Help me! Population of heights of college students is approximately normally distributed with a mean of 64.37 inches and standard deviation of 6.26 inches A random sample of 74 heights is obtained. Find the mean and standard error of the x bar distribution 2. Find P(xbar > 65.25)

Z = (mean1 - mean2)/SEm

Z = (65.25-64.37)/SEm

SEm = SD/√n

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

To find the mean and standard error of the x-bar distribution, you need to use the following formulas:

1. Mean of the x-bar distribution (μ_x̄):
The mean of the x-bar distribution is equal to the mean of the population (μ). So, in this case, the mean of the x-bar distribution would be 64.37 inches.

2. Standard error of the x-bar distribution (SE_x̄):
The standard error of the x-bar distribution is equal to the standard deviation of the population (σ) divided by the square root of the sample size (n). So, in this case, the standard error of the x-bar distribution would be calculated as follows:
SE_x̄ = σ / √n
SE_x̄ = 6.26 / √74
SE_x̄ ≈ 0.730 inches (rounded to three decimal places)

Now, to find P(xbar > 65.25), you need to use the properties of a normal distribution.

1. Standardizing:
P(xbar > 65.25) can be transformed into P(Z > (65.25 - μ) / SE_x̄), where Z is the standard normal random variable.

2. Calculating the Z-score:
Z = (65.25 - μ) / SE_x̄
Z = (65.25 - 64.37) / 0.730
Z ≈ 1.21 (rounded to two decimal places)

3. Finding the probability:
P(Z > 1.21) can be found using a Z-table or a statistical software. The area to the right of 1.21 under the standard normal curve represents the probability that x-bar is greater than 65.25.

Please note that I cannot provide the actual probability without using a Z-table or a statistical software, as it requires the cumulative distribution function for the standard normal distribution.