To calculate the mean and standard deviation of the sample proportion, we'll start by defining the sample proportion ($\hat{p}$):

$\hat{p} = \frac{x}{n}$

where:

• $x$ is the number of successes (people who liked the product), which is 320.
• $n$ is the total number of trials (total number of people surveyed), which is 400.

1. Mean of the Sample Proportion ($\mu_{\hat{p}}$)

The mean of the sample proportion is given by:

$\mu_{\hat{p}} = p$

Since $p$ is the sample proportion:

$\hat{p} = \frac{320}{400} = 0.8$

So, the mean is:

$\mu_{\hat{p}} = 0.8$

2. Standard Deviation of the Sample Proportion ($\sigma_{\hat{p}}$)

The standard deviation of the sample proportion is given by:

$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

Substituting the values:

$\sigma_{\hat{p}} = \sqrt{\frac{0.8 \times (1 - 0.8)}{400}} = \sqrt{\frac{0.8 \times 0.2}{400}} = \sqrt{\frac{0.16}{400}} = \sqrt{0.0004} = 0.02$

Final Answers:

• Mean of the sample proportion ($\mu_{\hat{p}}$): 0.8
• Standard deviation of the sample proportion ($\sigma_{\hat{p}}$): 0.02

Would you like more details or have any further questions?

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Here are 5 related questions you might be interested in:

1. How would the standard deviation change if the sample size were larger?
2. What is the significance of the standard deviation in understanding the spread of sample proportions?
3. How do you interpret the sample proportion in the context of the population proportion?
4. What if the number of successes (people who liked the product) was 280 instead of 320? How would that affect the mean and standard deviation?
5. How does the normal distribution relate to the sample proportion when the sample size is large?

Tip: The standard deviation of the sample proportion decreases as the sample size increases, meaning the estimate of the population proportion becomes more precise with a larger sample.