Respuesta :
Answer:
[tex] z =\frac{680-513}{81}=2.06[/tex]
The Eleanor score is 2.06 deviations above the mean.
[tex] z =\frac{27-22.7}{4}=1.08[/tex]
The Gerald score is 1.08 deviations above the mean.
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".
Part a
Let X the random variable that represent the SAT scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(513,81)[/tex]
Where [tex]\mu=513[/tex] and [tex]\sigma=81[/tex]
For this case we can calculate the z score with the following formula:
[tex] z=\frac{X-\mu}{\sigma}[/tex]
And if we replace we got:
[tex] z =\frac{680-513}{81}=2.06[/tex]
So the Eleanor score is 2.06 deviations above the mean.
Part b
Let X the random variable that represent the ACT scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(22.7,4)[/tex]
Where [tex]\mu=22.7[/tex] and [tex]\sigma=4[/tex]
For this case we can calculate the z score with the following formula:
[tex] z=\frac{X-\mu}{\sigma}[/tex]
And if we replace we got:
[tex] z =\frac{27-22.7}{4}=1.08[/tex]
So the Gerald score is 1.08 deviations above the mean.