Q. 16 - The Practice of Statistics for AP

Question

According to the U.S. census, the proportion of adults in a certain county who owned their own home was $0.71$. An SRS of $100$ adults in a certain section of the county found that $65$ owned their home. Which one
of the following represents the approximate probability of obtaining a sample of $100$ adults in which fewer than $65$own their home, assuming that this section of the county has the same overall proportion of adults who own their home as does the entire county?

(a) $\left(\begin{array}{c}100\\ 65\end{array}\right)(0.71{)}^{65}(0.29{)}^{35}$ 

(b) $\left(\begin{array}{c}100\\ 65\end{array}\right)(0.29{)}^{65}(0.71{)}^{35}$
(c) $P\left(z<\frac{0.65-0.71}{\sqrt{\frac{(0.71)(0.29)}{100}}}\right)$ 

(d) $P\left(z<\frac{0.65-0.71}{\sqrt{\frac{(0.65)(0.35)}{100}}}\right)$
(e) $P\left(z<\frac{0.65-0.71}{\left.\frac{(0.71)(0.29)}{\sqrt{100}}\right)}\right.$

Step-by-Step Solution

Answer

The correct answer is option (c)$P\left(z<\frac{0.65-0.71}{\sqrt{\frac{(0.71)(0.29)}{100}}}\right)$.

Step 1: Given information

The proportion of adults who owned their own home was $0.71.$ An SRS of $100$ adults of the county found that $65$ owned their home.

Step 2: Explanation

Let, sample size$n=100$
Number of successes$x=65$
Population proportion $p=0.71$
Requirements for a normal approximation of the binomial distribution: $np\ge 10$ and $nq\ge 10$.
$$\begin{gathered} np=100(0.71)=71\ge 10 \\ nq=n(1-p)=100(1-0.71)=29\ge 10 \end{gathered}$$

The proportion is:
$\hat{p}=\frac{x}{n}=\frac{65}{100}=0.65$
Then the mean is:
${\mu }_{\hat{p}}=p=0.71$

Step 3: Calculation

The standard deviation is:
$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.71(1-0.71)}{100}} \\ =\sqrt{\frac{0.71(0.29)}{100}} \end{gathered}$$

The z-score is:

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

To determine the probability that the sample proportion is less than $0.65$.

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

Hence, option (c) $P\left(z<\frac{0.65-0.71}{\sqrt{\frac{(0.71)(0.29)}{100}}}\right)$ is correct.