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.
The answer I got was .9536803695 is the correct?
SEm = 16/√48 = 16/6.9282 = 2.3
Z = (score-mean)/SEm = (100-96)/2.3 = 1.7
Z = (score-mean)/SEm = (90-96)/2.3 = -2.6
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.
.4554 + .4953 = .9507
Yep, .95 is close enough
Thank you so much
To find the probability that the mean of the group's age is between 90 and 100 months, we need to calculate the z-scores and then find the corresponding cumulative probability using a standard normal distribution table or a calculator.
The formula to calculate the z-score is:
z = (x - μ) / (σ / sqrt(n))
where:
x = mean age of the group
μ = average age of vehicles registered in the United States (96 months)
σ = standard deviation (16 months)
n = sample size (48 vehicles)
For the lower bound of 90 months:
z1 = (90 - 96) / (16 / sqrt(48))
= -6 / (16 / 6.9282)
= -2.1961
For the upper bound of 100 months:
z2 = (100 - 96) / (16 / sqrt(48))
= 4 / (16 / 6.9282)
= 2.1961
Now, we can calculate the cumulative probability:
P(90 ≤ x ≤ 100) = P(z1 ≤ Z ≤ z2)
Using a standard normal distribution table or calculator, we can find the probabilities corresponding to z-scores -2.1961 and 2.1961. Subtracting the cumulative probability of the lower bound from the cumulative probability of the upper bound will give us the result.
P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) - P(Z ≤ z1)
P(z1 ≤ Z ≤ z2) = 0.9851 - 0.0149
P(z1 ≤ Z ≤ z2) = 0.9702
Therefore, the probability that the mean of the group's age is between 90 and 100 months is approximately 0.9702. So, your answer of approximately 0.9537 is not correct.
To find the probability that the mean of the group's age is between 90 and 100 months, we need to calculate the z-scores for these two values and then use those z-scores to find the corresponding probabilities from a standard normal distribution table.
First, let's calculate the z-scores:
Z90 = (90 - 96) / (16 / √48) ≈ -1.71
Z100 = (100 - 96) / (16 / √48) ≈ 1.71
Next, we need to find the corresponding probabilities from a standard normal distribution table. These probabilities represent the area under the curve of the standard normal distribution between -1.71 and 1.71.
Using the standard normal distribution table, which provides probabilities between 0 and the given z-score, we find:
P(Z ≤ -1.71) ≈ 0.0436
P(Z ≤ 1.71) ≈ 0.9564
To find the probability between -1.71 and 1.71, we subtract the probability of Z ≤ -1.71 from the probability of Z ≤ 1.71:
P(-1.71 ≤ Z ≤ 1.71) = P(Z ≤ 1.71) - P(Z ≤ -1.71)
≈ 0.9564 - 0.0436
= 0.9128
Therefore, the probability that the mean of the group's age is between 90 and 100 months is approximately 0.9128, or 91.28%.
So, the answer you got (0.9536803695) is not correct. The correct answer is approximately 0.9128.