In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints this year involving computer stores. You plan to select a random sample of complaints to estimate the proportion of complaints the Bureau is able to settle. Assume the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above. Suppose your sample size is 209. What is the proba

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In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints this year involving computer stores. You plan to select a random sample of complaints to estimate the proportion of complaints the Bureau is able to settle. Assume the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above. Suppose your sample size is 209. What is the probability that th

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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 = 209p = 0.75As 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.

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