Choose an American household at random and let the random variable X be the number of cars (including SUVs and light trucks) they own. Given is the probability distribution if we ignore the few households that own more than 5 cars. Number of cars 0 1 2 3 4 5 Probability 0.09 0.36 0.35 0.13 0.05 0.02 About what percentage of households have a number of cars within 2 standard deviations of the mean?

66.67%

Step-by-step explanation:

given that an American household at random and let the random variable X be the number of cars (including SUVs and light trucks) they own. Given is the probability distribution if we ignore the few households that own more than 5 cars.

Number of cars 0 1 2 3 4 5

Prob 0.09 0.36 0.35 0.13 0.05 0.02

x 0 1 2 3 4 5

p 0.09 0.36 0.35 0.13 0.05 0.02 1

x*p 0 0.36 0.7 0.39 0.2 0.1 1.75

x^2*p 0 0.36 1.4 1.17 0.8 0.5 4.23

Mean 1.75

Var 1.1675

Std dev 1.080509139

We have within 2 std deviation form the mean the interval

(1.75-2 std dev, 1.75 + 2 std dev)

= (-0.411, 3.911)

We observe that 4 and 5 lie outside the interval

Percentage of households have a number of cars within 2 standard deviations of the mean = 4/6 = 66.67%