Respuesta :
Answer:
a) [tex]P(X>85)=P(\frac{X-\mu}{\sigma}>\frac{85-\mu}{\sigma})=P(Z>\frac{85-80.5}{9.9})=P(z>0.455)[/tex]
And we can find this probability using the complement rule and the normal standard table and we got:
[tex]P(z>0.455)=1-P(z<0.455)=1-0.675=0.325 [/tex]
b) [tex]z=-0.253<\frac{a-80.5}{9.9}[/tex]
And if we solve for a we got
[tex]a=80.5 -0.253*9.9=77.995[/tex]
So the value of height that separates the bottom 40% of data from the top 60% is 77.995.
c) [tex]P(\bar X >78.5)=P(Z>\frac{78.5-80.5}{\frac{9.9}{\sqrt{81}}}=-1.82)[/tex]
And using the complement rule a calculator, excel ir the normal standard table we have that:
[tex]P(Z>-1.82)=1-P(Z<-1.82) = 1-0.0344=0.966[/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 diastolic pressure of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(80.5,9.9)[/tex]
Where [tex]\mu=80.5[/tex] and [tex]\sigma=9.9[/tex]
We are interested on this probability
[tex]P(X>85)[/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>85)=P(\frac{X-\mu}{\sigma}>\frac{85-\mu}{\sigma})=P(Z>\frac{85-80.5}{9.9})=P(z>0.455)[/tex]
And we can find this probability using the complement rule and the normal standard table and we got:
[tex]P(z>0.455)=1-P(z<0.455)=1-0.675=0.325 [/tex]
Part b
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.6[/tex] (a)
[tex]P(X<a)=0.4[/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 0.4 of the area on the left and 0.6 of the area on the right it's z=-0.253. On this case P(Z<-0.253)=0.4 and P(z>-0.253)=0.6
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.4[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.4[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.253<\frac{a-80.5}{9.9}[/tex]
And if we solve for a we got
[tex]a=80.5 -0.253*9.9=77.995[/tex]
So the value of height that separates the bottom 40% of data from the top 60% is 77.995.
Part c
Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
We can find the individual probability like this:
[tex]P(\bar X >78.5)=P(Z>\frac{78.5-80.5}{\frac{9.9}{\sqrt{81}}}=-1.82)[/tex]
And using the complement rule a calculator, excel ir the normal standard table we have that:
[tex]P(Z>-1.82)=1-P(Z<-1.82) = 1-0.0344=0.966[/tex]
