The average age of a vehicle registered in the United States is 8 years, or 96 months. Assume the standard deviation is 16 months. If a random sample of a group of 48 vehicles is selected find the probability that the mean of the group’s age is between 90 and 100 months.
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.
This is a distribution of means, while your next post is a distribution of raw scores.
To find the probability that the mean of the group's age is between 90 and 100 months, we need to use the Central Limit Theorem and the concept of a standard normal distribution.
According to the Central Limit Theorem, if we take random samples of a sufficiently large size from any population with any distribution, the distribution of the sample means will be approximately normally distributed.
In this case, we have a random sample of 48 vehicles. Since the sample size is reasonably large (above 30), we can assume that the sample mean will follow a normal distribution with the same mean but a standard deviation equal to the population standard deviation divided by the square root of the sample size.
First, let's calculate the standard error of the mean (SEM), which is equal to the population standard deviation divided by the square root of the sample size:
SEM = 16 / sqrt(48)
≈ 2.3094
Next, we need to convert the values 90 and 100 to z-scores. The z-score measures how many standard deviations an observation or sample mean is from the mean of a distribution. We use the z-score to find the probability in a standard normal distribution.
To convert 90 to a z-score:
z1 = (90 - 96) / 2.3094
≈ -2.605
To convert 100 to a z-score:
z2 = (100 - 96) / 2.3094
≈ 1.738
Now, we can find the probability that the mean of the group's age is between 90 and 100 months by using the standard normal distribution table or a calculator that provides the cumulative probability.
P(90 ≤ x ≤ 100) = P(z1 ≤ z ≤ z2)
Using the standard normal distribution table or a calculator, we can find the cumulative probabilities corresponding to the z-scores:
P(z ≤ -2.605) ≈ 0.0040
P(z ≤ 1.738) ≈ 0.9608
Therefore, the probability that the mean of the group's age is between 90 and 100 months is approximately:
P(90 ≤ x ≤ 100) = P(z1 ≤ z ≤ z2) = P(z ≤ 1.738) - P(z ≤ -2.605)
≈ 0.9608 - 0.0040
≈ 0.9568
So, the probability is approximately 0.9568 or 95.68%.