Respuesta :
Answer:
The probability that the sample proportion will be within 6 percent of the population proportion is 0.9556.
Step-by-step explanation:
The question is:
What is the probability that the sample proportion will be within 6 percent of the population proportion?
Solution:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
The information provided is:
n = 209
p = 0.75
As n = 209 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by the Normal distribution.
[tex]\hat p\sim N(\mu_{\hat p}=0.75,\ \sigma_{\hat p}=0.0299)[/tex]
Compute the probability that the sample proportion will be within 6 percent of the population proportion as follows:
[tex]P(p-0.06<\hat p<p+0.06)=P(0.69<\hat p<0.81)[/tex]
[tex]=P(\frac{0.69-0.75}{0.0299}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.81-0.75}{0.0299})\\\\=P(-2.01<Z<2.01)\\\\=P(Z<2.01)-P(Z<-2.01)\\\\=0.97778-0.02222\\\\=0.95556\\\\\approx 0.9556[/tex]
Thus, the probability that the sample proportion will be within 6 percent of the population proportion is 0.9556.