91Ó°ÊÓ

Referring to question Exercise 5

It is known that when p=0.2, \(E\left( {{{\bar X}_n}} \right) = p = 0.2\) and \(Var\left( {{{\bar X}_n}} \right) = \left( {0.2} \right)\left( {0.8} \right)/n = 0.16/n\)

Therefore,\(Z = \left( {{{\bar X}_n} - 0.2} \right)/\left( {0.4/\sqrt n } \right)\)will have approximately a standard normal distribution. It now follows that

\(\begin{align}\Pr \left( {|{{\bar X}_n} - p| \le 0.1} \right) &= \Pr \left( {|Z| \le 0.25\sqrt n } \right)\\ & \approx 2\Phi \left( {0.25\sqrt n } \right) - 1\end{align}\)

Therefore, this value will be at least 0.95 if and only if\(\Phi \left( {0.25\sqrt n } \right) \ge 0.975\)or, equivalently, if and only if\(0.25\sqrt {n \ge 1.96} \).

This final relation is satisfied if and only if\(n \ge 61.5\).

Therefore, the sample size must be \(n \ge 62\)