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 65 or fewer 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. $\left(10065\right)(0.71)65(0.29)35\left(\begin{array}{c}100\\ 65\end{array}\right)(0.71{)}^{65}(0.29{)}^{35}$

b. $\left(10065\right)(0.29)65(0.71)35\left(\begin{array}{c}100\\ 65\end{array}\right)(0.29{)}^{65}(0.71{)}^{35}$

c.

$\mathrm{P}(\mathrm{z}\le 0.65-0.71(0.71\left)\right(0.29\left)100\right)P\left(z\le \frac{0.65-0.71}{\sqrt{\frac{(0.71)(0.29)}{100}}}\right)$

d.$\mathrm{P}(\mathrm{z}\le 0.65-0.71(0.65\left)\right(0.35\left)100\right)P\left(z\le \frac{0.65-0.71}{\frac{(0.65)(0.35)}{\sqrt{100}}}\right)$

e.$\mathrm{P}(\mathrm{z}\le 0.65-0.71(0.71\left)\right(0.29\left)100\right)P\left(z\le \frac{0.65-0.71}{\frac{(0.71)(0.29)}{\sqrt{100}}}\right)$

Short answer

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

Step-by-step solution

Given information

Given:

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

Let use the below formula:

${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

Explanation for correct option

The binomial distribution's normal approximation is as follows: $np\ge 10$and $nq\ge 10$

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

As a result, the conditions are met, and the normal distribution can then be applied to the binomial distribution.

Size of the sample

$$\begin{gathered} \hat{p}=\frac{x}{n}=\frac{65}{100}=0.65 \\ {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.71(1-0.71)}{100}} \\ =\sqrt{\frac{0.71(0.29)}{100}} \\ z=\frac{x-\mu }{\sigma } \\ =\frac{0.65-0.71}{\sqrt{\frac{0.71\left(029\right)}{100}}} \end{gathered}$$

The chance that the sample proportion is smaller than $0.65$must be calculated.

$P(\hat{p}<0.65)=P\left(Z<\frac{0.65-0.71}{\sqrt{\frac{0.7(02.2\pi }{100}}}\right)$

Therefore, the correct option is (c)

Explanation for incorrect option

Option a $\left(10065\right)(0.71)65(0.29)35\left(\begin{array}{c}100\\ 65\end{array}\right)(0.71{)}^{65}(0.29{)}^{35}$is not the correct option

Option b$\left(10065\right)(0.29)65(0.71)35\left(\begin{array}{c}100\\ 65\end{array}\right)(0.29{)}^{65}(0.71{)}^{35}$is not the correct option

Option d$\mathrm{P}(\mathrm{z}\le 0.65-0.71(0.65\left)\right(0.35\left)100\right)P\left(z\le \frac{0.65-0.71}{\frac{(0.65)(0.35)}{\sqrt{100}}}\right)$is not the correct option

Option e$\mathrm{P}(\mathrm{z}\le 0.65-0.71(0.71\left)\right(0.29\left)100\right)P\left(z\le \frac{0.65-0.71}{\frac{(0.71)(0.29)}{\sqrt{100}}}\right)$is not the correct option