Question
The theorem that states that the sampling distribution of the sample mean is approximately normal when the sample is large is called the
theorem (make sure that you spell it right).
According to this theorem, if the population had mean 200 and standard deviation 25, then the sampling distribution of the the sample mean of size 100 has
mean
and standard deviation
.
theorem (make sure that you spell it right).
According to this theorem, if the population had mean 200 and standard deviation 25, then the sampling distribution of the the sample mean of size 100 has
mean
and standard deviation
.
Answers
Answered by
GPT-4o mini
The theorem you're referring to is the Central Limit Theorem (CLT).
According to the Central Limit Theorem, if the population has a mean (\(\mu\)) of 200 and standard deviation (\(\sigma\)) of 25, then the sampling distribution of the sample mean (\(\bar{x}\)) for a sample size of \(n = 100\) has:
- Mean (\(\mu_{\bar{x}}\)): This is equal to the population mean, so \(\mu_{\bar{x}} = 200\).
- Standard Deviation (\(\sigma_{\bar{x}}\)): This is calculated using the formula \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\). Given that \(\sigma = 25\) and \(n = 100\):
\[
\sigma_{\bar{x}} = \frac{25}{\sqrt{100}} = \frac{25}{10} = 2.5.
\]
Therefore, the answers are:
- Mean: 200
- Standard Deviation: 2.5
According to the Central Limit Theorem, if the population has a mean (\(\mu\)) of 200 and standard deviation (\(\sigma\)) of 25, then the sampling distribution of the sample mean (\(\bar{x}\)) for a sample size of \(n = 100\) has:
- Mean (\(\mu_{\bar{x}}\)): This is equal to the population mean, so \(\mu_{\bar{x}} = 200\).
- Standard Deviation (\(\sigma_{\bar{x}}\)): This is calculated using the formula \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\). Given that \(\sigma = 25\) and \(n = 100\):
\[
\sigma_{\bar{x}} = \frac{25}{\sqrt{100}} = \frac{25}{10} = 2.5.
\]
Therefore, the answers are:
- Mean: 200
- Standard Deviation: 2.5