Respuesta :
Answer:
[tex]P(X<85)=P(\frac{X-\mu}{\sigma}<\frac{85-\mu}{\sigma})=P(Z<\frac{85-100}{15})=P(Z<-1)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(Z<-1)=0.1587[/tex]
So then we can conclude that the value of 85 is approximately the 16 percentile for this case.
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 scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(100,15)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]
We are interested on this probability
[tex]P(X<85)[/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<85)=P(\frac{X-\mu}{\sigma}<\frac{85-\mu}{\sigma})=P(Z<\frac{85-100}{15})=P(Z<-1)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(Z<-1)=0.1587[/tex]
So then we can conclude that the value of 85 is approximately the 16 percentile for this case.
