Let's work through the question from the image. Here's a breakdown of the problem:

Given:

• Sample size $n = 25$
• Population size $N = 600$
• Population mean $\mu = 108.4$
• Population variance $\sigma^2 = 812.25$

1. Expected Value of the Sampling Distribution:

The expected value (mean) of the sampling distribution of the sample mean is equal to the population mean:

$E(\bar{X}) = \mu = 108.4$

2. Standard Error of the Sampling Distribution:

Since sampling is done without replacement, we need to apply the finite population correction factor (FPC) to adjust the standard error.

The standard error (SE) of the sampling distribution of the sample mean is given by:

$SE(\bar{X}) = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N - n}{N - 1}}$

Where:

• $\sigma = \sqrt{812.25} = 28.5$ (standard deviation of the population)
• $n = 25$ (sample size)
• $N = 600$ (population size)

Now, plugging in the values:

$SE(\bar{X}) = \frac{28.5}{\sqrt{25}} \times \sqrt{\frac{600 - 25}{600 - 1}} = \frac{28.5}{5} \times \sqrt{\frac{575}{599}}$

Simplifying:

$SE(\bar{X}) = 5.7 \times \sqrt{0.960}$

$SE(\bar{X}) \approx 5.7 \times 0.98 = 5.59$

Final Answers:

• Expected value of the sampling distribution: $E(\bar{X}) = 108.4$
• Standard error of the sampling distribution: $SE(\bar{X}) \approx 5.59$

Let me know if you need further details or if you have any questions!

Related Questions:

1. How would the expected value change if the population mean were different?
2. What happens to the standard error if sampling is done with replacement?
3. How does the finite population correction factor influence the standard error?
4. What effect does increasing the sample size have on the standard error?
5. How would the calculation differ if the population size were infinite?

Tip:

When sampling from a finite population without replacement, always remember to apply the finite population correction factor to get a more accurate standard error.