If I toss 400 fair coins, estimate (to the nearest whole percentage point) the probability that more than 220 heads appear.
Here is my thought process:
This is a binomial variable. We know that $p = 0.5$. Therefore, the mean is 200 ($n*p$), and the variance is equal to $n*p*(1-p)$, meaning that variance is 100, and standard deviation is 10.
Therefore, would this mean the outcome of 400 flips conforms to $N(200, 100)$--that is, a normal distribution with mean 200 and variance 100?
To solve this, would I then simple solve for $P(X >= 220)$, which yields 2.28%?
Another approach would be to calculate your answer using the cumulative distribution function. Because it is binomial distribution, it makes more sense to use this kind of calculation. The calculation you used was for normal distribution. Hope this helps!
$k$ is the amount of heads, $n$ is the total amount of flips, $p$ is the probability of getting either a heads or tails.
$f(k,n,p) = \displaystyle\sum_{i=k}^{n}\binom{n}{i}p^i(1-p)^{n-i} $
And $\binom{n}{k} = \frac{n!}{k!(n-k)!}$
$Pr(220) =f(220,400,0.5)= Pr(X=220) = \displaystyle\sum_{i=220}^{400}\binom{400}{i}0.5^{i}(1-0.5)^{400-i}$
When I plugged this into a calculator I found that there is about a $2.55201$% chance of flipping heads more than 220 times. Here's some more reading on it if you would like