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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?

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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.

Therefore:

$\mu$ - $3\sqrt{np(1-p)} > 0 \hspace{2cm}$ 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):

$0.474<p<0.526$

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

$0.0826<p<0.9174$

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.

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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".

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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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.

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