Variance of a normal distribution for coin toss. I have difficulties constructing the normal distribution for (20) coin tosses. (Don't ask why, but I never had probability in school.)

What is the probability of getting at most 12 heads out of 20 tosses?

With the help of Wikipedia, I think that the CDF of a normal distribution gives the answer at $X=12$:
For that I need the mean, which is $10$, and the variance, which is
$$\frac{1}{20}\sum_{k=1}^{20}(k-10)^2=\frac{1}{20}\sum_{k=1}^{9}k^2+\frac{1}{20}\sum_{k=1}^{10}k^2=$$$$\frac{1}{20}\cdot\frac{9(9 + 1)(2\cdot9 + 1)}{6}+\frac{1}{20}\cdot\frac{10(10 + 1)(2\cdot10 + 1)}{6}=\frac{67}{2}$$
Now I tried to look up the normal distribution with mean $10$ and variance $33.5$ on wolframalpha, but at $X=12$ the CDF is much lower $(\sim 0.6352)$ than the result given in the book ($\sim 0.8$). The thing I was most unsure about was the variance, that's why I included it in the title, but maybe something else is wrong, too.
 A: Point (1): The distribution of heads from flipping a coin is a binomial distribution, not a normal distribution. Note that, for example, the former is discrete while the latter is continuous. However: the normal distribution can be used as an approximation, since the binomial is numerically much more intensive (esp. by hand calculation). 
Point (2): You're correct that your calculation of the variance is way off the reservation -- personally I can't tell what the reference or source for that calculation was. Perhaps that's vaguely in the direction of the variance of rolling a single 20-sided die? (Or maybe one with faces 0 to 20.) The correct variance for this binomial distribution is: $\sigma^2 = np(1-p) = 20(0.5)(0.5) = 5$; i.e., a standard deviation of $\sigma = \sqrt{5} \approx 2.24$. 
So, taking an upper limit of $x = 12.5$ (the halfway cutoff between 12 and 13) and standardizing, this gives us a standardized score of $z = (x - \mu)/\sigma = (12.5 - 10)/2.24 = 1.12$. And looking up the associated area on a standard normal curve table gives $\Phi(1.12) = 0.8686$ (with other technology varying a bit for rounding). 
As an additional hint, you can just type this word problem directly into Wolfram Alpha (and note the "More statistics" button there). 
