Respuesta :
Answer:
a) [tex]P(X<20)=P(\frac{X-\mu}{\sigma}<\frac{20-\mu}{\sigma})=P(Z<\frac{20-9}{5})=P(z<2.2)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<2.2)=0.9861[/tex]
b) [tex]P(X<5)=P(\frac{X-\mu}{\sigma}>\frac{5-\mu}{\sigma})=P(Z>\frac{5-9}{5})=P(z>-0.8)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel and we got:
[tex]P(z>-0.8)=1-P(z<-0.8)=1-0.2119= 0.7881[/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 variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(9,5)[/tex]
Where [tex]\mu=9[/tex] and [tex]\sigma=5[/tex]
Part a
We are interested on this probability
[tex]P(X<20)[/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<20)=P(\frac{X-\mu}{\sigma}<\frac{20-\mu}{\sigma})=P(Z<\frac{20-9}{5})=P(z<2.2)[/tex]
And we can find this probability using the normal standard table or excel:
[tex]P(z<2.2)=0.9861[/tex]
Part b
We are interested on this probability
[tex]P(X>5)[/tex]
If we apply the z score formula to our probability we got this:
[tex]P(X<5)=P(\frac{X-\mu}{\sigma}>\frac{5-\mu}{\sigma})=P(Z>\frac{5-9}{5})=P(z>-0.8)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel and we got:
[tex]P(z>-0.8)=1-P(z<-0.8)=1-0.2119= 0.7881[/tex]
