The average age of a vehicle registered in the United States is 8 years, or 96 months. Assume the standard deviation is 9 months. If a random sample of 36 card is selected, find the probability that the mean of their age is between 90 and 99 months.

Probability that the mean of their age is between 90 and 99 months is 0.97722 or 0.98.

Step-by-step explanation:

We are given that the average age of a vehicle registered in the United States is 8 years, or 96 months. Assume the standard deviation is 9 months.

Also, a random sample of 36 card is selected.

Firstly, Let = mean age of selected cars

The z score probability distribution for is given by;

         Z = ~ N(0,1)

where, = average age of a vehicle = 96 months

             = standard deviation = 9 months

            n = sample of cars = 36

Probability that the mean of their age is between 90 and 99 months is given by = P(90 < < 99) = P( < 99) - P( 90)

        P( < 99) = P( < ) = P(Z < 2) = 0.97725

        P( 90) = P( ) = P(Z -4) = 1 - P(Z < 4)

                                                       = 1 - 0.99997 = 0.00003

Therefore, P(90 < < 99) = 0.97725 - 0.00003 = 0.97722 ≈ 0.98

Hence, probability that the mean of their age is between 90 and 99 months is 0.98.

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1

answer is  

d.

3.09

hope you got this right.