Question 1175836: 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.
Hint: Find the probability function for X.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve this step-by-step.
**1. Probability Function for X**
We need to find the probability of each value of X.
* **P(X = 0):**
* From the table, P(X = 0, Y = 0) = 0.07.
* Since this is the only case where X = 0, P(X = 0) = 0.07.
* **P(X = 1):**
* From the table, P(X = 1, Y = 0) = 0.13 and P(X = 1, Y = 1) = 0.10.
* P(X = 1) = P(X = 1, Y = 0) + P(X = 1, Y = 1) = 0.13 + 0.10 = 0.23.
* **P(X = 2):**
* From the table, P(X = 2, Y = 0) = 0.15 and P(X = 2, Y = 2) = 0.44.
* P(X = 2) = P(X = 2, Y = 0) + P(X = 2, Y = 2) = 0.15 + 0.44 = 0.59.
**Probability Function for X:**
* P(X = 0) = 0.07
* P(X = 1) = 0.23
* P(X = 2) = 0.59
**2. Calculate the Expected Value (Mean) of X (E[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 Expected Value of X Squared (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
**4. Calculate the Variance of X (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
**Therefore, the variance of X is 0.6019.**
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