Chapter 7.5, Problem 70E

a.

To determine

Prove that $p+3\sqrt{pq/n}<1$ if and only if $n>9\left(p/q\right)$.

Explanation of Solution

Let

$\begin{array}{l}p+3\sqrt{pq/n}<1\iff 3\sqrt{pq/n}<1-p\\ \iff 3\sqrt{pq/n}<q\\ \iff \sqrt{pq/n}<\frac{q}{3}\end{array}$

Taking square on both sides

$\begin{array}{l}p+3\sqrt{pq/n}<1\iff pq/n<\frac{q^2}{9}\\ \iff \frac{p}{n}<\frac{q}{9}\\ \iff \frac{n}{p}>\frac{9}{q}\\ \iff n>9\left(\frac{p}{q}\right)\end{array}$

Hence proved

b.

To determine

Prove that $0<p-3\sqrt{pq/n}$ if and only if $n>9\left(q/p\right)$.

Explanation of Solution

Let

$\begin{array}{l}0<p-3\sqrt{pq/n}\iff 3\sqrt{pq/n}<p\\ \iff 3\sqrt{pq/n}<p\\ \iff \sqrt{pq/n}<\frac{p}{3}\end{array}$

Taking square on both sides

$\begin{array}{l}0<p-3\sqrt{pq/n}\iff pq/n<\frac{p^2}{9}\\ \iff \frac{q}{n}<\frac{p}{9}\\ \iff \frac{n}{q}>\frac{9}{p}\\ \iff n>9\left(\frac{q}{p}\right)\end{array}$

Hence proved

c.

To determine

Prove that the normal approximation to the binomial is adequate if $n>9\left(\frac{p}{q}\right)$ and $n>9\left(\frac{q}{p}\right)$ or, equivalently, $n>9\left(\frac{\text{larger}\text{of}p\text{and}q}{\text{smaller}\text{of}p\text{and}q}\right)$.

Explanation of Solution

From the part a, normal approximation to the binomial is adequate if $n>9\left(\frac{p}{q}\right)$ and from part b, normal approximation to the binomial is adequate if $n>9\left(\frac{q}{p}\right)$.

By combining the parts a and b, normal approximation to the binomial is adequate if $n>9\left(\frac{p}{q}\right)$ and $n>9\left(\frac{q}{p}\right)$.

Or,

$n>9\max \left(\frac{p}{q},\frac{q}{p}\right)$

If $p=q$, then

$n>9\max \left(\frac{p}{q},\frac{q}{p}\right)=n>9$

If $p>q$, then

$n>9\max \left(\frac{p}{q},\frac{q}{p}\right)=n>9\left(\frac{p}{q}\right)$

If $p<q$, then

$n>9\max \left(\frac{p}{q},\frac{q}{p}\right)=n>9\left(\frac{q}{p}\right)$

Then,

$n>9\max \left(\frac{p}{q},\frac{q}{p}\right)=n>9\left(\frac{\text{larger}\text{of}p\text{and}q}{\text{smaller}\text{of}p\text{and}q}\right)$

Hence proved