Adirector 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.

Probability that the proportion of no-shows in a sample would be less than 9% is 0.8289

Step-by-step explanation:

Given:

8% of ticketed passengers are no shows.

Number of ticket passengers = 664

To 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 = 59.76

So, Now we have to find probability that 59.76 people or less will be no shows.

Probability of No show ticketed passengers, p = 2% = 0.08

Probability of ticketed passengers who show tickets, q = 92% = 0.92

Mean Number of no shows = 8% of 664 = 0.08 × 664 = 53.12

The Standard Deviation for no show is as follows,

Now, using z-score

Using z- score table, we get

P( z < 0.95 ) = 0.8289

Therefore, Probability that the proportion of no-shows in a sample would be less than 9% is 0.8289

is only a line? i think so

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225