Respuesta :
Answer:
[tex]P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-75}{8})=P(Z>-0.75)[/tex]
And we can find this probability on this way:
[tex]P(Z>-0.75)=1-P(Z<-0.75)= 1-0.227= 0.773[/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 problm
Let X the random variable that represent the scores on an exam of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(75,8)[/tex]
Where [tex]\mu=75[/tex] and [tex]\sigma=8[/tex]
We are interested on this probability
[tex]P(X>69)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-75}{8})=P(Z>-0.75)[/tex]
And we can find this probability on this way:
[tex]P(Z>-0.75)=1-P(Z<-0.75)= 1-0.227= 0.773[/tex]
