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The question in the image is:

**Problem 2**:  
"Samples of size 25 are taken from a population of 600 values. The population has a mean of 100 and a variance equal to 812.25. Obtain the expected value and standard error of the sampling distribution of means if sampling is done without replacement."

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### Solution:

1. **Expected Value of the Sampling Distribution**:

The expected value (mean) of the sampling distribution of sample means is the same as the population mean, regardless of whether the sampling is with or without replacement.

- **Population Mean (μ)** = 100

Thus, the expected value (mean) of the sample means is:

\[
\mu_{\bar{x}} = \mu = 100
\]

2. **Standard Error of the Sampling Distribution**:

The standard error of the sample mean for **sampling without replacement** is given by the formula:

\[
SE_{\bar{x}} = \sqrt{\frac{\sigma^2}{n} \cdot \left(\frac{N - n}{N - 1}\right)}
\]

Where:
- \( \sigma^2 \) = Population variance = 812.25
- \( n \) = Sample size = 25
- \( N \) = Population size = 600
- \( \frac{N - n}{N - 1} \) is the finite population correction factor.

Let’s calculate the standard error:

\[
SE_{\bar{x}} = \sqrt{\frac{812.25}{25} \cdot \left(\frac{600 - 25}{600 - 1}\right)}
\]

First, calculate the variance divided by the sample size:

\[
\frac{812.25}{25} = 32.49
\]

Next, compute the finite population correction factor:

\[
\frac{600 - 25}{600 - 1} = \frac{575}{599} \approx 0.960
\]

Now calculate the standard error:

\[
SE_{\bar{x}} = \sqrt{32.49 \cdot 0.960} \approx \sqrt{31.1904} \approx 5.584
\]

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### Final Answers:
- **Expected value of the sampling distribution**: \( 100 \)
- **Standard error of the sampling distribution**: \( \approx 5.584 \)

Would you like more details on any of the steps, or do you have any other questions? Here are some related questions for further practice:

1. How does the standard error change when the sample size increases?
2. What is the impact of the finite population correction on the standard error?
3. What is the formula for the standard error when sampling **with** replacement?
4. Why does the expected value of the sample means remain the same as the population mean?
5. How would the calculation change if the population size was infinite?

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**Tip**: When the sample size is small compared to the population (typically \( n/N < 0.05 \)), the finite population correction factor can be ignored.

Samples of size 25 are taken from a population of 600 values. The population has a mean of 100 and a variance equal to 812.25. Obtain the expected value and standard error of the sampling distribution of means if sampling is done without replacement.

This problem involves calculating the expected value and standard error for a sampling distribution when sampling is done without replacement from a finite population. The population has a mean of 100 and a variance of 812.25, with a sample size of 25 from a total of 600 values.

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