62 475
Assignments Done
98,8%
Successfully Done
In June 2018

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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions