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 99% confidence interval is (\bar{x} \pm z \frac{\sigma}{\sqrt{n}}), where (z = 2.58), (\sigma = 5), (n = 36).
Standard error SE = 5 / sqrt(36) = 5/6.
Margin of error E = 2.58 * (5/6).
The lower bound is (\bar{x} - E = 1.2), so upper bound (a = \bar{x} + E = 1.2 + 2E).
Compute E exactly, then a, then 10a. [/THOUGHT] [PYTHON] import math
z = 2.58 sigma = 5 n = 36 SE = sigma / math.sqrt(n) E = z * SE a = 1.2 + 2 * E ten_a = 10 * a print(f"E = {E}") print(f"a = {a}") print(f"10a = {ten_a}") [/PYTHON]
[THOUGHT] The Python output gives 10a = 55.00000000000001, which is 55. [/THOUGHT] \boxed{55}
Explanation
The reasoning follows the standard formula for a confidence interval: (\bar{x} \pm z \frac{\sigma}{\sqrt{n}}), where (z = 2.58), (\sigma = 5), and (n = 36).
First, compute the standard error: (\text{SE} = \frac{5}{\sqrt{36}} = \frac{5}{6} \approx 0.8333).
Next, calculate the margin of error: (E = 2.58 \times \frac{5}{6} \approx 2.15).
The given lower bound is (\bar{x} - E = 1.2), so the upper bound is (a = \bar{x} + E = 1.2 + 2E).
Substitute (E \approx 2.15): (2E \approx 4.3), thus (a \approx 1.2 + 4.3 = 5.5).
Finally, compute (10a \approx 10 \times 5.5 = 55), matching the Python output of 55.00000000000001, which rounds to the exact integer 55.