# Answer to Question #16488 in Statistics and Probability for Vlatko

Question #16488

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 ?

Expert's answer

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.

So, there is a 2.28% risk that the time of delivery will be more than 50minutes.

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