Delta Airlines recently claimed that its on - time flight rate was 82 % . Find the probablity that among 500 randomly selected nights that fewer than 400 arrived on time.

Question

MTH243 Test # 3 Chapters 6 and 7 Instructor: Sandy Smithrevises

Answer 1

The mean expectation is 500×0.82=410.

If we assume a normal distribution Z=(400-410)/s where s is the standard deviation (not given). We can estimate its value using √(npq) where n=500, p=0.82, q=1-0.82=0.18, so s=√73.8=8.59 approx.

So Z=-10/8.59=-1.164. From tables this is equivalent to 0.1222 or about 12.2%, being the probability that fewer than 400 arrive on time.