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 know for a known σ and normal population, a 99% CI for m is: x̄ ± z * σ / √n with z = 2.58, σ = 5, n = 36.
Compute the margin of error E = 2.58 * 5 / √36, then use the fact that the interval is (x̄ - E, x̄ + E) = (1.2, a) to solve for a, and then compute 10a. [/THOUGHT] [PYTHON] import math
z = 2.58 sigma = 5 n = 36
E = z * sigma / math.sqrt(n)
The lower bound is xbar - E = 1.2
So xbar = 1.2 + E
xbar = 1.2 + E a = xbar + E print("E =", E) print("a =", a) print("10a =", 10*a) [/PYTHON]
[THOUGHT] Using the margin of error E = 2.15, the upper bound is a = 5.5, so 10a = 55. \boxed{55}