Respuesta :
Answer:
[tex]P(X>1600)=P(\frac{X-\mu}{\sigma}>\frac{1600-\mu}{\sigma})=P(Z>\frac{1600-1518}{325})=P(z>0.252)[/tex]
And we can find this probability using the z score formula and the complement rule and we got:
[tex]P(z>0.252)=1-P(z<0.252) =1-0.599= 0.401 [/tex]
[tex] z =\frac{1600-1518}{\frac{325}{\sqrt{81}}}= 2.27[/tex]
And we can find this probability using the z score formula and the complement rule and we got:
[tex]P(z>2.27)=1-P(z<2.27) =1-0.988=0.012[/tex]
Step-by-step explanation:
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(1518,325)[/tex]
Where [tex]\mu=1518[/tex] and [tex]\sigma=325[/tex]
We want to find this probability:
[tex]P(X>1600)[/tex]
And we can use the z score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using this formula we got:
[tex]P(X>1600)=P(\frac{X-\mu}{\sigma}>\frac{1600-\mu}{\sigma})=P(Z>\frac{1600-1518}{325})=P(z>0.252)[/tex]
And we can find this probability using the z score formula and the complement rule and we got:
[tex]P(z>0.252)=1-P(z<0.252) =1-0.599= 0.401 [/tex]
For the other part we need to take in count that the distribution for the sampel mean if the sample size is large (n>30) is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we can use the z score formula given by:
[tex]z=\frac{x-\mu}{\frac{sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z =\frac{1600-1518}{\frac{325}{\sqrt{81}}}= 2.27[/tex]
And we can find this probability using the z score formula and the complement rule and we got:
[tex]P(z>2.27)=1-P(z<2.27) =1-0.988=0.012[/tex]
