The table you've provided gives the probability distribution for the number of cars owned by a college student.

To address the questions:

a) Is this a valid discrete probability distribution?

For a distribution to be valid:

1. All probabilities must be between 0 and 1.
2. The sum of all probabilities must equal 1.

Let's check:

• $P(0) = 0.46$
• $P(1) = 0.33$
• $P(2) = 0.12$
• $P(3) = 0.09$

Summing these probabilities: $0.46 + 0.33 + 0.12 + 0.09 = 1$ Since the sum equals 1 and all probabilities are between 0 and 1, yes, this is a valid discrete probability distribution.

b) Find the mean number of cars owned.

The mean (expected value) of a discrete random variable is calculated as: $E(x) = \sum [x \cdot P(x)]$ Let's calculate this step by step: $E(x) = (0 \cdot 0.46) + (1 \cdot 0.33) + (2 \cdot 0.12) + (3 \cdot 0.09)$ $E(x) = 0 + 0.33 + 0.24 + 0.27$ $E(x) = 0.84$ So, the mean number of cars owned is 0.84.

Would you like more details on any of these calculations, or have any questions?

Here are some related questions you could explore:

1. What is the variance of the number of cars owned?
2. What is the standard deviation of the number of cars owned?
3. How would the mean change if the probabilities were different?
4. What is the probability that a student owns at least 2 cars?
5. How can the probability distribution be used to model other student-owned assets?

Tip: The expected value (mean) gives you the average outcome over many trials, not necessarily what will happen in a single trial.