A director of reservations believes that 8% of the ticketed passengers are no-shows. If the director is accurate, what is the probability that the proportion of no-shows in a sample of 664 ticketed passengers would be less than 9%? Round your answer to four decimal places.

Answer:Probability that the proportion of no-shows in a sample would be less than 9% is 0.8289Step-by-step explanation:Given:8% of ticketed passengers are no shows.Number of ticket passengers = 664To find: Probability that the proportion of no-shows in a sample would be less than 9%.Number of no shows ticketed passenger = 9% of 664 = 9% × 664 = 59.76So, Now we have to find probability that 59.76 people or less will be no shows.Probability of No show ticketed passengers, p = 2% = 0.08Probability of ticketed passengers who show tickets, q = 92% = 0.92Mean Number of no shows = 8% of 664 = 0.08 × 664 = 53.12The Standard Deviation for no show is as follows,[tex]\sigma=\sqrt{n\times p\times q}[/tex][tex]\sigma=\sqrt{664\times0.08\times0.92}[/tex][tex]\sigma=\sqrt{48.8704}[/tex][tex]\sigma=6.9907[/tex]Now, using z-score [tex]\implies\:z=\frac{X-\mu}{\sigma}=\frac{59.76-53.12}{6.9907}=0.9498=0.95[/tex]Using z- score table, we getP( z < 0.95 ) = 0.8289Therefore, Probability that the proportion of no-shows in a sample would be less than 9% is 0.8289