Respuesta :
Answer:
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(70.2,2.12)[/tex]
Where [tex]\mu=70.2[/tex] and [tex]\sigma=2.12[/tex]
Since the disgtribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
We can find the probability of interest we got:
[tex]P(\bar X >71.2)=P(Z>\frac{71.2-70.2}{\frac{2.12}{\sqrt{36}}}=2.83)[/tex]
And using the complement rule and a calculator, excel or the normal standard table we got this:
[tex]P(Z>2.83)=1-P(Z<2.83) = 1-0.9977=0.0023 [/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 heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(70.2,2.12)[/tex]
Where [tex]\mu=70.2[/tex] and [tex]\sigma=2.12[/tex]
Since the disgtribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
We can find the probability of interest we got:
[tex]P(\bar X >71.2)=P(Z>\frac{71.2-70.2}{\frac{2.12}{\sqrt{36}}}=2.83)[/tex]
And using the complement rule and a calculator, excel or the normal standard table we got this:
[tex]P(Z>2.83)=1-P(Z<2.83) = 1-0.9977=0.0023 [/tex]
