Answer:
![a=150.7 -0.385*10.2=146.773](https://tex.z-dn.net/?f=a%3D150.7%20-0.385%2A10.2%3D146.773)
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:
Where
and ![\sigma=10.2](https://tex.z-dn.net/?f=%5Csigma%3D10.2)
We want to find a value a, such that we satisfy this condition:
(a)
(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:
![P(z](https://tex.z-dn.net/?f=P%28z%3C%5Cfrac%7Ba-%5Cmu%7D%7B%5Csigma%7D%29%3D0.35)
But we know which value of z satisfy the previous equation so then we can do this:
![Z=-0.385](https://tex.z-dn.net/?f=Z%3D-0.385%3C%5Cfrac%7Ba-150.7%7D%7B10.2%7D)
And if we solve for a we got
![a=150.7 -0.385*10.2=146.773](https://tex.z-dn.net/?f=a%3D150.7%20-0.385%2A10.2%3D146.773)
So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.