A director of reservations believes that 2% of the ticketed passengers are no-shows. If the director is correct, what is the probability that the proportion of no-shows in a sample of 507 ticketed passengers would be greater than 3%? Round your answer to four decimal places
+118654 A director of reservations believes that 2% of the ticketed passengers are no-shows. If the director is correct, what is the probability that the proportion of no-shows in a sample of 507 ticketed passengers would be greater than 3%? Round your answer to four decimal places
3% of 507=15.21
So the question is ... what is the probability that 15.21 people or more will be no shows.
p=0.02 q=0.98
The mean number of no shows would be 0.02*507 = 10.14
The standard deviation fo no shows is
\(\sigma=\sqrt{n*p*q}\\ \sigma=\sqrt{507*0.02*0.98}\\ \sigma=3.15233247\)
I used this site but you should be able to do more of it yourself is you need to. Just work out the z score for a raw score of 15.21 and it is the same as any other normal curve.
So the probability is 0.0539
which is 5.39%
