[THOUGHT] To find the value of $a$ in the 99% confidence interval $1.2 \leq m \leq a$, we use the formula for the confidence interval of the population mean:

Given:

• The lower bound of the interval is $1.2$.
• The population standard deviation $\sigma = 5$.
• The sample size $n = 36$.
• The Z-score for 99% confidence is $Z = 2.58$.

The margin of error (ME) is calculated as:

Since the confidence interval is symmetric around the sample mean $\bar{x}$, the sample mean is the midpoint of the interval:

The lower bound of the interval is:

Substituting $\bar{x}$:

Solving for $a$:

Thus, $10a = 10 \times 5.5 = 55$. [/THOUGHT] [PYTHON]

Calculate the margin of error

Z = 2.58 sigma = 5 n = 36 ME = Z * (sigma / (n ** 0.5))

Lower bound of the interval

lower_bound = 1.2

Calculate the sample mean (midpoint)

sample_mean = lower_bound + ME

Calculate the upper bound 'a'

a = 2 * sample_mean - lower_bound

Compute 10a

result = 10 * a print(result) [/PYTHON]

[THOUGHT] The Python output confirms that $10a = 55.00000000000001$, which rounds to the integer $55$. This matches our earlier derivation. [/THOUGHT] \boxed{55}