, 24.01.2020dashavasilisk

# 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?

The mean would increase by 3

Step-by-step explanation:

The number cars and their probability is shown on the first uploaded image

The mean for the number of cars[X] owned can be mathematically represented as

From the question we are told that each household purchased additional three cars

Let Z be the random variable for the number of cars when the the additional car where added

So Mathematically

Z  = X + 3

The mean for the number of cars[X + 3] owned can be mathematically represented as

From the above equation we can see that the mean would increase by factor of  3

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:

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

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