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Sagot :
Let's break down the problem step-by-step to determine the distribution, mean, and standard deviation of the sample mean ([tex]\(\overline{x}\)[/tex]).
### Given Data:
1. Population mean ([tex]\(\mu\)[/tex]) = 156 lb
2. Population standard deviation ([tex]\(\sigma\)[/tex]) = 20 lb
3. Sample size ([tex]\(n\)[/tex]) = 4
### Step-by-Step Solution:
1. Determine the Mean of the Sample Mean ([tex]\(\overline{x}\)[/tex]):
The mean of the sample mean ([tex]\(\mu_{\overline{x}}\)[/tex]) is equal to the population mean ([tex]\(\mu\)[/tex]):
[tex]\[ \mu_{\overline{x}} = \mu = 156 \text{ lb} \][/tex]
2. Determine the Standard Deviation of the Sample Mean ([tex]\(\overline{x}\)[/tex]):
The standard deviation of the sample mean ([tex]\(\sigma_{\overline{x}}\)[/tex]), also known as the standard error, is calculated by dividing the population standard deviation ([tex]\(\sigma\)[/tex]) by the square root of the sample size ([tex]\(n\)[/tex]):
[tex]\[ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{4}} = \frac{20}{2} = 10 \text{ lb} \][/tex]
3. Determine the Distribution Shape:
For a normally distributed population with any sample size, the distribution of the sample mean ([tex]\(\overline{x}\)[/tex]) is also normally distributed. Given the population is normally distributed, the distribution of the sample mean will be approximately normal for small samples.
### Conclusion:
Given the calculated mean and standard deviation of the sample mean ([tex]\(\overline{x}\)[/tex]), we review the options:
- Option A: Normal, mean = 156 lb, standard deviation = 20 lb
- This is incorrect because the standard deviation should be 10 lb.
- Option B: Normal, mean = 156 lb, standard deviation = 10 lb
- This is correct.
- Option C: Approximately normal, mean = 156 lb, standard deviation = 5 lb
- This is incorrect because the standard deviation should be 10 lb.
- Option D: Approximately normal, mean = 156 lb, standard deviation = 10 lb
- This is correct and acceptable since the sample size is small and the population distribution is normal.
From the above reasoning, both Option B and Option D are correct.
### Final Answer:
- Both B and D are the correct choices, with:
- Mean ([tex]\(\mu_{\overline{x}}\)[/tex]) = 156 lb
- Standard Deviation ([tex]\(\sigma_{\overline{x}}\)[/tex]) = 10 lb
### Given Data:
1. Population mean ([tex]\(\mu\)[/tex]) = 156 lb
2. Population standard deviation ([tex]\(\sigma\)[/tex]) = 20 lb
3. Sample size ([tex]\(n\)[/tex]) = 4
### Step-by-Step Solution:
1. Determine the Mean of the Sample Mean ([tex]\(\overline{x}\)[/tex]):
The mean of the sample mean ([tex]\(\mu_{\overline{x}}\)[/tex]) is equal to the population mean ([tex]\(\mu\)[/tex]):
[tex]\[ \mu_{\overline{x}} = \mu = 156 \text{ lb} \][/tex]
2. Determine the Standard Deviation of the Sample Mean ([tex]\(\overline{x}\)[/tex]):
The standard deviation of the sample mean ([tex]\(\sigma_{\overline{x}}\)[/tex]), also known as the standard error, is calculated by dividing the population standard deviation ([tex]\(\sigma\)[/tex]) by the square root of the sample size ([tex]\(n\)[/tex]):
[tex]\[ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{4}} = \frac{20}{2} = 10 \text{ lb} \][/tex]
3. Determine the Distribution Shape:
For a normally distributed population with any sample size, the distribution of the sample mean ([tex]\(\overline{x}\)[/tex]) is also normally distributed. Given the population is normally distributed, the distribution of the sample mean will be approximately normal for small samples.
### Conclusion:
Given the calculated mean and standard deviation of the sample mean ([tex]\(\overline{x}\)[/tex]), we review the options:
- Option A: Normal, mean = 156 lb, standard deviation = 20 lb
- This is incorrect because the standard deviation should be 10 lb.
- Option B: Normal, mean = 156 lb, standard deviation = 10 lb
- This is correct.
- Option C: Approximately normal, mean = 156 lb, standard deviation = 5 lb
- This is incorrect because the standard deviation should be 10 lb.
- Option D: Approximately normal, mean = 156 lb, standard deviation = 10 lb
- This is correct and acceptable since the sample size is small and the population distribution is normal.
From the above reasoning, both Option B and Option D are correct.
### Final Answer:
- Both B and D are the correct choices, with:
- Mean ([tex]\(\mu_{\overline{x}}\)[/tex]) = 156 lb
- Standard Deviation ([tex]\(\sigma_{\overline{x}}\)[/tex]) = 10 lb
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