Question

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

number of cars x	0	1	2	3	4	5	6

a housing company builds houses with two - car garages. what percent of households can be accommodated in the houses they built?

33%
22%
36%
78%

Explanation:

Step1: Identify relevant probabilities

We need to find the probability of households having 0, 1, or 2 cars since the houses have two - car garages. The probabilities for 0, 1, and 2 cars are 0.09, 0.33, and 0.36 respectively.

Step2: Calculate the total probability

We sum up the relevant probabilities: $P(X\leq2)=0.09 + 0.33+0.36$.
$0.09 + 0.33+0.36=0.78$.

Step3: Convert to percentage

To convert the decimal to a percentage, we multiply by 100. $0.78\times100 = 78\%$.

Answer:

78%