Respuesta :
Answer:
a) We can find the z score for the value of 225 and we got:
[tex] z = \frac{225-204}{55}=0.382[/tex]
And we can find this probability with the complement rule:
[tex]P(z>0.382)=1-P(z<0.382)=1-0.6488=0.3512[/tex]
b) [tex] z = \frac{140-204}{55}=-1.164[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-1.164)=0.1223[/tex]
c) [tex]P(200<X<300)=P(\frac{200-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{300-\mu}{\sigma})=P(\frac{200-204}{55}<Z<\frac{300-204}{55})=P(-0.072<z<1.74)[/tex]
And we can find this probability with this difference:
[tex]P(-0.072<z<1.74)=P(z<1.74)-P(z<-0.072)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.072<z<1.74)=P(z<1.74)-P(z<-0.072)=0.9595-0.4710=0,4885 [/tex]
d) For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.2[/tex] (a)
[tex]P(X<a)=0.8[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.8 of the area on the left and 0.2 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.8[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.8[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.842<\frac{a-204}{55}[/tex]
And if we solve for a we got
[tex]a=204 +0.842*55=250.31[/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 costs of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(204,55)[/tex]
Where [tex]\mu=204[/tex] and [tex]\sigma=55[/tex]
We are interested on this probability
[tex]P(X>225)[/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]
We can find the z score for the value of 225 and we got:
[tex] z = \frac{225-204}{55}=0.382[/tex]
And we can find this probability with the complement rule:
[tex]P(z>0.382)=1-P(z<0.382)=1-0.6488=0.3512[/tex]
Part b
[tex]P(X<140)[/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] z = \frac{140-204}{55}=-1.164[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-1.164)=0.1223[/tex]
Part c
[tex]P(200<X<300)[/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(200<X<300)=P(\frac{200-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{300-\mu}{\sigma})=P(\frac{200-204}{55}<Z<\frac{300-204}{55})=P(-0.072<z<1.74)[/tex]
And we can find this probability with this difference:
[tex]P(-0.072<z<1.74)=P(z<1.74)-P(z<-0.072)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.072<z<1.74)=P(z<1.74)-P(z<-0.072)=0.9595-0.4710=0,4885 [/tex]
Part d
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.2[/tex] (a)
[tex]P(X<a)=0.8[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.8 of the area on the left and 0.2 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.8[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.8[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.842<\frac{a-204}{55}[/tex]
And if we solve for a we got
[tex]a=204 +0.842*55=250.31[/tex]
