Respuesta :
Answer:
a) [tex]P(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)[/tex]
And we can find this probability using the complement rule:
[tex]P(z>-1.25)=1-P(z<-1.25)= 1-0.1057=0.8944 [/tex]
b) [tex]P(X<6)=P(\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(Z<\frac{6-5}{1.2})=P(z<0.83)[/tex]
And we can find this probability uing the normal standard table:
[tex]P(z<0.83)=0.7967 [/tex]
c) [tex]P(3.5<X<6)=P(\frac{3.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{3.5-5}{1.2}<Z<\frac{6-5}{1.2})=P(-1.25<z<0.83)[/tex]
And we can find this probability with this difference:
[tex]P(-1.25<z<0.83)=P(z<0.83)-P(z<-1.25)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.25<z<0.83)=P(z<0.83)-P(z<-1.25)=0.7967-0.1057= 0.6910[/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 variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(5,1.2)[/tex]
Where [tex]\mu=5[/tex] and [tex]\sigma=1.2[/tex]
We are interested on this probability
[tex]P(X>3.5)[/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>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)[/tex]
And we can find this probability using the complement rule:
[tex]P(z>-1.25)=1-P(z<-1.25)= 1-0.1057=0.8944 [/tex]
Part b
We are interested on this probability
[tex]P(X<6)[/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<6)=P(\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(Z<\frac{6-5}{1.2})=P(z<0.83)[/tex]
And we can find this probability uing the normal standard table:
[tex]P(z<0.83)=0.7967 [/tex]
Part c
[tex]P(3.5<X<6)=P(\frac{3.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{3.5-5}{1.2}<Z<\frac{6-5}{1.2})=P(-1.25<z<0.83)[/tex]
And we can find this probability with this difference:
[tex]P(-1.25<z<0.83)=P(z<0.83)-P(z<-1.25)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.25<z<0.83)=P(z<0.83)-P(z<-1.25)=0.7967-0.1057= 0.6910[/tex]
