Given a normal distribution with $\mu=50$ and $\sigma=4$, and given you select a sample of $n=100$, complete parts (a) through (d). a. What is the probability that $\bar{X}$ is less than 49 ? \[ P(\bar{X}<49)= \] (Type an integer or decimal rounded to four decimal places as needed.)

Step 1 ：We are given a normal distribution with a mean (\(\mu\)) of 50 and a standard deviation (\(\sigma\)) of 4. We are also given a sample size (\(n\)) of 100 and asked to find the probability that the sample mean (\(\bar{X}\)) is less than 49.

Step 2 ：To solve this, we need to standardize the sample mean and then use the standard normal distribution (Z-distribution) to find the probability. The formula to standardize a sample mean is: \[Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}\] 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 ：Substituting the given values into the formula, we get: \[Z = \frac{49 - 50}{4 / \sqrt{100}} = -2.5\]

Step 4 ：After calculating the Z-score, we can use a Z-table or a statistical function to find the probability. The probability corresponding to a Z-score of -2.5 is approximately 0.0062.

Step 5 ：Final Answer: The probability that \(\bar{X}\) is less than 49 is approximately \(\boxed{0.0062}\).