Answer :
### (a) Shape of the Sampling Distribution of Means
Given that the sample size \( n = 183 \) is large (greater than 30), we can apply the Central Limit Theorem. The Central Limit Theorem states that for a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution.
Therefore, the shape of the sampling distribution of means is:
normal
### (b) Center of the Sampling Distribution of Means
The center of the sampling distribution of the sample mean is the same as the population mean. This means that the expected value (mean) of the sample means is equal to the population mean.
Given:
[tex]\[ \mu = 108.9 \][/tex]
The center of the sampling distribution of means is:
[tex]\[ \mu_{\bar{x}} = 108.9 \][/tex]
So, the answer is:
[tex]\[ \mu_{\bar{x}} = 108.9 \][/tex]
### (c) Standard Error of the Sampling Distribution of Means
The standard error of the sampling distribution of means is calculated as the population standard deviation divided by the square root of the sample size.
Given:
[tex]\[ \sigma = 82.2 \][/tex]
[tex]\[ n = 183 \][/tex]
The formula for the standard error \( \sigma_{\bar{x}} \) is:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
By plugging in the given values:
[tex]\[ \sigma_{\bar{x}} = \frac{82.2}{\sqrt{183}} \][/tex]
After performing the calculation:
[tex]\[ \sigma_{\bar{x}} = 6.0764 \][/tex]
Therefore, the standard error of the sampling distribution of means is:
[tex]\[ \sigma_{\bar{x}} = 6.0764 \][/tex]
### Summary
- (a) The shape of the sampling distribution of means is normal.
- (b) The center of the sampling distribution of means is \( \mu_{\bar{x}} = 108.9 \).
- (c) The standard error of the sampling distribution of means is [tex]\( \sigma_{\bar{x}} = 6.0764 \)[/tex].
Given that the sample size \( n = 183 \) is large (greater than 30), we can apply the Central Limit Theorem. The Central Limit Theorem states that for a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution.
Therefore, the shape of the sampling distribution of means is:
normal
### (b) Center of the Sampling Distribution of Means
The center of the sampling distribution of the sample mean is the same as the population mean. This means that the expected value (mean) of the sample means is equal to the population mean.
Given:
[tex]\[ \mu = 108.9 \][/tex]
The center of the sampling distribution of means is:
[tex]\[ \mu_{\bar{x}} = 108.9 \][/tex]
So, the answer is:
[tex]\[ \mu_{\bar{x}} = 108.9 \][/tex]
### (c) Standard Error of the Sampling Distribution of Means
The standard error of the sampling distribution of means is calculated as the population standard deviation divided by the square root of the sample size.
Given:
[tex]\[ \sigma = 82.2 \][/tex]
[tex]\[ n = 183 \][/tex]
The formula for the standard error \( \sigma_{\bar{x}} \) is:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
By plugging in the given values:
[tex]\[ \sigma_{\bar{x}} = \frac{82.2}{\sqrt{183}} \][/tex]
After performing the calculation:
[tex]\[ \sigma_{\bar{x}} = 6.0764 \][/tex]
Therefore, the standard error of the sampling distribution of means is:
[tex]\[ \sigma_{\bar{x}} = 6.0764 \][/tex]
### Summary
- (a) The shape of the sampling distribution of means is normal.
- (b) The center of the sampling distribution of means is \( \mu_{\bar{x}} = 108.9 \).
- (c) The standard error of the sampling distribution of means is [tex]\( \sigma_{\bar{x}} = 6.0764 \)[/tex].
