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

x012345  

p0.090.360.350.130.050.021

x*p00.360.70.390.20.11.75

x^2*p00.361.41.170.80.54.23

Mean1.75      

Var1.1675      

Std dev1.080509139      

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%

answer: i can’t tell you that

step-by-step explanation:

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)