\[ X\sim N\big(np, np(1-p)\big) = N(288, 144) \]

The standard deviation of the normal approximation is \(\sigma=12\). Hence, we can get:

\[ 312 - 288 = 24 = 2 \sigma \]

We know that the probability of getting an observation of at least 2 standard deviations from the mean of a normal distribution is roughly 0.025, and hence \(P(X > 311) = 0.025\).

Taken from the above link: “This approximation, known as de Moivre–Laplace theorem, is a huge time-saver when undertaking calculations by hand (exact calculations with large n are very onerous); historically, it was the first use of the normal distribution, introduced in Abraham de Moivre’s book The Doctrine of Chances in 1738. Nowadays, it can be seen as a consequence of the central limit theorem since B(n, p) is a sum of n independent, identically distributed Bernoulli variables with parameter p. This fact is the basis of a hypothesis test, a proportion z-test, for the value of p using x/n, the sample proportion and estimator of p, in a common test statistic.”