Setting up normal approximations for coin tosses

sillybuffalo said:

We usually treat tosses of a coin as equally-likely outcomes, with exactly the same probability ½ for the outcomes "Heads" and "Tails". Suppose the coin is actually biased, with P[Heads]=0.51. Find the exact or approximate (using the normal approximation to four decimal places, being careful about the ±½s) probability of observing strictly more Heads than Tails in

1. N=3 tosses (Give the exact answer)?
2. N=3 tosses (Give the approximate answer, using normal)?
3. N=100 tosses (Give the approximate answer, using normal)?

I understand part 1 easily, but I am having difficulty setting up the parameters for parts 2 and 3.

To find the normal approximation for more heads than tails in 3 tosses, I'm not sure if I should set it up as P [1.5 </= X </= 3.5] (taking into account the continuity correction), or if I should do 1 - P[x<2]. If I use 1 - P[x<2], do I need to take into account the continuity factor? (as in, do I need to use 1.5 instead of 2 to find the value of z?

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I would have liked to see your work for part 1. Did you use a Binomial Distribution, and find P(0), P(1), P(2), and P(3) to determine P(more heads than tails)?

Do you know how to find the mean (mu) and standard deviation (sigma) of the Binomial Distribution? Use a normal distribution with that same mu and sigma as the approximation. Show us your work!