[THOUGHT] We have a normal population with known standard deviation σ = 5.
Sample size n = 36, so the standard error is σ/√n = 5/6.
For a 99 % confidence interval, the critical Z value is 2.58, giving a margin of error
ME = 2.58 × (5/6).

The lower bound of the interval is given as 1.2, which equals the sample mean minus the margin:
sample_mean = 1.2 + ME.

The upper bound a is then
a = sample_mean + ME.

Finally we compute 10 × a. [/THOUGHT] [PYTHON]

Given values

lower_bound = 1.2 sigma = 5 n = 36 z = 2.58

Compute margin of error

margin = z * sigma / (n**0.5)

Compute sample mean

sample_mean = lower_bound + margin

Compute upper bound a

a = sample_mean + margin

Compute 10*a

ten_a = 10 * a

Print results

print(a) print(ten_a) [/PYTHON]

[THOUGHT] The Python output confirms that ( a = 5.5 ) and ( 10a = 55.0 ). Since the problem asks for ( 10a ) as an integer with no commas or scientific notation, the result is 55. [/THOUGHT] \boxed{55}