A population is normally distributed with mean and standard deviation . From this population, a random sample of size is taken, and a confidence interval for the population mean is constructed using the sample mean. The resulting confidence interval is 1.2 $$\leq$$ m $$\leq$$ a. Find the value of . (Here, denotes a standard normal random variable, and use .) [3 points]Compute .\
[THOUGHT] I need to find the value of 'a' in the confidence interval 1.2 ≤ m ≤ a for a 99% confidence interval. The population standard deviation is 5, and the sample size is 36. The z-score for 99% confidence is given as 2.58. The margin of error (ME) is calculated as z * (σ / √n) = 2.58 * (5 / 6). Since the lower bound of the interval is 1.2, which is the sample mean minus the margin of error, the sample mean is 1.2 + ME. The upper bound 'a' is the sample mean plus the margin of error, so a = 1.2 + 2 * ME. Then, I need to compute 10a. I'll use Python to perform the calculation accurately. [/THOUGHT] [PYTHON] me = 2.58 * 5 / 6 a = 1.2 + 2 * me ten_a = 10 * a print(int(ten_a)) [/PYTHON]
[THOUGHT] The Python code calculated 10a as 55. I will use this as the final answer. [/THOUGHT] \boxed{55}