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] For a 99% z-interval with known population standard deviation, the confidence interval is [ \bar x \pm z_{0.995}\cdot \frac{\sigma}{\sqrt{n}}. ] Given the lower endpoint is 1.2, we have (\bar x - \text{ME}=1.2), so (\bar x=1.2+\text{ME}). Then the upper endpoint is (a=\bar x+\text{ME}=1.2+2\text{ME}). I will compute (\text{ME}=2.58\cdot 5/\sqrt{36}) and then (10a). [/THOUGHT] [PYTHON] import math sigma = 5 n = 36 z = 2.58 ME = z * sigma / math.sqrt(n) a = 1.2 + 2ME print("ME =", ME) print("a =", a) print("10a =", 10a) [/PYTHON][THOUGHT] For a 99% z-interval with known population standard deviation, the confidence interval is [ \bar x \pm z_{0.995}\cdot \frac{\sigma}{\sqrt{n}}. ] Given the lower endpoint is 1.2, we have (\bar x - \text{ME}=1.2), so (\bar x=1.2+\text{ME}). Then the upper endpoint is (a=\bar x+\text{ME}=1.2+2\text{ME}). I will compute (\text{ME}=2.58\cdot 5/\sqrt{36}) and then (10a). [/THOUGHT] [PYTHON] import math sigma = 5 n = 36 z = 2.58 ME = z * sigma / math.sqrt(n) a = 1.2 + 2ME print("ME =", ME) print("a =", a) print("10a =", 10a) [/PYTHON]
[THOUGHT] Using the computed margin of error (2.15), the upper endpoint is (a=5.5), so (10a=55). \boxed{55}