Respuesta :
Answer:
[tex]a=150.7 -0.385*10.2=146.773[/tex]
So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.
Step-by-step explanation:
1) 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".
2) Solution to the problem
Let X the random variable that represent the scores on the LSAT of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(150.7,10.2)[/tex]
Where [tex]\mu=150.7[/tex] and [tex]\sigma=10.2[/tex]
We want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.65[/tex] (a)
[tex]P(X<a)=0.35[/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 0.35 of the area on the left and 0.65 of the area on the right it's z=-0.385. On this case P(Z<-0.385)=0.35 and P(Z>-0.385)=0.65
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.35[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.35[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]Z=-0.385<\frac{a-150.7}{10.2}[/tex]
And if we solve for a we got
[tex]a=150.7 -0.385*10.2=146.773[/tex]
So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.
