Probability of an unbalanced coin This was an optional practice problem in statistics: An unbalanced coin has a 0.3 probability of being heads. The coin is tossed 20 times, let Y be the number of heads recorded.
a) Find the distribution of Y.
b) Find E(Y) and V(Y).
c) Find probability that Y falls within 2 standard deviations of the mean.
What I've got is as follows:
a) I'm guessing it's binomial distribution, since we're looking for the number of successes.
b) (since it's binomial): $E(Y)=np=0.3*20 = 6$ and $V(Y)=np(1-p)=4.2$
c) What do I compare to 2 standard deviations, or [0,14.4]? I'm guessing I take $p*14.4$, which is 4.32, and take $4.32/6$, so the answer is that 2 standard deviations is within 72% of the mean.
I appreciate any pointers.

The standard deviation is the square root of the variance $\sigma \approx 2.04$
you are looking for the probability that $6- 2\sigma < Y < 6+2\sigma$
since Y has to be an integer you are interested in values between 2 and 10.
So your answer will be ...
$$ P = \sum_{k=2}^{10}\binom {20} k (0.3)^k(0.7)^{20-k} $$
