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.

Answer 1

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.