Question 1176040: A car dealership sells 0, 1 or 2 luxury vehicles each 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. Let Y denote the number of extended warranties sold.
X Y Probability
0 0 0.07
1 0 0.13
1 1 0.1
2 0 0.15
2 1 0.11
2 2 0.44
Use the table to find the variance of X.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to calculate the variance of X using the given table:
**1. Calculate the Probabilities for X**
* P(X = 0) = 0.07
* P(X = 1) = 0.13 + 0.10 = 0.23
* P(X = 2) = 0.15 + 0.44 = 0.59
**2. Calculate the Expected Value (Mean) of X**
* E(X) = Σ [x * P(X = x)]
* E(X) = (0 * 0.07) + (1 * 0.23) + (2 * 0.59)
* E(X) = 0 + 0.23 + 1.18
* E(X) = 1.41
**3. Calculate the Variance of X**
* Var(X) = E(X²) - [E(X)]²
* First, calculate E(X²):
* E(X²) = Σ [x² * P(X = x)]
* E(X²) = (0² * 0.07) + (1² * 0.23) + (2² * 0.59)
* E(X²) = 0 + 0.23 + 2.36
* E(X²) = 2.59
* Now, calculate Var(X):
* Var(X) = E(X²) - [E(X)]²
* Var(X) = 2.59 - (1.41)²
* Var(X) = 2.59 - 1.9881
* Var(X) = 0.6019
**Alternative Method**
* Var(X) = Σ [(x - E(X))² * P(X = x)]
* Var(X) = [(0 - 1.41)² * 0.07] + [(1 - 1.41)² * 0.23] + [(2 - 1.41)² * 0.59]
* Var(X) = [(1.9881) * 0.07] + [(0.1681) * 0.23] + [(0.3481) * 0.59]
* Var(X) = 0.139167 + 0.038663 + 0.205379
* Var(X) = 0.383209
**Therefore, the variance of X is 0.383209**
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