For a large n, let k$beapproximatelyequalto\(np+t√npq, and use Stirling's formula to estimate the probability \prob(Sn=k), as follows: \begin{eqnarray*} \prob(S_n = k) &=& \bin...For a large n, let k$beapproximatelyequalto\(np+t√npq, and use Stirling's formula to estimate the probability \prob(Sn=k), as follows: \begin{eqnarray*} \prob(S_n = k) &=& \binom{n}{k} p^k q^{n-k} = (1+o(1)) \frac{\sqrt{2\pi n} (n/e)^n p^k q^{n-k}}{ \sqrt{2\pi k} (k/e)^k \sqrt{2\pi (n-k)} ((n-k)/e)^{n-k}} \\ &=& \frac{1+o(1)}{\sqrt{2\pi n p q}} \left(\frac{np}{k}\right)^k \left( \frac{nq}{n-k}\right)^{n-k} \\ &=& \frac{1+o(1)}{\sqrt{2\pi n p q}} \left(1+\frac{t\sqrt{q}}{…