91Ó°ÊÓ

Referring to question Exercise 5

When p=0.2, the random variable \({Z_n} = n{\bar X_n}\) will have a binomial distribution with parameters n and p=0.2, and

\(\Pr \left( {|{{\bar X}_n} - p| \le 0.1} \right) = \Pr \left( {0.1n \le {Z_n} \le 0.3n} \right)\)

The value of n for which this probability will be at least 0.75 must be determined by trial and error from the binomial distribution table at the back of the book. For n=8, the probability becomes

\(\begin{align}\Pr \left( {0.8 \le {Z_8} \le 2.4} \right) &= \Pr \left( {{Z_8} = 1} \right) + \Pr \left( {{Z_8} = 2} \right)\\ &= 0.3355 + 0.2639\\ &= 0.6291\end{align}\)

For n=9, they have

\(\begin{align}\Pr \left( {0.9 \le {Z_9} \le 2.7} \right) &= \Pr \left( {{Z_9} = 1} \right) + \Pr \left( {{Z_9} = 2} \right)\\ &= 0.3020 + 0.3020\\ &= 0.6040\end{align}\)

For n=10, they have

\(\begin{align}\Pr \left( {1 \le {Z_{10}} \le 3} \right) &= \Pr \left( {{Z_{10}} = 1} \right) + \Pr \left( {{Z_{10}} = 2} \right) + \Pr \left( {{Z_{10}} = 3} \right)\\ &= 0.2684 + 0.3020 + 0.2013\\ &= 0.7717\end{align}\)

Hence n=10 is sufficient.

It should be noted that although a sample size of n=10 will meet the required conditions, a sample size of n=11 will not meet the required conditions. For n=11, we would have

\(\Pr \left( {1.1 \le {Z_{11}} \le 3.3} \right) = \Pr \left( {{Z_{11}} = 2} \right) + \Pr \left( {{Z_{11}} = 3} \right)\)

Thus, only two terms of the binomial distribution for n=11 are included, whereas three of the binomial distribution for n=10 were included.