[THOUGHT] The problem asks for the upper bound $a$ of a 99% confidence interval for the population mean $m$. The population standard deviation is $\sigma = 5$. The sample size is $n = 36$. The confidence interval is given as $1.2 \leq m \leq a$. The formula for a confidence interval for the mean with known $\sigma$ is $\bar{x} \pm z^* \frac{\sigma}{\sqrt{n}}$. Here, $z^*$ corresponds to the 99% confidence level. The problem states to use $P(|Z| \leq 2.58) = 0.99$, so $z^* = 2.58$. The lower bound is $\bar{x} - z^* \frac{\sigma}{\sqrt{n}} = 1.2$. The upper bound is $a = \bar{x} + z^* \frac{\sigma}{\sqrt{n}}$. The margin of error is $E = z^* \frac{\sigma}{\sqrt{n}}$. So, $1.2 = \bar{x} - E$ and $a = \bar{x} + E$. This implies $a - 1.2 = 2E$, or $a = 1.2 + 2E$. I need to calculate $E$ and then $a$, and finally $10a$.

Steps:

1. Define $\sigma = 5$, $n = 36$, $z^* = 2.58$.
2. Calculate standard error $SE = \sigma / \sqrt{n}$.
3. Calculate margin of error $E = z^* \times SE$.
4. Use the lower bound to find $\bar{x}$ (though not strictly necessary if I just add $2E$ to the lower bound). Actually, Lower Bound = $\bar{x} - E = 1.2$. Upper Bound $a = \bar{x} + E = (\bar{x} - E) + 2E = 1.2 + 2E$.
5. Calculate $a$.
6. Calculate $10a$. [/THOUGHT] [PYTHON] sigma = 5 n = 36 z_star = 2.58 lower_bound = 1.2

Standard Error

se = sigma / (n**0.5)

Margin of Error

E = z_star * se

Upper bound a

a = lower_bound + 2 * E

a = lower_bound + 2 * E

result = 10 * a print(f"{se=}") print(f"{E=}") print(f"{a=}") print(f"{result=}") [/PYTHON] [THOUGHT] The code calculated $a = 5.5$ and $10a = 55.0$. Let's double check the logic. Confidence Interval: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$. Given: $1.2 \le m \le a$. Lower bound: $\bar{x} - 2.58 \cdot \frac{5}{\sqrt{36}} = 1.2$. $\frac{5}{6} \approx 0.8333$. $E = 2.58 \cdot \frac{5}{6} = 2.58 \cdot 0.8333... = \frac{12.9}{6} = 2.15$. Lower bound: $\bar{x} - 2.15 = 1.2 \implies \bar{x} = 3.35$. Upper bound $a$: $\bar{x} + 2.15 = 3.35 + 2.15 = 5.5$. Then $10a = 10 \cdot 5.5 = 55$.

The calculation holds up. $SE = 5/6$. $E = 2.58 * (5/6) = 1.29 * (5/3) = 0.43 * 5 = 2.15$. $a = 1.2 + 2 * 2.15 = 1.2 + 4.3 = 5.5$. $10a = 55$.

The problem asks for $10a$. The value is 55. [/THOUGHT] \boxed{55}

The confidence interval is given by $\bar{x} \pm E$, where $E = z^* \frac{\sigma}{\sqrt{n}}$. Given: $\sigma = 5$ $n = 36$ $z^* = 2.58$ (for 99% confidence) Lower bound = 1.2

First, we calculate the margin of error $E$: $E = 2.58 \times \frac{5}{\sqrt{36}} = 2.58 \times \frac{5}{6} = 2.58 \times 0.8333\dots = 2.15$

The confidence interval is $[\bar{x} - E, \bar{x} + E]$. We know the lower bound is $1.2$, so: $\bar{x} - E = 1.2$ The upper bound $a$ is: $a = \bar{x} + E$ We can express $a$ in terms of the lower bound and $E$: $a = (\bar{x} - E) + 2E = 1.2 + 2(2.15) = 1.2 + 4.3 = 5.5$

The problem asks for the value of $10a$: $10a = 10 \times 5.5 = 55$

\boxed{55}