The mean height of women in a country (ages $20-29$ ) is 63.8 inches. A random sample of 55 women in this age group is selected. What is the probability that the mean height for the sample is greater than 65 inches? Assume $\sigma=2.63$. The probability that the mean height for the sample is greater than 65 inches is (Round to four decimal places as needed.)

Step 1 ：We are given a problem of normal distribution. The population mean (\(\mu\)) is 63.8 inches, the population standard deviation (\(\sigma\)) is 2.63 inches, and the sample size (n) is 55. We are asked to find the probability that the sample mean (\(\bar{x}\)) is greater than 65 inches.

Step 2 ：We can use the formula for the z-score to solve this problem. The formula is \[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\] where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and n is the sample size.

Step 3 ：Substitute the given values into the formula: \(\bar{x}\) = 65 inches, \(\mu\) = 63.8 inches, \(\sigma\) = 2.63 inches, and n = 55. The calculated z-score is approximately 3.3838.

Step 4 ：We can use the z-score to find the probability that the sample mean is greater than 65 inches. This is equivalent to finding the area under the standard normal curve to the right of the calculated z-score. The calculated probability is approximately 0.0004.

Step 5 ：Final Answer: The probability that the mean height for the sample is greater than 65 inches is \(\boxed{0.0004}\).