Respuesta :
Answer:
a) [tex]P(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]
[tex]P(z<-0.882)=0.189[/tex]
b) [tex]P(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]
[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]
c) [tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]
[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/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".
Part a
Let X the random variable that represent the thickness of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(4.5,1.7)[/tex]
Where [tex]\mu=4.5[/tex] and [tex]\sigma=1.7[/tex]
We are interested on this probability
[tex]P(X<3.0)[/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<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]
And we can find this probability using excel or the normal standard table:
[tex]P(z<-0.882)=0.189[/tex]
Part b
We are interested on this probability
[tex]P(X>3.0)[/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>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]
And we can find this probability using excel or the normal standard table:
[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]
Part c
[tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]
And we can find this probability using excel or the normal standard table liek this:
[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/tex]
