for a sample of n = 75, find the probaBILITY OF A SAMPLE MEAN BEING BEIGN GREATER THAN 221 IF U = 220 AND O = 3.9

First, please do not use all capitals. Online it is like SHOUTING. Not only is it rude, but it is harder to understand. Thank you.

I assume "U" = mean and "O" = standard deviation.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since only one SD is provided, you can use just that to determine SEdiff.

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

To find the probability of a sample mean being greater than a certain value, we can use the Central Limit Theorem and the standard normal distribution.

First, let's calculate the standard error (SE), which represents the standard deviation of the sample mean. The formula for the standard error is given by SE = σ / √n, where σ is the population standard deviation and n is the sample size. In this case, σ = 3.9 and n = 75, so the standard error is SE = 3.9 / √75.

Next, we need to find the z-score corresponding to the given sample mean of 221. The z-score measures the number of standard errors the sample mean is away from the population mean. The formula to calculate the z-score is z = (x - μ) / SE, where x is the sample mean, μ is the population mean, and SE is the standard error. In this case, x = 221, μ = 220, and SE is calculated above.

Using these values, we can now calculate the z-score. z = (221 - 220) / (3.9 / √75)

Next, we can use a standard normal distribution table or a calculator to find the probability corresponding to this z-score. The probability will represent the area under the curve to the right of the z-score.

Finally, we can interpret the result as the probability of obtaining a sample mean greater than 221, assuming a population mean of 220 and a population standard deviation of 3.9.

Please note that this calculation assumes that the distribution of the sample means is approximately normal due to the Central Limit Theorem.