We can use the Central Limit Theorem and the Approximating the Binomial probability.

We’re given

It is know that the expected value of flipping a coin n times is \(E[x] = np = 400*0.5 = 200\)

The standard deviation \(\sigma = \sqrt{npq} = \sqrt{400*0.5*0.5} = \sqrt{100} = 10\)

As these are 400 trials of a binomial distribution, we standardize the above equation so we can

find the z-score that corresponds to the probability \(\approx 0.8\) that is \[ \begin{aligned} P(200 - x \leq x \leq 200 + x) \approx P((200 - x - 0.5 - 200)10 \leq x \leq (200 + x + 0.5 - 200)/10) \\ = P((-0.5 - x)/10 \leq x \leq (0.5 + x)/10) \\ \approx NA(-(0.5+x)/10, (0.5+x)/10) \\ \text{For a standard normal distribution as it is symmetric around 0, we can re-write} \\ NA(-(0.5+x)/10, (0.5+x)/10) \\ \text{as} \\ 2*NA(0, (0.5+x)/10)) \approx 0.8 \end{aligned} \] Dividing 2 on both sides of the equation gives \(NA(0, (0.5+x)/10) \approx 0.4\)

As the area under the normal curve is \(NA(0, (0.5+x)/10) \approx 0.4\) we can let \(z = (0.5 + x) /10\) and look up the z value that is \(\approx 0.8\)

and using a standard probability table (there are plenty of websites and books that show it), we find the closest z value

with probability \(\approx 0.8\) is \(z =1.3\).

Finally, we can solve for x by solving

\(z = 1.3 =(0.5 + x) / 10\) which gives \(x = 12.5\) so this means \(x = 12.5\) to have the probability that the number of heads be

about 0.8 for 400 fair coin flips.