Answer :
To solve this problem, we need to find the probability that the mean age of a random sample of 47 residents is greater than 37.1 years, given that the population mean age is 36.8 years with a standard deviation of 8.2 years.
Step-by-Step Solution:
1. Identify the Population Parameters:
- Mean ([tex]\(\mu\)[/tex]) = 36.8 years
- Standard Deviation ([tex]\(\sigma\)[/tex]) = 8.2 years
2. Identify the Sample Size:
- Sample Size (n) = 47 residents
3. Identify the Sample Mean:
- Sample Mean ([tex]\(\bar{x}\)[/tex]) = 37.1 years
4. Compute the Standard Error of the Mean (SEM):
- The standard error of the mean is calculated using the formula:
[tex]\[
SEM = \frac{\sigma}{\sqrt{n}}
\][/tex]
- Plugging in the values:
[tex]\[
SEM = \frac{8.2}{\sqrt{47}}
\][/tex]
5. Calculate the Z-Score:
- The Z-score helps us determine how many standard deviations the sample mean is from the population mean. It is calculated using the formula:
[tex]\[
Z = \frac{\bar{x} - \mu}{SEM}
\][/tex]
- Substituting the known values:
[tex]\[
Z = \frac{37.1 - 36.8}{\frac{8.2}{\sqrt{47}}}
\][/tex]
6. Find the Probability:
- To find the probability that the sample mean is greater than 37.1, we need to look up the Z-score in a standard normal distribution table, or use a cumulative distribution function (CDF).
- Since we are interested in the probability that the mean is greater than 37.1, we need to calculate:
[tex]\[
P(\bar{x} > 37.1) = 1 - P(Z \leq Z \text{-score})
\][/tex]
After performing these calculations, we find:
- The Z-score is approximately 0.25.
- The probability that the sample mean is greater than 37.1 years is about 0.401 or 40.1%.
This means there is a 40.1% chance that the mean age of the residents in the sample is greater than 37.1 years.
Step-by-Step Solution:
1. Identify the Population Parameters:
- Mean ([tex]\(\mu\)[/tex]) = 36.8 years
- Standard Deviation ([tex]\(\sigma\)[/tex]) = 8.2 years
2. Identify the Sample Size:
- Sample Size (n) = 47 residents
3. Identify the Sample Mean:
- Sample Mean ([tex]\(\bar{x}\)[/tex]) = 37.1 years
4. Compute the Standard Error of the Mean (SEM):
- The standard error of the mean is calculated using the formula:
[tex]\[
SEM = \frac{\sigma}{\sqrt{n}}
\][/tex]
- Plugging in the values:
[tex]\[
SEM = \frac{8.2}{\sqrt{47}}
\][/tex]
5. Calculate the Z-Score:
- The Z-score helps us determine how many standard deviations the sample mean is from the population mean. It is calculated using the formula:
[tex]\[
Z = \frac{\bar{x} - \mu}{SEM}
\][/tex]
- Substituting the known values:
[tex]\[
Z = \frac{37.1 - 36.8}{\frac{8.2}{\sqrt{47}}}
\][/tex]
6. Find the Probability:
- To find the probability that the sample mean is greater than 37.1, we need to look up the Z-score in a standard normal distribution table, or use a cumulative distribution function (CDF).
- Since we are interested in the probability that the mean is greater than 37.1, we need to calculate:
[tex]\[
P(\bar{x} > 37.1) = 1 - P(Z \leq Z \text{-score})
\][/tex]
After performing these calculations, we find:
- The Z-score is approximately 0.25.
- The probability that the sample mean is greater than 37.1 years is about 0.401 or 40.1%.
This means there is a 40.1% chance that the mean age of the residents in the sample is greater than 37.1 years.
