Respuesta :
Answer:
a)[tex] P( X <40) =P(Z< \frac{40-42}{5.5}) =P(Z<-0.363)=0.3583[/tex]
We want this probability:
[tex] P( X >64)[/tex]
And using the z score formula given by:
[tex] z = \frac{x -\mu}{\sigma}[/tex]
We got:
[tex] P( X >64) =P(Z> \frac{64-42}{5.5}) =P(Z>4)=0.0000316[/tex]
b) For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.25[/tex] (a)
[tex]P(X<a)=0.75[/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 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.75[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.75[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.674<\frac{a-42}{5.5}[/tex]
And if we solve for a we got
[tex]a=42 +0.674*5.5=45.707[/tex]
So the value of height that separates the bottom 75% of data from the top 25% is 45.707.
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 heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(42,25.5)[/tex]
Where [tex]\mu=42[/tex] and [tex]\sigma=5.5[/tex]
And we want this probability:
[tex] P( X <40)[/tex]
And using the z score formula given by:
[tex] z = \frac{x -\mu}{\sigma}[/tex]
We got:
[tex] P( X <40) =P(Z< \frac{40-42}{5.5}) =P(Z<-0.363)=0.3583[/tex]
We want this probability:
[tex] P( X >64)[/tex]
And using the z score formula given by:
[tex] z = \frac{x -\mu}{\sigma}[/tex]
We got:
[tex] P( X >64) =P(Z> \frac{64-42}{5.5}) =P(Z>4)=0.0000316[/tex]
Part b
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.25[/tex] (a)
[tex]P(X<a)=0.75[/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 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.75[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.75[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.674<\frac{a-42}{5.5}[/tex]
And if we solve for a we got
[tex]a=42 +0.674*5.5=45.707[/tex]
So the value of height that separates the bottom 75% of data from the top 25% is 45.707.
