Suppose the ages of the residents of Georgetown are normally distributed with a mean of 38.2 years and a standard deviation of 8.6 years. If a random sample of 30 residents is selected, what is the probability that the mean age of the residents in the sample is less than 38 years?

To solve this problem, we need to determine the probability that the mean age of a sample of 30 residents is less than 38 years when the ages of the residents follow a normal distribution with a mean of 38.2 years and a standard deviation of 8.6 years.

Here's how we can figure this out step by step:

1. Identify the Given Values:
- Population mean ([tex]$\mu$[/tex]) = 38.2 years
- Population standard deviation ([tex]$\sigma$[/tex]) = 8.6 years
- Sample size ([tex]$n$[/tex]) = 30
- Target sample mean = 38 years

2. Calculate the Standard Error of the Mean:
- The standard error of the mean is calculated using the formula:
[tex]\[
\text{Standard Error} = \frac{\sigma}{\sqrt{n}}
\][/tex]
- Plugging in the values:
[tex]\[
\text{Standard Error} = \frac{8.6}{\sqrt{30}} \approx 1.5701
\][/tex]

3. Calculate the Z-Score:
- The z-score measures how many standard errors the target sample mean is from the population mean. It is calculated using the formula:
[tex]\[
z = \frac{\text{Target Sample Mean} - \mu}{\text{Standard Error}}
\][/tex]
- Using the given values:
[tex]\[
z = \frac{38 - 38.2}{1.5701} \approx -0.1274
\][/tex]

4. Find the Probability:
- The probability that the sample mean is less than 38 is found using a standard normal distribution table or a calculator that provides cumulative distribution function values.
- A z-score of approximately [tex]$-0.1274$[/tex] translates to a probability of about 0.4493.

Therefore, the probability that the mean age of the sample is less than 38 years is approximately 0.4493, or 44.93%.