= Solution
The number $R$ of successes has a <binomial distribution>. Each arrangement of $r$ successes and $n-r$ failures has <probability> $p^rq^{n-r}$ by <independence>, and the <binomial coefficient> counts the arrangements. Therefore
$$
\boxed{p_r=\binom nr p^rq^{n-r},\quad 0\leq r\leq n;\qquad\sum_{r=0}^n p_r=(p+q)^n=1.}
$$
The final equality is the <binomial theorem>. Writing $R$ as a sum of <independent> <Bernoulli random variables> gives its <expectation> and <variance>: $\mathbb E R=np$, $\operatorname{Var}(R)=npq$. Thus the matching <normal distribution> has
$$
\boxed{\mu=np,\qquad\sigma=\sqrt{npq}.}
$$
The <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 $\sigma^2=45/4$, $\sigma=3\sqrt5/2$. The inclusive integer range $13,14,15,16,17$ becomes $12.5\leq X\leq17.5$ after the <continuity correction>. Standardizing the <normal distribution> gives
$$
\begin{aligned}
\mathbb P(13\leq R\leq17)&\approx\mathbb P(12.5\leq X\leq17.5)\\
&=\frac1{\sqrt{2\pi}}\int_{-\sqrt5/3}^{\sqrt5/3}e^{-z^2/2}\,dz\\
&=\frac2{\sqrt{2\pi}}\int_0^{\sqrt5/3}e^{-z^2/2}\,dz>2\cdot\frac14=\boxed{\frac12}.
\end{aligned}
$$
Here $(5/2)/(3\sqrt5/2)=\sqrt5/3$, 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.
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