Question

Homework: Section 8.2 Homework Score: 0 of 1 pt 7 of 11 (5 complete) 8.2.13-T HW Score: 30%, 3.3 of Question Help Suppose a simple random sample of size n 1000 is obtained from a population whose size is N=1,000,000 and whose population proportion with a specified characteristic is p=0.58 Complete parts (a) through (c) below. (a) Describe the sampling distribution of p A. Approximately normal, 0.58 and O. NO 0156 OB. Approximately normal, a 0.58 and 1000 OC. Approximately normal, A 0.58 and 30.0002 (b) What is the probability of obtaining x 610 or more individuals with the characteristic? P(x2610) - (Round to four decimal places as needed)

Answer 1

Solution

$\small \dpi{100} \mathbf{given:}\;n=1000,\;N=1000000,\;p=0.58$

${\color{Blue} \mathbf{(a)}}$

A. Approximately normal = 0.58 and p = 0.0156

mean of the sampling distribution of p

$\mu_{\hat p}=p=\mathbf{{\color{Red} 0.58}}$

standard deviation of the sampling distribution of

$\sigma_{\hat p}=\sqrt{\frac{p*(1-p)}{n}}=\sqrt{\frac{0.58*(1-0.58)}{1000}}=\sqrt{0.0002436}=\mathbf{{\color{Red} 0.0156}}$

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${\color{Blue} \mathbf{(b)}}\;\mathbf{given}:\;x=610$

$Sample\;proportion\;\left (\hat p \right )=\frac{x}{n}=\frac{610}{1000}=\mathbf{0.61}$

P(0.61 or more) = PP > 0.61) =?

$\mathbf{formula:}\;z=\frac{\hat p-p}{\sqrt{\frac{p*(1-p)}{n}}}$

$\Rightarrow P\left ( \frac{\hat p-p}{\sqrt{\frac{p*(1-p)}{n}}}\geq \frac{0.61-0.58}{\sqrt{\frac{0.58*(1-0.58)}{1000}}} \right )$

$\Rightarrow P\left (z\geq 1.92 \right )$

${\color{Blue} \mathbf{Note:}}\;P(z\geq a)=1-P(z\leq a)$

$\Rightarrow 1-P\left (z\leq 1.92 \right )$

Refer Z-table to find the probability or use excel formula "=NORM.S.DIST(1.92, TRUE)" to find the probability.

$\Rightarrow 1-0.9727$

$\Rightarrow \mathbf{{\color{Red} 0.0273}}$

$\mathbf{\therefore P(x\geq 610)=\mathbf{{\color{Red} 0.0273}}}$