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Answer :
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%.
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%.
Thank you for reading the article 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. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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Rewritten by : Jeany
