2\" (30%) Consider a plane with a maximum capacity of 50 passengers. Suppose it

Question

2" (30%) Consider a plane with a maximum capacity of 50 passengers. Suppose it is known that on a particular the discrete random variable X (which counts the number of passengers who actually show up for the flight) has the following pmf: 45 46 47 48 49 50 5152 53 54 55 PX = z) 0.05T0.1T 0.12 0.14 0.25 0.17 0.06 0.05 0.03 0.02 0.01 a) b) With what probability will all passengers who show up for the flight have a seat on this flight? What is the expected number of passengers who show up for this flight? What is the associated variance?

Explanation / Answer

a) P(X < 50) = P(X = 45) + P(X = 46) + P(X = 47) + P(X = 48) + P(X = 49) + P(X = 50)

                     = 0.05 + 0.1 + 0.12 + 0.14 + 0.25 + 0.17

                     = 0.83

b) Expected number = E(X) = 45 * 0.05 + 46 * 0.1 + 47 * 0.12 + 48 * 0.14 + 49 * 0.25 + 50 * 0.17 + 51 * 0.06 + 52 * 0.05 + 53 * 0.03 + 54 * 0.02 + 55 * 0.01 = 48.84 or 49 (approx.)

E(X2) = 452 * 0.05 + 462 * 0.1 + 472 * 0.12 + 482 * 0.14 + 492 * 0.25 + 502 * 0.17 + 512 * 0.06 + 522 * 0.05 + 532 * 0.03 + 542 * 0.02 + 552 * 0.01 = 2389.84

Variance = E(X2) - (E(X))2 = 2389.84 - 48.842 = 4.4944 or 4 (approx.)

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