A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to buy an extended warranty for the car. Let X denote the number of luxury cars sold in a given day, and let Y denote the number of extended warranties sold.
P(X = 0, Y = 0) = 1 / 6
P(X = 1, Y = 0) = 1/12
P(X = 1, Y = 1) = 1 /6
P(X = 2, Y = 0) = 1 /12
P(X = 2, Y = 1) = 1 /3
P(X = 2, Y = 2) = 1/6
What is the variance of X?
- 0.47
- 0.58 Correct Answer
- 0.83
- 1.42
- 2.58
Solution: B
Note
P(X = 0) = 1/6
P(X = 1) = 1/12 + 1/6 = 3/12
P(X = 2) = 1/12 + 1/3 + 1/6 = 7/12 .
E[X] = (0)(1/6) + (1)(3/12) + (2)(7/12) = 17/12
E[X2] = (0)2(1/6) + (1)2(3/12) + (2)2(7/12) = 31/12
Var[X] = 31/12 – (17/12)2 = 0.58 .