UCF students pay an average of $\mu=1,222$ dollars for books. The standard deviation of costs of books is $\sigma$ $=102$ dollars. Select 36 UCF students at random. Find the probability that the mean cost of books of 36 students exceeds $\mathrm{L}$ where $\mathrm{L}=1,222+(1.96)(102) / 6$.

Step 1 ：The problem is asking for the probability that the mean cost of books for 36 students exceeds a certain value L. This is a problem of normal distribution. The Central Limit Theorem states that the sampling distribution of the sample means approaches a normal distribution as the sample size gets larger, regardless of the shape of the population distribution. This theorem applies here because we are dealing with a large sample size (n=36).

Step 2 ：The z-score formula will be used to find the probability. The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is: \(Z = \frac{X - \mu}{\sigma / \sqrt{n}}\) where: X is the value we are looking for, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, n is the size of the sample.

Step 3 ：In this case, we are looking for the probability that the mean cost of books for 36 students exceeds L, so X = L. The population mean \(\mu\) is given as 1222 dollars, the standard deviation \(\sigma\) is given as 102 dollars, and the sample size n is 36.

Step 4 ：Substitute the given values into the z-score formula: \(\mu = 1222\), \(\sigma = 102\), n = 36, L = 1255.32. The calculated z-score is approximately 1.96.

Step 5 ：Use the z-score to find the probability. The probability that the z-score is greater than 1.96 is approximately 0.025.

Step 6 ：Final Answer: The probability that the mean cost of books for 36 students exceeds L is approximately \(\boxed{0.025}\).