Exercises 27 to 29 refer to the following setting: Choose an American household at random and let the random variable X be the number of cars (including SUVs and light trucks) they own; Here is the probability model if we ignore the few households that Own more than 5 cars: Number of cars X: Probability: 0.09 0.36 0.35 0.13 0.05 0.02
The Correct Answer and Explanation is:
Let’s start by organizing the given data into a probability distribution for the number of cars XXX owned by a randomly selected American household:
| Number of Cars (X) | Probability |
|---|---|
| 0 | 0.09 |
| 1 | 0.36 |
| 2 | 0.35 |
| 3 | 0.13 |
| 4 | 0.05 |
| 5 | 0.02 |
Exercise 27: Verify this is a valid probability distribution.
To verify a probability distribution, two conditions must be satisfied:
- Each probability must be between 0 and 1.
- The sum of all probabilities must equal 1.
Checking each value:
- All listed probabilities are between 0 and 1.
- Sum = 0.09 + 0.36 + 0.35 + 0.13 + 0.05 + 0.02 = 1.00
Answer: Yes, this is a valid probability distribution.
Exercise 28: What is the probability that a randomly chosen household has at least 2 cars?
We sum the probabilities for X=2,3,4,5X = 2, 3, 4, 5X=2,3,4,5:
P(X ≥ 2) = 0.35 + 0.13 + 0.05 + 0.02 = 0.55
Answer: The probability is 0.55.
Exercise 29: What is the expected number of cars a household owns?
The expected value E(X)E(X)E(X) is calculated as:E(X)=∑X⋅P(X)=(0)(0.09)+(1)(0.36)+(2)(0.35)+(3)(0.13)+(4)(0.05)+(5)(0.02)E(X) = \sum X \cdot P(X) = (0)(0.09) + (1)(0.36) + (2)(0.35) + (3)(0.13) + (4)(0.05) + (5)(0.02)E(X)=∑X⋅P(X)=(0)(0.09)+(1)(0.36)+(2)(0.35)+(3)(0.13)+(4)(0.05)+(5)(0.02)E(X)=0+0.36+0.70+0.39+0.20+0.10=∗∗1.75∗∗E(X) = 0 + 0.36 + 0.70 + 0.39 + 0.20 + 0.10 = **1.75**E(X)=0+0.36+0.70+0.39+0.20+0.10=∗∗1.75∗∗
Answer: The expected number of cars is 1.75.
Explanation
Probability distributions describe how likely each outcome of a random variable is. In this setting, the random variable XXX represents the number of cars an American household owns. The distribution includes all households with zero through five cars. First, we confirm it is a valid distribution. This means all the probabilities must fall between zero and one and add up to one. In this case, they do, confirming the model is mathematically valid.
Next, we determine the probability that a household has at least two cars. This includes all probabilities for households with two or more cars. By summing these values, we find that more than half of households—specifically 55 percent—own at least two vehicles. This result aligns with trends in suburban and rural areas where multiple cars may be necessary due to limited public transportation.
Lastly, we compute the expected value, or the average number of cars per household. This value is found by multiplying each possible number of cars by its associated probability and summing the products. The result is 1.75 cars. While no household can own exactly 1.75 cars, this average represents the central tendency of car ownership. It implies that car ownership is fairly common and that many families own one or two vehicles. These insights help policymakers, marketers, and urban planners understand transportation needs, infrastructure planning, and market demand for vehicles across the population.
