Final answer:
The probability that 2 or fewer of these 25 households own a dog as a pet is 0.0195, rounded to four decimal places.
Step-by-step explanation:
According to the problem statement, we are given that 37.7% of households in the United States own a dog as a pet. Suppose that a company that sells dog food would like to establish a focus group to gather input on a new dog food marketing campaign.
The company plans to contact 25 randomly selected households to invite people to join the focus group.
To compute the probability that 2 or fewer of these 25 households own a dog as a pet, we can use the binomial distribution formula. The probability of getting exactly x successes in n trials is given by:
P(x) = (n choose x) * p^x * (1 - p)^(n - x)
where n is the number of trials, p is the probability of success, and x is the number of successes we want to find. In this case, we want to find the probability of getting 0, 1, or 2 households that own a dog as a pet.
Since we know that 37.7% of households in the United States own a dog as a pet, we can set p = 0.377. We also know that we are selecting 25 households, so n = 25.
Using the binomial distribution formula, we can compute the probability of getting 0, 1, or 2 households that own a dog as a pet:
P(0) + P(1) + P(2) = (25 choose 0) * 0.377^0 * (1 - 0.377)^(25 - 0) + (25 choose 1) * 0.377^1 * (1 - 0.377)^(25 - 1) + (25 choose 2) * 0.377^2 * (1 - 0.377)^(25 - 2)
P(0) + P(1) + P(2) = 0.0001 + 0.0021 + 0.0173
P(0) + P(1) + P(2) = 0.0195
Therefore, the probability that 2 or fewer of these 25 households own a dog as a pet is 0.0195, rounded to four decimal places.