Question

Because many pastors who make reservation do not sure airlines often overbook flights( sell more tickets than there are seats). A certain airplane holds 295 passengers. If the airline believes the rate of passenger no shows is 8% and sells 321 tickets is it likely they will not have enough seats and someone will get bumped? Use the normal model to approximate the binomial to determinethe probability of at least 296 passengers showing up.

Answer #1

**Answer:**

**1)**

Hence, the probability of showing up is p = 1-0.08 = 0.92

n = 321 here.

mu = mean = np = 321*0.92 =
**295.32 **

s = standard deviation = sqrt(np(1-p)) =
**4.8606
**

Thus, the corresponding z score is

z = (x-u)/s = (296 - 295.32)/4.8606

= 0.1399

Thus, the right tailed area is

**P(z > 0.139 ) = 0.551 [ANSWER]**

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