The ages of five randomly chosen cars in the parking lot are determined to be 7,9,3,4 and 6 years old. If we consider this sample in groups of 3, what is the probability of the true mean falling between 5.5 and 6.5?

To find the probability of the true mean falling between 5.5 and 6.5, we can use the concept of sampling distributions and the central limit theorem. Here's a step-by-step explanation of how to calculate the probability:

1. Identify the sample size and the population size:
- Sample size: 3 (since we are considering the ages of cars in groups of 3)
- Population size: 5 (since there are 5 randomly chosen cars)

2. Calculate the mean and standard deviation of the given sample:
- Mean: (7 + 9 + 3 + 4 + 6) / 5 = 5.8
- Standard deviation: Calculate the sample standard deviation using the formula:
SD = sqrt( (∑(x - mean)^2) / (n - 1) ), where x is each individual value and n is the sample size.
Plugging in the values, we get:
SD = sqrt((4.8^2 + 3.8^2 + 2.8^2 + 1.8^2 + 0.8^2) / 4) ≈ 1.29

3. Compute the standard error of the sample mean:
- The standard error is given by the formula: SE = SD / sqrt(n), where SD is the standard deviation and n is the sample size.
Plugging in the values, we get:
SE = 1.29 / sqrt(3) ≈ 0.744

4. Calculate the z-scores for the lower and upper bounds:
- The z-score can be calculated using the formula: z = (x - mean) / SE, where x is the value we want to find the probability for.
For the lower bound (5.5), the z-score is: z = (5.5 - 5.8) / 0.744 ≈ -0.40
For the upper bound (6.5), the z-score is: z = (6.5 - 5.8) / 0.744 ≈ 0.93

5. Look up the probability in the standard normal distribution table:
- Using the z-scores obtained, we can find the probabilities associated with them. The probability of the true mean falling between 5.5 and 6.5 is the difference between these probabilities.
P(5.5 < true mean < 6.5) = P(-0.40 < z < 0.93)
Look up the z-scores in the standard normal distribution table to find the corresponding probabilities.
Let's say P(Z < -0.40) = A and P(Z < 0.93) = B, then the required probability can be calculated as:
P(-0.40 < z < 0.93) = B - A

And that's how you can calculate the probability of the true mean falling between 5.5 and 6.5 using the given sample.

To determine the probability of the true mean falling between 5.5 and 6.5, we can use the concept of sampling distribution. Since we are considering the sample in groups of 3, we can find the mean of each group and create a sampling distribution of means.

Step 1: Find all possible combinations of groups of 3 cars from the given sample of 5 cars. In this case, we have:

Group 1: {7, 9, 3}
Group 2: {7, 9, 4}
Group 3: {7, 9, 6}
Group 4: {7, 3, 4}
Group 5: {7, 3, 6}
Group 6: {7, 4, 6}
Group 7: {9, 3, 4}
Group 8: {9, 3, 6}
Group 9: {9, 4, 6}
Group 10: {3, 4, 6}

Step 2: Calculate the mean of each group:

Group 1: (7 + 9 + 3) / 3 = 6.33
Group 2: (7 + 9 + 4) / 3 = 6.67
Group 3: (7 + 9 + 6) / 3 = 7.33
Group 4: (7 + 3 + 4) / 3 = 4.67
Group 5: (7 + 3 + 6) / 3 = 5.33
Group 6: (7 + 4 + 6) / 3 = 5.67
Group 7: (9 + 3 + 4) / 3 = 5.33
Group 8: (9 + 3 + 6) / 3 = 6.00
Group 9: (9 + 4 + 6) / 3 = 6.33
Group 10: (3 + 4 + 6) / 3 = 4.33

Step 3: Determine how many of these means fall between 5.5 and 6.5:

From the calculated means, we can see that the following groups fall between 5.5 and 6.5:
Group 2: 6.67
Group 5: 5.33
Group 6: 5.67
Group 8: 6.00
Group 9: 6.33

Therefore, there are 5 out of 10 possible groups whose means fall between 5.5 and 6.5.

Step 4: Calculate the probability:

The probability of the true mean falling between 5.5 and 6.5 is the ratio of the number of groups whose means fall between 5.5 and 6.5 to the total number of possible groups.

Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 5/10
Probability = 0.5

Therefore, the probability of the true mean falling between 5.5 and 6.5 is 0.5 or 50%.

I'm not sure what you mean by considering "in groups of 3."

Find the mean first = sum of scores/number of scores

Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

Standard deviation = square root of variance

Z = (score-mean)/SEm

SEm = SD/√n (Are you saying n = 3? If not, n = 5.)

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z scores.

I'll let you do the calculations.