Respuesta :
Answer:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [/tex]
The mean for this case is given by:
[tex] \mu_{\bar X} = 2300.0[/tex]
And the deviation would be:
[tex]\sigma_{\bar X}= \frac{500}{\sqrt{25}}= 100.0[/tex]
Step-by-step explanation:
Using the following problem info since is incomplete the data: "The living spaces of all homes in a city have a mean of 2300 square feet and a standard deviation of 500 square feet. Let x be the mean living space for a random sample of 25 homes selected from this city. Find the mean and standard deviation of the sampling distribution of x."
We know the following info given:
[tex]\mu = 2300[/tex] represent the mean for tehe living spaces of all homes in a city
[tex]\sigma = 500[/tex] represent the population deviation for the data
We select a sample size of n=25 so then this size is large enough in order to use the central limit theorem and we can use for the distribution of the sampel mean:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [/tex]
The mean for this case is given by:
[tex] \mu_{\bar X} = 2300.0[/tex]
And the deviation would be:
[tex]\sigma_{\bar X}= \frac{500}{\sqrt{25}}= 100.0[/tex]
