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
probability 0.09 0.33 0.36 0.14 0.05 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?
78%
42%
22%
14%
question 2
1 pts
ch 12 a randomly selected sample of 100 horse owners found that 72 of them feed two flakes of grass hay in the morning and one flake of alfalfa plus one flake of grass hay in the evening to their horses. the estimated probability that horse owners feed grass hay in the morning and alfalfa plus grass hay in the afternoon is
0.75.
0.72.
0.25.
0.50.

Explanation:

Step1: Identify relevant probabilities

We want to find the probability that a household has 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: $P(X > 2)=0.14 + 0.05+0.02 + 0.01$.
$P(X > 2)=0.22$ or 22%.

for Question 2:

Step1: Recall probability formula

The probability of an event is given by the number of favorable outcomes divided by the total number of outcomes. Here, the total number of horse - owners is $n = 100$, and the number of horse - owners who feed grass hay in the morning and alfalfa plus grass hay in the afternoon is $m = 72$.

Step2: Calculate the probability

The probability $P=\frac{m}{n}=\frac{72}{100}=0.72$.

Answer:

The percentage of households that have more cars than the two - car garage can hold is 22%.