condition $np > 5$ is not the condition, merely a rough estimate of what should be true in order for the normal distribution approximation to be "good enough".

From Wikipedia:

One rule is that both $x=np$ and $n(1 − p)$ must be greater than 5. However, the specific number varies from source to source, and depends on how good an approximation one wants.

There you can also find a list of other "rules".

  • $\begingroup$ Thank you for the answer. However this condition must have been derived empirically. And i am looking for source of it. What if mean of binomial distribution is 0 and prob of success is 0.5. In this case the prob mass func will still look like normal distribution. But it violates np> 5. Does that mean that using normal distribution to approximate answer in this case will be incorrect? $\endgroup$ – Durin Apr 20 '14 at 8:40
  • 1
    $\begingroup$ @Ani: The mean of a binomial distribution is $np$ so if it is $0$ then either $n=0$ or $p=0$, neither of which is suitable for a normal approximation. $\endgroup$ – Henry Apr 20 '14 at 10:13
  • $\begingroup$ So I plotted binomial distribution chart using an online tool for some random values of n and p. It was really interesting to see that for np> 5 prob mass func looked like normal distribution curve. I am new to statistics. I am still curious how do statisticians come to these generic values? @Henry Thank you for pointing out my mistake. $\endgroup$ – Durin Apr 20 '14 at 10:33
  • $\begingroup$ @Henry and naslundx I have submitted a new answer below. Please let me know if I am thinking in right direction. $\endgroup$ – Durin Apr 22 '14 at 15:09

So I did some experiments. I think np>5 condition is not correct at all. It depends on Excess Kurtosis value for a given binomial distribution. If it is Mesokurtic then approximation will give accurate results.

Check following table enter image description here

for n=11 and p=0.5 kurtosis will be around 0.18. That is platykurtic and so I don't think approximation will give accurate results, even though n*p=5.5 > 5. The table shows results which manifests what I am trying to say.

  • 2
    $\begingroup$ For your table, you need a continuity correction, e.g. asking for at most half an even number of tosses. Take the 10 tosses example. You are asking for "at most 5 heads" or "fewer than 6 heads", which might lead to different approximations. So take 5.5 as the cut-off, i.e. 55% of 10. The normal approximation with this continuity correction would give this a probability of about $0.6240852$ compared with the binomial probability of about $0.6230469$. Quite close in my mind. $\endgroup$ – Henry Apr 22 '14 at 20:29
  • $\begingroup$ When I take 5 i am calculating area from X=5 to 0. But if I take 5.5 ares under that extra strip of 0.5 is also added. I don't need that. I might say I will integrate upto 5.001 instead for continuity correction. Secondly prob at a given point in dist will be almost zero. So isn't <5 and <=5 same? $\endgroup$ – Durin Apr 23 '14 at 2:16
  • $\begingroup$ So i read a bit more about continuity correction and I agree that i have missed out on it. But now my question is how is 0.5 considered a good enough value for continuity correction. In fact the more I am reading statistics the more I am finding such rules. But somehow I am unable to find the source and reason of making the rule. $\endgroup$ – Durin Apr 23 '14 at 4:46
  • 1
    $\begingroup$ The point is that you want both $\le 5$ and $\lt 6$, which are equivalent for a discrete distribution on integers but are different for a continuous distribution. $5.5$ is simply halfway between them. $\endgroup$ – Henry Apr 23 '14 at 7:51
  • $\begingroup$ Thank you very much for patiently explaining me. Your comments helped me think and explore more about this topic. $\endgroup$ – Durin Apr 23 '14 at 8:23

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.


For $np$ and $nq$ to increase $n$ must increase. $n$ is the number of independent trials, so it should be clear that the more independent trials made, the more accurate your approximation is. The probability histogram approximates a normal curve pretty accurately when $np$ and $nq$ are greater(or equal to) $5$. However bigger is better! If $np$ and $nq$ were greater than $10$ the probability histogram would approximate the normal curve even more.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.