The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual.
For a sample of $n=67$, find the probability of a sample mean being less than 22.3 if $\mu=22$ and $\sigma=1.28$.
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For a sample of $n=67$, the probability of a sample mean being less than 22.3 if $\mu=22$ and $\sigma=1.28$ is (Round to four decimal places as needed)

Step 1 ：We are given the population mean (\(\mu\)), standard deviation (\(\sigma\)), and the sample size (\(n\)). We are asked to find the probability of the sample mean being less than a certain value. This is a problem of finding the probability under a normal distribution.

Step 2 ：We can use the formula for the z-score to standardize our variable. The z-score is given by: \[Z = \frac{X - \mu}{\sigma / \sqrt{n}}\] where \(X\) is the value for which we want to find the probability, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size.

Step 3 ：Given that \(\mu = 22\), \(\sigma = 1.28\), \(n = 67\), and \(X = 22.3\), we can substitute these values into the z-score formula to get \(z = 1.91844205590761\).

Step 4 ：After finding the z-score, we can use the standard normal distribution table to find the probability. The probability corresponding to the z-score is approximately 0.9724725086361877.

Step 5 ：The probability of a sample mean being less than 22.3 is approximately 0.9725. This means that about 97.25% of the time, we would expect a sample mean of 67 observations to be less than 22.3, given that the population mean is 22 and the standard deviation is 1.28.

Step 6 ：Final Answer: The probability of a sample mean being less than 22.3 is approximately \(\boxed{0.9725}\).