[THOUGHT] The confidence interval for the population mean ( m ) is given by: [ \bar{x} - Z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) \leq m \leq \bar{x} + Z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) ] where:

• (\bar{x}) is the sample mean,
• (Z_{\alpha/2}) is the critical value from the standard normal distribution,
• (\sigma) is the population standard deviation,
• (n) is the sample size.

Given:

• The confidence interval is (1.2 \leq m \leq a),
• The confidence level is 99%,
• The critical value (Z_{\alpha/2} = 2.58),
• The population standard deviation (\sigma = 5),
• The sample size (n = 36).

From the lower bound of the confidence interval: [ \bar{x} - 2.58 \left(\frac{5}{\sqrt{36}}\right) = 1.2 ]

First, calculate the standard error: [ \frac{5}{\sqrt{36}} = \frac{5}{6} \approx 0.8333 ]

Then, substitute the standard error into the equation: [ \bar{x} - 2.58 \times 0.8333 = 1.2 ]

Calculate (2.58 \times 0.8333): [ 2.58 \times 0.8333 \approx 2.125 ]

So: [ \bar{x} - 2.125 = 1.2 ]

Solving for (\bar{x}): [ \bar{x} = 1.2 + 2.125 = 3.325 ]

Now, use the sample mean to find the upper bound (a): [ a = \bar{x} + 2.58 \left(\frac{5}{\sqrt{36}}\right) ]

Substitute the sample mean and standard error: [ a = 3.325 + 2.125 = 5.45 ]

Finally, compute (10a): [ 10a = 10 \times 5.45 = 54.5 ]

Since the final answer must be an integer: [ \boxed{545} ] [/THOUGHT] [PYTHON]

Define the values

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

Calculate the standard error

standard_error = sigma / (n ** 0.5)

Calculate the sample mean

sample_mean = lower_bound + Z * standard_error

Calculate the upper bound a

a = sample_mean + Z * standard_error

Calculate 10a

result = 10 * a print(int(result))

[THOUGHT] Based on the Python output, the value of (10a) is (55). Therefore, the final answer is: \boxed{55}