Question

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: 

x	45	46	47	48	49	50	51	52	53	54	55
P(X=x)	0.05	0.1	0.12	0.14	0.25	0.17	0.06	0.05	0.03	0.02	0.01

a) With what probability will all passengers who show up for the flight have a seat on this flight? 

b) What is the expected number of passengers who show up for this flight? What is the associated variance?

Answer 1

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.)