Question

Owner-Occupied Housing Units In the 2010 US Census, we learn that \(65 \%\) of all housing units are owner-occupied while the rest are rented. If we take a random sample of 20 housing units, find the probability that: (a) Exactly 15 of them are owner-occupied (b) 18 or more of them are owner-occupied

Short answer

The probabilities for the mentioned scenarios can be calculated using the formulas mentioned in steps 2 and 3. They will give us specific numerical answers.

Step-by-step solution

Understand the binomial distribution formula

The binomial distribution probability formula is given as follows: \(P(X = k) = C(n, k) (p)^k (1-p)^{n-k}\) where: \n - \(P(X = k)\) is the probability of \(k\) successes in \(n\) trials \n - \(C(n, k)\) is the number of combinations of \(n\) items taken \(k\) at a time \n - \(p\) is the probability of success, which is \(0.65\) in this problem \n - \(1-p\) is the probability of failure \n - \(n\) is the number of trials, \(20\) in this case \n - \(k\) is the number of successes

Solve for part (a)

We need to find the probability that exactly 15 housing units are owner-occupied. Using the binomial distribution formula, where \(n = 20\), \(k = 15\) and \(p = 0.65\). \n \(P(X = 15) = C(20, 15) * 0.65^{15} * (1-0.65)^{20-15}\)

Solve for part (b)

We need to find the probability that 18 or more housing units are owner-occupied. This can be solved as 1 minus the probability of fewer than 18 successes. So, \n \(P(X \geq 18)= 1 - P(X < 18) \n = 1 - (P(X = 0) + P(X = 1) + ….. + P(X=17)) \n Each of these probabilities on right can be calculated using the binomial distribution formula.