To solve this, we’ll calculate the probability related to the sample mean distribution, $\bar{X}$, for samples of size $n = 82$ drawn from a population where $\mu = 84$ and $\sigma = 13$.

Since the sample size $n = 82$ is large enough (greater than 30), we can apply the Central Limit Theorem, which tells us that the distribution of the sample mean, $\bar{X}$, will be approximately normal with:

1. Mean: $\mu_{\bar{X}} = \mu = 84$
2. Standard deviation (standard error): $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{13}{\sqrt{82}}$

Now let’s calculate $\sigma_{\bar{X}}$:

$\sigma_{\bar{X}} = \frac{13}{\sqrt{82}} \approx 1.4373$

With this information, the sample mean distribution $\bar{X}$ is approximately normal with $\mu_{\bar{X}} = 84$ and $\sigma_{\bar{X}} \approx 1.4373$.

Could you specify the probability you would like calculated (e.g., $P(\bar{X} < a)$, $P(\bar{X} > b)$, or $P(a < \bar{X} < b)$)?