Respuesta :
Answer:
a) [tex]P(X<40000)=P(\frac{X-\mu}{\sigma}<\frac{40000-\mu}{\sigma})=P(Z<\frac{40000-72000}{12000})=P(z<-2.67)[/tex]
[tex]P(z<-2.67)=0.0038[/tex]
b) [tex]P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)[/tex]
[tex]P(z>-0.583)=1-P(Z<-0.583)=1-0.280=0.720[/tex]
c) [tex]P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)[/tex]
[tex]P(z>2.33)=1-P(Z<2.33)=1-0.990=0.01[/tex]
Sicne this probability just represent 1% of the data we can consider this value as unusual.
d) [tex]z=1.28<\frac{a-72000}{12000}[/tex]
And if we solve for a we got
[tex]a=72000 +1.28*12000=87360[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 87360. Â
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". Â
Part a
Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(72000,12000)[/tex] Â
Where [tex]\mu=72000[/tex] and [tex]\sigma=12000[/tex]
We are interested on this probability
[tex]P(X<40000)[/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<40000)=P(\frac{X-\mu}{\sigma}<\frac{40000-\mu}{\sigma})=P(Z<\frac{40000-72000}{12000})=P(z<-2.67)[/tex]
And we can find this probability using excel or the normal standard table and we got:
[tex]P(z<-2.67)=0.0038[/tex]
Part b
[tex]P(X>65000)[/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>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)[/tex]
And we can find this probability using the complement rule and excel or the normal standard table and we got:
[tex]P(z>-0.583)=1-P(Z<-0.583)=1-0.280=0.720[/tex]
Part c
[tex]P(X>100000)[/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>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)[/tex]
And we can find this probability using the complement rule and excel or the normal standard table and we got:
[tex]P(z>2.33)=1-P(Z<2.33)=1-0.990=0.01[/tex]
Sicne this probability just represent 1% of the data we can consider this value as unusual.
Part d
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] Â (a)
[tex]P(X<a)=0.9[/tex] Â (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a. Â
As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9[/tex] Â
[tex]P(z<\frac{a-\mu}{\sigma})=0.9[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-72000}{12000}[/tex]
And if we solve for a we got
[tex]a=72000 +1.28*12000=87360[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 87360. Â
