Chapter 7, Problem AP2.16CPT

To determine

To find: the one of the represents the approximate probability of obtaining a sample of same overall proportion of adults who owns their home as does the entire country.

Answer to Problem AP2.16CPT

  $P\left(Z<\frac{0.65-0.71}{\sqrt{\frac{0.71\left(0.29\right)}{100}}}\right)$

Explanation of Solution

Given:

  $\begin{array}{c}n=100\\ x=65\\ p=0.71\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

Normal approximation of the binomial distribution: $np\ge 10$ and $nq\ge 10$

  $\begin{array}{c}np=100\left(0.71\right)=71\ge 10\\ nq=n\left(1-p\right)=100\left(1-0.71\right)=29\ge 10\end{array}$

Therefore the requirements are satisfied and it can then about the binomial distribution by the normal distribution.

Sample size

  $\widehat{p}=\frac{x}{n}=\frac{65}{100}=0.65$

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.71\left(1-0.71\right)}{100}}\\ =\sqrt{\frac{0.71\left(0.29\right)}{100}}\end{array}$

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.65-0.71}{\sqrt{\frac{0.71\left(0.29\right)}{100}}}\end{array}$

It is required to calculate the probability that the sample proportion is less then $0.65$

  $P\left(\widehat{p}<0.65\right)=P\left(Z<\frac{0.65-0.71}{\sqrt{\frac{0.71\left(0.29\right)}{100}}}\right)$

Hence, the correct option is (c)