Yes the CLT is a bit of a confusion here because as we know, The central limit theorem states that the sampling distribution of any statistic will be normal or nearly normal, if the sample size is large enough. That's where the problem comes in, sample size should be at least 30, some say 40. That would lead you to believe that you cannot use apply CLT here, BUT. the more closely the original population resembles a normal distribution, the fewer sample points will be required.
But what we can do is use the normal approximation to the binomial if
this condition holds true. If np > 5 and n(1-p)> 5, then it can
be used. We know p = .5 and if you consider 4 trials, then np < 5,
as is n(1-p). BUT if you use all 12 tosses, then n = 12 and np = 6 and
n(1-p) = 6. But doing it this was we would get the average number of
heads expected in 12 tosses to be 6 and the average in all 12 tosses to
be 8, instead of 2 in 3 tosses.
npq = (12)(.5)(.5) = 3
sqrt(npq) = 1.732
Formula is z = (x-bar - np)/(sqrt(npq)
z= (8 - 6)/(1.732) = 1.15
z(1.15) = .8749
1- .8749 = .1251
If you notice when you go through a coin tossing experiment tossing
three times, there are 8 outcomes and the only way to get greater the 2
heads is if you get HHH, which is 1 out of 8, which is 12.5%.