Question 892720
Tchebycheff's inequality states that {{{P(abs(X - mu)<=k*sigma)>=1-1/k^2}}}, where {{{mu}}} is the mean of the random variable and  {{{sigma}}} is its standard deviation.
Incidentally, if X = ratio of number of heads to n tosses of a fair coin, then the mean {{{mu = 1/2}}}, and variance is {{{sigma^2 = 1/(4n)}}}. (Remember, E(X) = p, while {{{Var(X) = (pq)/n}}}.)

==> standard deviation is {{{sigma = 1/(2*sqrt(n))}}}

Now from the given {{{P(0.45<=X<=0.55)>=0.95}}}

<==>{{{P(abs(X - 0.50)<=0.05)>=0.95}}}

Now let {{{0.05 = k*sigma}}}, and {{{1 - 1/k^2 = 0.95}}}

The first equation is equivalent to {{{k/(2*sqrt(n)) = 0.05}}}, or {{{sqrt(n) = 10k}}}.
==> {{{n = 100k^2}}}

But the second equation gives {{{k^2 = 20}}}.
==> n = 100*20 = 2,000 tosses.