Respuesta :
Answer:
[tex] \bar X \sim N (\mu ,\frac{\sigma}{\sqrt{n}})[/tex]
Then the mean would be:
[tex] \mu_{\bar X}= 2.02[/tex]
And the deviation:
[tex] \sigma_{\bar X}= \frac{3.00}{\sqrt{45}}= 0.447[/tex]
And the distribution would be given by:
[tex] \bar X \sim N(2.02, 0.447)[/tex]
Step-by-step explanation:
For this case we can define the variable of interest X as the amount of change in UF student's pockets and we know the following parameters:
[tex] \mu = 2.02, \sigma =3.00[/tex]
And we select a sample of n=45. Since this sample size is higher than 30 we can apply the central limit theorem and the distirbution for the sample size would be given by:
[tex] \bar X \sim N (\mu ,\frac{\sigma}{\sqrt{n}})[/tex]
Then the mean would be:
[tex] \mu_{\bar X}= 2.02[/tex]
And the deviation:
[tex] \sigma_{\bar X}= \frac{3.00}{\sqrt{45}}= 0.447[/tex]
And the distribution would be given by:
[tex] \bar X \sim N(2.02, 0.447)[/tex]
