posted by Karen .
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.
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.