Question

Choose an American household at random, and let the random variable X be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 6 cars:
Number of cars X Probability 0 1 2 3 4 5 6

0.07 0.31 0.43 0.12 0.04 0.02 0.01

A housing company builds houses with two-car garages. What percent of households have more cars than the garage can hold?

Answer 1

Answer:

19% of households have more cars than the garage can hold

Step-by-step explanation:

We are given the following distribution for the number of cars owned by a family.

Number of cars X:        0        1           2        3           4          5          6

Probability:                0.07    0.31     0.43    0.12     0.04     0.02    0.01

We have to find the percentage of households have more cars than the garage can hold.

A garage can hold two cars. Thus, the household with more than two cars are the households that  have more cars than the garage can hold.

The given distribution is a discrete probability distribution.

Thus, we evaluate:

$P(x\geq 3) = P(x+3) + P(x+4) + P(x+5) + P(x+6)\\P(x\geq 3) = 0.12 +0.04+ 0.02 +0.01 = 0.19\\P(x\geq 3) = 19\%$

Thus, 19% of households have more cars than the garage can hold.