Question

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.

Answer 1

To find the probability, we can use the Central Limit Theorem and the standard normal distribution.

Step-by-step explanation:

To find the probability that the mean of their age is between 90 and 99 months, we need to use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means approaches a normal distribution as the sample size increases.

The distribution can be approximated by a normal distribution with a mean equal to the population mean (96 months) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (9 / sqrt(36) = 1.5 months).

To find the probability, we can use the standard normal distribution and calculate the z-scores for the lower and upper boundaries of the interval:

z1 = (90 - 96) / 1.5 ≈ -4

z2 = (99 - 96) / 1.5 ≈ 2

Using a standard normal table or calculator, we can find the corresponding probabilities:

Prob(Z < -4) ≈ 0

Prob(Z < 2) ≈ 0.977

Therefore, the probability that the mean of their age is between 90 and 99 months is approximately 0.977 - 0 ≈ 0.977, or 97.7%.

Answer 2

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
`\bar X` = mean age of selected cars

The z score probability distribution for is given by;

Z =
`( \bar X - \mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = average age of a vehicle = 96 months

`\sigma` = 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 <
`\bar X` < 99) = P(
`\bar X` < 99) - P(
`\bar X`
`\leq` 90)

P(
`\bar X` < 99) = P(
`( \bar X - \mu)/((\sigma)/(√(n) ) )` <
`( 99-96)/((9)/(√(36) ) )` ) = P(Z < 2) = 0.97725

P(
`\bar X`
`\leq` 90) = P(
`( \bar X - \mu)/((\sigma)/(√(n) ) )`
`\leq`
`( 90-96)/((9)/(√(36) ) )` ) = P(Z
`\leq` -4) = 1 - P(Z < 4)

= 1 - 0.99997 = 0.00003

Therefore, P(90 <
`\bar X` < 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.