Tests on a Chevy Cobalt incdicate a population mean of 32 mpg and a population standard deviation of 3.5 mpg. You take an SRS (sample random sample) of 10 cars to get a sample mean. What is the probability that your sample mean is greater than 34 mpg?

I am not sure how to go about this problem. Can someone help?

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 score.

What is SEm??

Nevermind, thanks for the help!

To solve this problem, we will use the Central Limit Theorem (CLT) since the sample size is relatively large (n = 10). The Central Limit Theorem states that if the sample size is sufficiently large, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.

The mean of the sampling distribution of the sample mean (μx̄) will be the same as the population mean (μ) and the standard deviation of the sampling distribution of the sample mean (σx̄) will be the population standard deviation (σ) divided by the square root of the sample size (n).

Here's how we can calculate the probability that the sample mean is greater than 34 mpg:

1. Find the standard deviation of the sampling distribution of the sample mean (σx̄):

σx̄ = σ / √n = 3.5 / √10 = 1.11 mpg (approximated to two decimal places)

2. Calculate the z-score using the formula:

z = (x̄ - μ) / σx̄

where x̄ is the sample mean (34 mpg), μ is the population mean (32 mpg), and σx̄ is the standard deviation of the sampling distribution of the sample mean (1.11 mpg):

z = (34 - 32) / 1.11 = 1.80

3. Now, we need to find the probability that the z-score is greater than 1.80. We can use a standard normal distribution table or a calculator to find this probability.

Using a standard normal distribution table, we find that the probability corresponding to a z-score of 1.80 (P(Z > 1.80)) is approximately 0.0359.

Therefore, the probability that the sample mean is greater than 34 mpg is approximately 0.0359 or 3.59%.

Note that this probability represents how likely we are to obtain a sample mean greater than 34 mpg if we repeatedly sample 10 cars under the given conditions.