You are the new owner of a restaurant. You read that a sample survey by the National Restaurant Association shows that 40% of adults are committed to eating nutritious food when eating away from home. To help plan your menu, you decide to conduct a sample survey in your own area. You will use random digit dialing to contact an SRS of 200 households by telephone. What is the probability that X lies between 75 and 85?

Question

prettygirl_shynice · Answer

a restaurant owner would ka) yes - central limit theorem justifies normality for a SRS of 200 people - that's a reasonably large sample 
b) expected mean = .4(200)=80 
c) use normal approx to binomial - std dev (number in sample) = sqrt(n p (1-p)) 
= sqrt(200(.4)(.6))=6.928 
75 is (75-80)/6.928=-0.72 std dev below mean (85 is +0.72 std dev above mean) 
from normal table, prob is .5284 

d) 100 is (100-80)/6.928=2.89 std dev above mean; so there is reason to believe area is highernow how to do his own homework

anonymous · Answer

Thank you. If I knew how to do it I wouldn't be asking. I needed further help than what my textbook could provide