# Normal approximation for binomial distribution: probabilities don't match up between 2 models [duplicate]

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).

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?

• density is not probability – Glen_b Nov 20 '17 at 6:16