Respuesta :
Answer:
[tex]a=620 -0.842*30=594.74[/tex]
So the value of height that separates the bottom 20% of data from the top 80% is 594.74.
[tex]a=620 +0.842*30=645.26[/tex]
So the value of height that separates the bottom 80% of data from the top 20% is 645.26.
[tex] (594.74,645.26)[/tex] we will have the 60% of the weigths
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 weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(620,30)[/tex]
Where [tex]\mu=620[/tex] and [tex]\sigma=30[/tex]
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
We want the 60% middle values so we need to have 20% of the values on each tail, and we want to find the limits so we can do this:
LOWER LIMIT
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.8[/tex] (a)
[tex]P(X<a)=0.2[/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 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(z>-0.842)=0.8
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.2[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.2[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.842<\frac{a-620}{30}[/tex]
And if we solve for a we got
[tex]a=620 -0.842*30=594.74[/tex]
So the value of height that separates the bottom 20% of data from the top 80% is 594.74.
UPPER LIMIT
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-620}{30}[/tex]
And if we solve for a we got
[tex]a=620 +0.842*30=645.26[/tex]
So the value of height that separates the bottom 80% of data from the top 20% is 645.26.
So then the answer for this case would be:
[tex] (594.74,645.26)[/tex] we will have the 60% of the weigths