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# According to Runzheimer International, a typical business traveler spends an average of \$281 per day in Chicago. This co...

According to Runzheimer International, a typical business traveler spends an average of \$281 per day in Chicago. This cost includes hotel, meals, car rental, and incidentals. A survey of 50 randomly selected business travelers who have been to Chicago on business recently is taken. For the population mean of \$281 per day, what is the probability of getting a sample average of more than \$267 per day if the population standard deviation is \$47?

We have given that

Mean(µ) = 281
Population Standard deviations (σ) = 47

Sample size (n) = 50

We have to find probability of getting a sample average of more than \$267 per day

P(x̅ >267) = P[(x̅ -µ)/(σ/√n) >(267-281)(47/√50)]

= P( Z > -2.106)

= 1 - P(Z < -2.106)

= 1 - 0.0174( from standard normal table)

P(x̅ >267) = 0.9826

Thus, probability of getting sample average of more than \$281 per day is 0.9826

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