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 $268 per day if the population standard deviation is $47?

The correct answer is - 97.74%.

Step-by-step explanation:

Given:

sample mean = 281

Std Dev = 47 ==> Sample

Std Dev (s) = std dev/sqrt(n) = 6.0677

Solution:

Find : P( x > 268 )

z = ( x - u )/s = ( 268 - 281 )/6.0677

= -1.9558

then, P( x > 270) = P( Z > -1.9558)

= P( Z <1.9558)

= 0.9747

change into percentage:

= 97.471%

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