[THOUGHT] We need to find a, the upper endpoint of the 99% confidence interval for m, given that the interval is 1.2 ≤ m ≤ a. The population standard deviation σ = 5, sample size n = 36, so the standard error = σ/√n = 5/√36 = 5/6 ≈ 0.833333... The confidence interval for the mean m is given by: sample mean ± z* × (σ/√n), where z* is the critical value for 99% confidence: given P(|Z| ≤ 2.58) = 0.99, so z* = 2.58.

The interval is (sample mean - 2.58*(5/6), sample mean + 2.58*(5/6)). The lower bound is given as 1.2, so: sample mean - 2.58*(5/6) = 1.2 → sample mean = 1.2 + 2.58*(5/6).

Then the upper bound a = sample mean + 2.58*(5/6) = (1.2 + 2.58*(5/6)) + 2.58*(5/6) = 1.2 + 2 * 2.58 * (5/6).

Compute 2 * 2.58 * (5/6) = (2 * 2.58 * 5) / 6 = (25.8) / 6 = 4.3. Then a = 1.2 + 4.3 = 5.5.

But we must compute 10a = 10 * 5.5 = 55.

Let's verify with Python.

[/PYTHON] [PYTHON] import math

given

sigma = 5 n = 36 z_star = 2.58

se = sigma / math.sqrt(n) margin = z_star * se lower = 1.2

sample_mean = lower + margin a = sample_mean + margin ten_a = 10 * a

print("se =", se) print("margin =", margin) print("sample_mean =", sample_mean) print("a =", a) print("10a =", ten_a) [/PYTHON]

[THOUGHT] The Python output shows a = 5.500000000000001 and 10a = 55.00000000000001. The problem asks to compute 10a, and the exact value is 55. Since the confidence interval is constructed from the sample mean and margin of error, and the given lower bound is 1.2, the calculation yields a = 5.5 exactly (within floating-point precision). Thus 10a = 55. The final answer should be the integer digits 55.

\boxed{55}