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] The margin of error E is calculated as z * (σ / sqrt(n)) = 2.58 * (5 / 6) = 2.15. The sample mean is then 1.2 + 2.15 = 3.35. The upper bound a is 3.35 + 2.15 = 5.5. Therefore, 10a = 55. [/THOUGHT] \boxed{55}
[THOUGHT] We are given a confidence interval for the population mean m: 1.2 ≤ m ≤ a. The confidence level is 99%, and the z-score for 99% confidence is 2.58. The population standard deviation σ is 5, and the sample size n is 36. The margin of error E is calculated as z * (σ / sqrt(n)). The sample mean is the midpoint of the confidence interval, so sample_mean = (1.2 + a) / 2. The margin of error is also given by E = sample_mean - 1.2. We can compute E and then find a.
We will use Python to compute E and then a, and then compute 10a. [/THOUGHT] [PYTHON] import math z = 2.58 sigma = 5 n = 36 E = z * (sigma / math.sqrt(n)) a = 1.2 + 2 * E ten_a = 10 * a print(ten_a) [/PYTHON]
\boxed{55}