Exercise 5.14

$P(\sqrt{(|X∕n-p|)}\sqrt{(a)})\le E|X∕n-p|∕a$.It follows from the Markov’s inequality. By the concavity of the function of taking squared root, $E|X∕n-p|∕a\le \sqrt{E{(X∕n-p)}^2}∕a$. For another inequality, $P({(X∕n-p)}^2{a}^2)\le E{(X∕n-p)}^2∕a^2$.Then the fact that $E{(X∕n-p)}^2=p(1-p)∕n$ completes the proof.