Respuesta :
Answer:
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(60,13.5)[/tex]
Where [tex]\mu=60[/tex] and [tex]\sigma=13.5[/tex]
And for this case we select a sample size of n= 81. Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And the standard error of the mean would be:
[tex]\sigma_{\bar X} =\frac{13.5}{\sqrt{81}}= 1.5[/tex]
4.1.5
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(60,13.5)[/tex]
Where [tex]\mu=60[/tex] and [tex]\sigma=13.5[/tex]
And for this case we select a sample size of n= 81. Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And the standard error of the mean would be:
[tex]\sigma_{\bar X} =\frac{13.5}{\sqrt{81}}= 1.5[/tex]
4.1.5
