Asked by Clifford
An airline reports that if has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows.
Answers
Answered by
MathGuru
Mean = np = 150 * .15 = ?
Standard deviation = √npq = √(150 * .15 * .85) = ?
Note: q = 1 - p
I'll let you finish the calculations.
Once you have the mean and standard deviation, use z-scores:
z = (x - mean)/sd
Note: x = 20
After you calculate z, check a z-table to find your probability. (Remember that the problem is asking for "fewer than 20" when you check the table.)
I hope this will help get you started.
Standard deviation = √npq = √(150 * .15 * .85) = ?
Note: q = 1 - p
I'll let you finish the calculations.
Once you have the mean and standard deviation, use z-scores:
z = (x - mean)/sd
Note: x = 20
After you calculate z, check a z-table to find your probability. (Remember that the problem is asking for "fewer than 20" when you check the table.)
I hope this will help get you started.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.