ch 12* choose an american household at random, and let the random variable x be the number of cars…

ch 12* choose an american household at random, and let the random variable x be the number of cars (including suvs and light trucks) the residents own. here is the probability model if we ignore the few households that own more than six cars\n\nnumber of cars x 0 1 2 3 4 5 6\nprobability 0.09 0.33 0.36 0.14 0.05 0.02 0.01\n\na housing company builds houses with two - car garages. what percent of households have more cars than the garage can hold?\n\n22%\n78%\n14%\n42%

ch 12* choose an american household at random, and let the random variable x be the number of cars (including suvs and light trucks) the residents own. here is the probability model if we ignore the few households that own more than six cars\n\nnumber of cars x 0 1 2 3 4 5 6\nprobability 0.09 0.33 0.36 0.14 0.05 0.02 0.01\n\na housing company builds houses with two - car garages. what percent of households have more cars than the garage can hold?\n\n22%\n78%\n14%\n42%

Answer

Explanation:

Step1: Identify relevant probabilities

We want to find the probability of having more than 2 cars. The probabilities for 3, 4, 5 and 6 cars are 0.14, 0.05, 0.02 and 0.01 respectively.

Step2: Sum the probabilities

We sum the probabilities of having 3, 4, 5 and 6 cars: $P(X > 2)=0.14 + 0.05+0.02 + 0.01$. $P(X > 2)=0.22$.

Step3: Convert to percentage

To convert the probability to a percentage, we multiply by 100. So $0.22\times100 = 22%$.

Answer:

22%