To solve this problem, let's go through each part.

Given Data:

The table shows the number of dogs per household in a small town along with their corresponding probabilities:

(a) Finding the Mean (μ) of the Probability Distribution

The mean (expected value) of a probability distribution can be calculated using the formula: $\mu = E(X) = \sum X \cdot P(X)$

Let's calculate each term:

1. $0 \cdot 0.691 = 0$
2. $1 \cdot 0.195 = 0.195$
3. $2 \cdot 0.077 = 0.154$
4. $3 \cdot 0.019 = 0.057$
5. $4 \cdot 0.012 = 0.048$
6. $5 \cdot 0.007 = 0.035$

Now, summing these results: $\mu = 0 + 0.195 + 0.154 + 0.057 + 0.048 + 0.035 = 0.489$

So, the mean (μ) is approximately 0.5 when rounded to one decimal place.

(b) Finding the Variance (σ²) and Standard Deviation (σ) of the Probability Distribution

The variance of a probability distribution is calculated as: $\sigma^2 = E(X^2) - \mu^2$ where $E(X^2) = \sum X^2 \cdot P(X)$.

Let's calculate $E(X^2)$:

1. $0^2 \cdot 0.691 = 0$
2. $1^2 \cdot 0.195 = 0.195$
3. $2^2 \cdot 0.077 = 0.308$
4. $3^2 \cdot 0.019 = 0.171$
5. $4^2 \cdot 0.012 = 0.192$
6. $5^2 \cdot 0.007 = 0.175$

Summing these: $E(X^2) = 0 + 0.195 + 0.308 + 0.171 + 0.192 + 0.175 = 1.041$

Now, we can find the variance: $\sigma^2 = E(X^2) - \mu^2 = 1.041 - (0.5)^2 = 1.041 - 0.25 = 0.791$

The standard deviation (σ) is the square root of the variance: $\sigma = \sqrt{0.791} \approx 0.89$

Summary of Answers:

• Mean (μ) ≈ 0.5
• Variance (σ²) ≈ 0.79
• Standard Deviation (σ) ≈ 0.89

Would you like further clarification on any steps?

Here are five additional questions to help deepen your understanding:

1. What would happen to the mean if a higher probability was assigned to larger numbers of dogs?
2. How would you calculate the expected number of households with exactly two dogs?
3. How does variance help in understanding the spread of dog ownership in households?
4. What is the probability of having at least two dogs in a household?
5. How would the standard deviation change if there were more households with 0 dogs?

Tip: Remember that the mean in a probability distribution is the "center" of the distribution, while the variance and standard deviation tell us how spread out the values are around this center.