The average age of vehicles registered in the united states is 96 months, with a standard deviation of 16 months. If a random sample of 36 vehicles is selected, find the probability that the mean of their ages is between 90 and 100 months. Provide four places past the decimal in your answer
To find the probability that the mean of the ages is between 90 and 100 months, we can use the Central Limit Theorem, which states that the distribution of sample means approaches a normal distribution as the sample size increases.
First, we need to calculate the standard error of the mean (SE), which is the standard deviation divided by the square root of the sample size. In this case, the standard deviation is 16 months and the sample size is 36:
SE = standard deviation / √sample size
SE = 16 / √36
SE = 16 / 6
SE = 2.6667 (rounded to four decimal places)
Next, we need to convert the two boundary values, 90 and 100 months, into z-scores. We use the formula:
z = (x - μ) / SE
where x is the boundary value, μ is the population mean, and SE is the standard error of the mean.
For the lower boundary (90 months):
z1 = (90 - 96) / 2.6667
z1 = -2.2499 (rounded to four decimal places)
For the upper boundary (100 months):
z2 = (100 - 96) / 2.6667
z2 = 1.4999 (rounded to four decimal places)
Now, we need to find the probabilities corresponding to these z-scores using a standard normal distribution table.
Using the table, we find that the probability corresponding to z = -2.2499 is 0.0121, and the probability corresponding to z = 1.4999 is 0.9332.
To find the probability that the mean of the ages is between 90 and 100 months, we subtract the probability of the lower boundary from the probability of the upper boundary:
P(90 ≤ x ≤ 100) = P(z1 ≤ z ≤ z2)
P(90 ≤ x ≤ 100) = P(z ≤ z2) - P(z ≤ z1)
P(90 ≤ x ≤ 100) = 0.9332 - 0.0121
P(90 ≤ x ≤ 100) = 0.9211 (rounded to four decimal places)
Therefore, the probability that the mean of the ages is between 90 and 100 months is approximately 0.9211.