Question

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?

Answer

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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