Questions: Question 11, 4.1.21 Part 2 of 4 Use the probability distribution to complete parts (a) through (d) below. The probability distribution of number of televisions per household in a small town x 0 1 2 3 P(x) 0.03 0.14 0.31 0.52 (a) Find the probability of randomly selecting a household that has one or two televisions The probability is 0.45 (Type an integer or a decimal. Do not round.) (b) Find the probability of randomly selecting a household that has two or more televisions. The probability is (Type an integer or a decimal. Do not round.)

Solution Steps

To find the probability of randomly selecting a household that has one or two televisions, we sum the probabilities of having one television and two televisions:

\[ P(1 \text{ or } 2) = P(1) + P(2) = 0.14 + 0.31 = 0.45 \]

Next, we calculate the probability of randomly selecting a household that has two or more televisions by summing the probabilities of having two televisions and three televisions:

\[ P(2 \text{ or more}) = P(2) + P(3) = 0.31 + 0.52 = 0.83 \]

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

Mean (\(\mu\)): \[ \mu = 0 \times 0.03 + 1 \times 0.14 + 2 \times 0.31 + 3 \times 0.52 = 2.32 \]

Variance (\(\sigma^2\)): \[ \sigma^2 = (0 - 2.32)^2 \times 0.03 + (1 - 2.32)^2 \times 0.14 + (2 - 2.32)^2 \times 0.31 + (3 - 2.32)^2 \times 0.52 = 0.678 \]

Standard Deviation (\(\sigma\)): \[ \sigma = \sqrt{\sigma^2} = \sqrt{0.678} \approx 0.823 \]

Final Answer