Respuesta :
Answer:
[tex]X \sim N(112,20)[/tex]
Where [tex]\mu=112[/tex] and [tex]\sigma=20[/tex]
Since th distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is also normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar x} = 112[/tex]
[tex]\sigma_{\bar x}= \frac{20}{\sqrt{200}}=1.414[/tex]
[tex] \bar X \sim N( 112,1.414)[/tex]
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 IQ scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(112,20)[/tex]
Where [tex]\mu=112[/tex] and [tex]\sigma=20[/tex]
Since th distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is also normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar x} = 112[/tex]
[tex]\sigma_{\bar x}= \frac{20}{\sqrt{200}}=1.414[/tex]
[tex] \bar X \sim N( 112,1.414)[/tex]
