91Ó°ÊÓ

Given:

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

Let use the below formula:

The binomial distribution's normal approximation is as follows: $npâ‰\yen 10$and $nqâ‰\yen 10$

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

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

Size of the sample

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

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

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

Therefore, the correct option is (c)

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

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