11. The mean triglyceride level for U.S. adults is 97 milligrams per deciliter. Assume the triglyceride level for U.S. adults is normally distributed, with standard deviation of 25 milligrams per deciliter. You are randomly select a U.S. adult. What is the probability that the person's triglyceride level is less than 77 ? (4 points) $P(x<77)=$

Step 1 ：Given that the mean (\(\mu\)) triglyceride level is 97 milligrams per deciliter, the standard deviation (\(\sigma\)) is 25 milligrams per deciliter, and we are interested in the value (X) of 77 milligrams per deciliter.

Step 2 ：We can use the Z-score formula to find how many standard deviations the value we are interested in is from the mean. The Z-score formula is \(Z = \frac{X - \mu}{\sigma}\).

Step 3 ：Substitute the given values into the Z-score formula: \(Z = \frac{77 - 97}{25} = -\frac{20}{25} = -0.8\).

Step 4 ：The Z-score of -0.8 means that the value we are interested in is 0.8 standard deviations below the mean.

Step 5 ：To find the probability that a randomly selected U.S. adult has a triglyceride level less than 77 milligrams per deciliter, we need to find the probability that the Z-score is less than -0.8.

Step 6 ：We can look up the value for Z = -0.8 in a standard normal distribution table, or use a calculator or software that can calculate it. The value from the standard normal distribution table for Z = -0.8 is approximately 0.2119.

Step 7 ：Therefore, the probability that a randomly selected U.S. adult has a triglyceride level less than 77 milligrams per deciliter is approximately 0.2119, or 21.19%.

Step 8 ：So, \(\boxed{0.2119}\) or \(\boxed{21.19\%}\) is the final answer.