Answer to Question #16488 in Statistics and Probability for Vlatko
The restaurant Exclusive advertises delivery of ready prepared food. The announcement states that ordered food will be served for less than 50 minutes from receiving the order. Based on a random sample of 100 orders the average time for delivery of orders is calculated to be 49 minutes, and standard deviation of 5 minutes. What can be concluded on this random sample about the time of delivery with acceptable risk ?
Z(50) = (50 - 49)/(5/sqrt(100)) = 2 => p(time > 50) = p(z > 2) =0.5 – 0.4772 = 0.0228 So, there is a 2.28% risk that the time of delivery will be more than 50minutes.