Q:

According to a 2009 Reader's Digest article, people throw away approximately 10% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 100 grocery shoppers to investigate their behavior. What is the probability that the sample proportion exceeds 0.02?

Accepted Solution

A:
Answer: 0.9962Step-by-step explanation:Given : According to a 2009 Reader's Digest article, people throw away approximately 10% of what they buy at the grocery store.i.e. the proportion of the people throw away what they buy at the grocery store [tex]p=0.10[/tex]Test statistic for population proportion : -[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]For [tex]\hat{p}=0.02[/tex][tex]z=\dfrac{0.02-0.1}{\sqrt{\dfrac{0.1(1-0.1)}{100}}}\approx-2.67[/tex]Now by using the standard normal distribution table , the probability that the sample proportion exceeds 0.02 will be :[tex]P(p>0.02)=P(z>-2.67)=1-P(z<-2.67)=1-0.0037925\\\\=0.9962075\approx0.9962[/tex]Hence, the probability that the sample proportion exceeds 0.02 =0.9962