Respuesta :
Answer:
By definition [tex] P(X= 6076)=0[/tex] since in a continuous probability distribution the area below a line is 0
If the real question is:
[tex] P(X<6076)[/tex]
We can use the z score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<6076)=P(\frac{X-\mu}{\sigma}<\frac{6076-\mu}{\sigma})=P(Z<\frac{6076-6098}{103})=P(z<-0.214)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-0.214)=0.415[/tex]
And if the question is [tex] P(X>6076)[/tex] we apply the complement rule and we got:
[tex] P(X>6076) = 1-P(X<6076) = 1-0.415=0.585[/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 strengths of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(6098,103)[/tex]
Where [tex]\mu=6098[/tex] and [tex]\sigma=103[/tex]
For this case we want [tex] P(X= 6076) [/tex]
And by definition [tex] P(X= 6076)=0[/tex] since in a continuous probability distribution the area below a line is 0
If the real question is:
[tex] P(X<6076)[/tex]
We can use the z score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<6076)=P(\frac{X-\mu}{\sigma}<\frac{6076-\mu}{\sigma})=P(Z<\frac{6076-6098}{103})=P(z<-0.214)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<-0.214)=0.415[/tex]
And if the question is [tex] P(X>6076)[/tex] we apply the complement rule and we got:
[tex] P(X>6076) = 1-P(X<6076) = 1-0.415=0.585[/tex]
