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 60 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

Given,sample size(n)=60mean(μ)=281standard deviation(σ)=47

Required probability is P(x¯>267)=Px¯-2814760>267-2814760=PZ>-2.31=0.9896  (from Z-table)The probability of getting a sample average of more than $267 per day is 0.9896

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