Respuesta :
Answer:
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,
[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 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
