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I have a binomial distribution of n = 400 and p = 0.2. With this distribution, the probability of getting k = 88 is: $$ \binom{400}{88} \ 0.2^{88} \ 0.8^{400 - 88} \ =\ 0.0295 $$

When I approximate this binomial distribution with a normal distribution, the mean of my normal distribution would be $400\ *\ 0.2\ =\ 80$, and the standard deviation would be $\sqrt{400\ *\ 0.2\ *\ 0.8} \ =\ 8$.

With this mean and standard deviation, the z-value of k = 88 would be: $$z\ =\frac{88\ -\ 80}{8} \ =\ 1$$

In a standard normal distribution, this z-value would have a probability of $0.24$ (I think this is called the pdf, but I'm still new to statistics).

Standard normal distribution

This probability ($0.24$) doesn't match up with the probability I'd expected from the binomial distribution ($0.0295$). How can I explain this discrepancy?

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    $\begingroup$ density is not probability $\endgroup$ – Glen_b Nov 20 '17 at 6:16
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The normal approximation only works for cumulative probabilities. E.g. Greater than or less than. If you want to find a normal approximation to a binomial density for a specific count, use two cumulative probabilities. For y=88 calculate the difference of the cumulative normal density for 87.5 and 88.5.

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  • $\begingroup$ Thanks @AdamO! I tried your method and it worked (both came out to around 0.03). I'm still wondering why using the exact probability wouldn't work, since that way I'm sort of finding the area of the rectangle of width = 1 around 88 (i.e. between 87.5 and 88.5) and height = probability. $\endgroup$ – seismatica Nov 20 '17 at 6:25
  • $\begingroup$ @AmdamO, I don't agree with you, why chose 87.5 and 88.5? , how about 87.6 and 88.4, then 87.9 to 88.3, then 87.9999999999 to 88.00000001? by your approach,you will ultimately reach to probability eqaul to zero. I think it will be OK to approximate CDF but not a point of probaiblity. $\endgroup$ – Deep North Nov 20 '17 at 8:31

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