Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 4

The number $R$ of successes has a binomial distribution. Each arrangement of $r$ successes and $n-r$ failures has probability $p^r{q}^{n-r}$ by independence, and the binomial coefficient counts the arrangements. ThereforeThe final equality is the binomial theorem. Writing $R$ as a sum of independent Bernoulli random variables gives its expectation and variance: $\mathbb{E}R=np$, $\mathrm{Var}(R)=npq$. Thus the matching normal distribution hasThe normal approximation is appropriate when both expected counts $np$ and $nq$ are sufficiently large; large $n$ alone is insufficient if $p$ is extremely close to zero or one.

For the guessing student, $n=60$, $p=1/4$, $q=3/4$. Therefore the expected number of correct answers is $\boxed{15}$, and ${\sigma }^2=45/4$, $\sigma =3\sqrt{5}/2$. The inclusive integer range $13,14,15,16,17$ becomes $12.5\le X\le 17.5$ after the continuity correction. Standardizing the normal distribution givesHere $(5/2)/(3\sqrt{5}/2)=\sqrt{5}/3$, and the last step uses the supplied bound. The approximate probability is $0.54394$. The normal approximation provides the requested conclusion; as a separate check, summing the five exact binomial probabilities gives approximately $0.54370$, also exceeding one half.