Let's break down your problem step by step.

a) Is this a valid discrete probability distribution?

To determine if a set of values represents a valid discrete probability distribution, we need to check two conditions:

1. All probabilities must be between 0 and 1, inclusive (i.e., $0 \leq P(x) \leq 1$ for all $x$).
2. The sum of all probabilities must equal 1.

Given the table:

$x$	$P(x)$
0	0.46
1	0.35
2	0.12
3	0.07

1. Probabilities between 0 and 1: All probabilities are within the range of 0 and 1 (0.46, 0.35, 0.12, 0.07).

2. Sum of probabilities: $0.46 + 0.35 + 0.12 + 0.07 = 1.00$

Since both conditions are met, this is a valid discrete probability distribution.

b) Find the mean number of cars owned.

The mean (or expected value) $E(x)$ of a discrete probability distribution is calculated as: $E(x) = \sum (x \cdot P(x))$ where $x$ represents the number of cars and $P(x)$ is the probability of owning $x$ cars.

Using the given data: $E(x) = (0 \cdot 0.46) + (1 \cdot 0.35) + (2 \cdot 0.12) + (3 \cdot 0.07)$ $E(x) = 0 + 0.35 + 0.24 + 0.21 = 0.8$

Thus, the mean number of cars owned is $0.8$.

c) Find the standard deviation of the number of cars owned.

The standard deviation $\sigma$ of a discrete probability distribution is found using the formula: $\sigma = \sqrt{\sum \left( (x - E(x))^2 \cdot P(x) \right)}$ Where $E(x)$ is the mean, and $x$ represents the number of cars owned.

We already know that $E(x) = 0.8$, so let's calculate the squared deviations and multiply by the probabilities:

• For $x = 0$: $(0 - 0.8)^2 \cdot 0.46 = (0.64) \cdot 0.46 = 0.2944$
• For $x = 1$: $(1 - 0.8)^2 \cdot 0.35 = (0.04) \cdot 0.35 = 0.014$
• For $x = 2$: $(2 - 0.8)^2 \cdot 0.12 = (1.44) \cdot 0.12 = 0.1728$
• For $x = 3$: $(3 - 0.8)^2 \cdot 0.07 = (4.84) \cdot 0.07 = 0.3388$

Now, sum these values: $\sum \left( (x - E(x))^2 \cdot P(x) \right) = 0.2944 + 0.014 + 0.1728 + 0.3388 = 0.820$

Finally, take the square root to find the standard deviation: $\sigma = \sqrt{0.820} \approx 0.905$

Thus, the standard deviation of the number of cars owned is approximately 0.905.

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Summary:

• a) Yes, it is a valid discrete probability distribution.
• b) The mean number of cars owned is $\mathbf{0.8}$.
• c) The standard deviation of the number of cars owned is approximately $\mathbf{0.905}$.

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Would you like more details on any part of this? Here are a few follow-up questions that might interest you:

1. How do you interpret the meaning of the standard deviation in this context?
2. What would happen if the sum of probabilities didn't equal 1?
3. Could this distribution represent something else besides cars (e.g., number of pets)?
4. How would the mean change if the probabilities were adjusted?
5. Can we calculate the variance from this distribution, and how?

Tip: To calculate the mean, always use the formula $E(x) = \sum (x \cdot P(x))$.