The weights of adult individuals in a certain country are normally distributed with a population mean of [tex]\mu = 172[/tex] pounds and a population standard deviation of [tex]\sigma = 29[/tex] pounds. Suppose [tex]n = 36[/tex] individuals are sampled.

1. What is the mean of the sampling distribution of the means?
[tex]\square[/tex]

Let's explore how to determine the mean of the sampling distribution of sample means when we have a sample size of [tex]$ n = 36 $[/tex] individuals from a population with known parameters.

1. Understanding the population parameters and sample size:
- Population mean ([tex]$\mu$[/tex]): 172 pounds
- Population standard deviation ([tex]$\sigma$[/tex]): 29 pounds
- Sample size ([tex]$n$[/tex]): 36 individuals

2. Concepts of sampling distribution of the sample mean:
According to the central limit theorem, the sampling distribution of the sample mean will tend to be a normal distribution if the sample size is large enough. In this case, the sample size is 36, which is generally considered to be sufficiently large for the central limit theorem to apply.

3. Mean of the sampling distribution:
One key property of the sampling distribution of the sample mean is that the mean of this distribution (denoted as [tex]$\mu_{\bar{x}}$[/tex]) is equal to the population mean ([tex]$\mu$[/tex]). This means that:

[tex]\[
\mu_{\bar{x}} = \mu
\][/tex]

4. Substituting the known values:
Given that the population mean ([tex]$\mu$[/tex]) is 172 pounds,

[tex]\[
\mu_{\bar{x}} = 172
\][/tex]

Therefore, the mean of the sampling distribution of the means is [tex]$ \boxed{172} $[/tex].