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] We have a normally distributed population with unknown mean m and known standard deviation σ = 5. A sample of size n = 36 is taken. The sample mean x̄ is used to construct a 99% confidence interval for m. Given: P(|Z| ≤ 2.58) = 0.99, so the critical value z* = 2.58 for a two-sided 99% CI. The confidence interval formula: x̄ ± z* * (σ / √n). Here σ = 5, n = 36, so σ/√n = 5/√36 = 5/6. Thus margin of error E = z* * (σ/√n) = 2.58 * (5/6). The CI is [x̄ - E, x̄ + E]. Given that the CI is 1.2 ≤ m ≤ a, so lower bound = 1.2 = x̄ - E. Thus x̄ = 1.2 + E. Then a = x̄ + E = (1.2 + E) + E = 1.2 + 2E. We need to compute a, then compute 10a. Let's compute E first. [/THOUGTH] [PYTHON] import math z_star = 2.58 sigma = 5 n = 36 se = sigma / math.sqrt(n) E = z_star * se print("Standard error:", se) print("Margin of error E:", E) a = 1.2 + 2 * E print("a:", a) print("10a:", 10 * a) [/PYTHON]
[THOUGHT] From the Python output: Standard error = 5/6 ≈ 0.8333333333333334 Margin of error E = 2.58 * (5/6) ≈ 2.1500000000000004 a = 1.2 + 2E ≈ 1.2 + 4.300000000000001 ≈ 5.500000000000001 10a ≈ 55.00000000000001
The result is essentially 55 (since 5.5 * 10 = 55). The slight floating-point imprecision gives 55.00000000000001, which is effectively 55. We need to compute 10a, which is 55. The problem asks: "Compute 10a." So the answer is 55. We should output as integer digits: 55. [/THOUGTH] \boxed{55}