Question

American households increasingly rely on cell phones as their exclusive telephone service. It is reported that 49.6% of American households still have landline phone service. We decide to randomly call eight households and ask if the home has a landline phone.

a-1. What is the random variable? Answer is complete and correct. The random variable is the count of the 8 households that have a landline.
a-2. How is the random variable distributed? Answer is complete and correct. The random variable follows a binomial probability distribution.
b. What is the probability that none of the households in the sampled group have landline phone service? (Round your answer to 4 decimal places.) Probability
c. What is the probability that exactly five of the households in the sampled group have a landline phone service? (Round your answer to 4 decimal places.) Probability
d. Given the probability distribution, what is the mean number of households with landline service? (Round your answer to 3 decimal places.) Mean
e. What is the variance of the probability distribution of the number of households with landline service? (Round your answer to 3 decimal places.) Variance

Answer 1

The random variable is the count of households with a landline phone out of the randomly selected eight households. The probability of none of the households having a landline phone is approximately 0.017. The mean number of households with landline service is approximately 3.968 and the variance is approximately 1.976.

Step-by-step explanation:

The random variable, in this case, is the count of households that have a landline phone out of the randomly selected eight households. This random variable follows a binomial probability distribution since each household has a fixed probability of having a landline phone.

For part c, the probability that none of the households in the sampled group have a landline phone service can be calculated using the binomial probability formula: P(X = 0) = (n choose x) * p^x * (1-p)^(n-x), where n is the number of trials, x is the number of successful outcomes, and p is the probability of success. In this case, n = 8, x = 0, and p = 0.496. Plugging in these values, we get P(X = 0) = (8 choose 0) * 0.496^0 * (1-0.496)^(8-0) ≈ 0.017.

For part d, the mean number of households with landline service can be calculated using the formula: mean = n * p, where n is the number of trials and p is the probability of success. In this case, n = 8 and p = 0.496. Plugging in these values, we get mean = 8 * 0.496 ≈ 3.968.

For part e, the variance of the probability distribution of the number of households with landline service can be calculated using the formula: variance = n * p * (1-p), where n is the number of trials and p is the probability of success. In this case, n = 8 and p = 0.496. Plugging in these values, we get variance = 8 * 0.496 * (1-0.496) ≈ 1.976.