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 .\
# Calculate margin of error E
z_score = 2.58
sigma = 5.0
n = 36
e = z_score * sigma / n**0.5
# Confidence interval lower bound is 1.2. We express a as the upper bound given the relationship:
lower_bound = 1.2
upper_bound = lower_bound + 2*e
ten_times_a = int(round(upper_bound *10))
ten_times_a
[THOUGHT] We construct the 99% confidence interval with the given parameters. The margin of error ( E ) is calculated as ( Z \times \frac{\sigma}{\sqrt{n}} ). Using ( Z = 2.58 ), ( \sigma = 5 ), and ( n = 36 ): [ E = 2.58 \times \frac{5}{6} \approx 2.155 ] Since the lower bound is 1.2, the upper bound ( a ) becomes: [ a = 1.2 + 2E ] Calculating numerically and rounding appropriately yields ( a ). Finally, ( 10a ) is found, leading to the answer.
\boxed{55} [/THOUGHT][THOUGHT] We construct the 99% confidence interval with the given parameters. The margin of error ( E ) is calculated as ( Z \times \frac{\sigma}{\sqrt{n}} ). Using ( Z = 2.58 ), ( \sigma = 5 ), and ( n = 36 ): [ E = 2.58 \times \frac{5}{6} \approx 2.155 ] Since the lower bound is 1.2, the upper bound ( a ) becomes: [ a = 1.2 + 2E ] Calculating numerically and rounding appropriately yields ( a ). Finally, ( 10a ) is found, leading to the answer.
\boxed{55} [/THOUGHT]