Respuesta :
Answer:
[tex]P( \bar X >4985)[/tex]
And the z score is given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using this formula we got:
[tex] z = \frac{4985-5500}{\frac{100}{\sqrt{9}}}= -15.45) [/tex]
And using the complement rule and the normal tandard distribution or excel we got:
[tex]P(z>-15.45) = 1-P(z<-15.45) \approx 1[/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 comprehensive strength of a population, and for this case we know the following info:
Where [tex]\mu=5500[/tex] and [tex]\sigma=100[/tex]
We select a sample size of n=9. Assuming that the distribution for X is normal then we can use 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 we want this probability:
[tex]P( \bar X >4985)[/tex]
And the z score is given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using this formula we got:
[tex] z = \frac{4985-5500}{\frac{100}{\sqrt{9}}}= -15.45) [/tex]
And using the complement rule and the normal tandard distribution or excel we got:
[tex]P(z>-15.45) = 1-P(z<-15.45) \approx 1[/tex]
