Questions: Homework - Homework #17 - Discrete Probability Distributions Question 11, 6.4.28 HW Score: 69.58%, Part 2 of 3 Points: 0 of 1 Use the probability distribution to find probabilities in parts (a) through (c). The probability distribution of number of dogs per household in a small town Dogs 0 1 2 3 4 5 Households 0.696 0.191 0.072 0.025 0.013 0.003 (a) Find the probability of randomly selecting a household that has fewer than two dogs. .887 (Round to three decimal places as needed.) (b) Find the probability of randomly selecting a household that has at least one dog. (Round to three decimal places as needed.)

Solution Steps

To find the probability of randomly selecting a household that has fewer than two dogs, we sum the probabilities of having 0 dogs and 1 dog:

\[ P(X < 2) = P(0) + P(1) = 0.696 + 0.191 = 0.887 \]

To find the probability of randomly selecting a household that has at least one dog, we subtract the probability of having 0 dogs from 1:

\[ P(X \geq 1) = 1 - P(0) = 1 - 0.696 = 0.304 \]

We calculate the mean, variance, and standard deviation of the distribution.

Mean (\(\mu\)): \[ \mu = E(X) = \sum_{i=0}^{5} x_i \cdot P(x_i) = 0 \times 0.696 + 1 \times 0.191 + 2 \times 0.072 + 3 \times 0.025 + 4 \times 0.013 + 5 \times 0.003 = 0.477 \]

Variance (\(\sigma^2\)): \[ \sigma^2 = E(X^2) - \mu^2 \] Calculating \(E(X^2)\): \[ E(X^2) = 0^2 \times 0.696 + 1^2 \times 0.191 + 2^2 \times 0.072 + 3^2 \times 0.025 + 4^2 \times 0.013 + 5^2 \times 0.003 = 0.759 + 0.477^2 \] Thus, \[ \sigma^2 = 0.759 \]

Standard Deviation (\(\sigma\)): \[ \sigma = \sqrt{\sigma^2} = \sqrt{0.759} \approx 0.871 \]

Final Answer