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The Zimmerman Agency conducted a study for Residence Inn by Marriott of business

ID: 3291241 • Letter: T

Question

The Zimmerman Agency conducted a study for Residence Inn by Marriott of business travelers who take trips of five nights or more. According to this study, 37% of these travelers enjoy sightseeing more than any other activity that they do not get to do as much at home. Suppose 120 randomly selected business travelers who take trips of five nights or more are contacted. What is the probability that fewer than 40 enjoy sightseeing more than any other activity that they do not get to do as much at home?

Explanation / Answer

p=37%=0.37

q=1-0.37=0.63

n=120

probability fewer than 40

p(x<40) = ncr*p^n *q^n-r =120c40 0.37^40 *0.63^(120-40)= 0.1774

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