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The problem in the image asks for the expected value and standard error of the sampling distribution of means for samples of size 15 drawn from a population of 200 with a mean (\(\mu\)) of 35.8 and a standard deviation (\(\sigma\)) of 4.8. The task is to solve this for both cases: (i) sampling with replacement and (ii) sampling without replacement.

### (i) **Sampling with replacement:**
- **Expected value (mean) of the sampling distribution**:
  The expected value of the sampling distribution of the sample mean is the same as the population mean, regardless of the sample size. So,
  \[
  \text{Expected value} = \mu = 35.8.
  \]

- **Standard error (SE) of the sampling distribution**:
  For sampling with replacement, the standard error is given by the formula:
  \[
  SE = \frac{\sigma}{\sqrt{n}},
  \]
  where \(\sigma = 4.8\) is the population standard deviation and \(n = 15\) is the sample size.
  \[
  SE = \frac{4.8}{\sqrt{15}} \approx \frac{4.8}{3.873} \approx 1.239.
  \]

### (ii) **Sampling without replacement:**
- **Expected value (mean)**:
  The expected value remains the same as the population mean, which is 35.8, even when sampling without replacement.

- **Standard error (SE)**:
  When sampling without replacement, the standard error is adjusted by the finite population correction (FPC). The formula for the standard error in this case is:
  \[
  SE = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N - n}{N - 1}},
  \]
  where \(N = 200\) is the population size, \(n = 15\), and \(\sigma = 4.8\).
  \[
  SE = \frac{4.8}{\sqrt{15}} \times \sqrt{\frac{200 - 15}{200 - 1}} = \frac{4.8}{3.873} \times \sqrt{\frac{185}{199}}.
  \]
  First, calculate the finite population correction:
  \[
  \sqrt{\frac{185}{199}} \approx \sqrt{0.9296} \approx 0.964.
  \]
  Now, calculate the standard error:
  \[
  SE \approx 1.239 \times 0.964 \approx 1.194.
  \]

### Summary:
- **With replacement**: Expected value = 35.8, SE ≈ 1.239.
- **Without replacement**: Expected value = 35.8, SE ≈ 1.194.

Would you like a more detailed explanation or clarification on any part?

Here are 5 related questions for further practice:
1. What happens to the standard error if the sample size increases?
2. How does the standard error change if the population size decreases?
3. Why does the finite population correction reduce the standard error when sampling without replacement?
4. How would the formulas change if the population standard deviation were unknown?
5. Can the sampling distribution of means be approximated by a normal distribution in this case? Why?

**Tip:** The standard error decreases as the sample size increases, which means larger samples provide more accurate estimates of the population mean.

Obtain the expected value and standard error of the sampling distribution of means for samples of size 15, drawn from a population of 200 having mean = 35.8 and standard deviation = 4.8 when sampling is done (i) with replacement, and (ii) without replacement.

Calculate the expected value and standard error of the sampling distribution for samples of size 15 drawn from a population of 200 with mean = 35.8 and standard deviation = 4.8. This solution covers cases for sampling with and without replacement, applying the finite population correction.

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