Question

The Undergraduate grade point average (UGPA) of students taking the Law School Admissions Test in recent year can be approximated with a normal distribution with mean=3.36 and standard deviation=.18

what is the minimum UGPA that will place a student in the top 10%?

Answer 1

Answer:

$a=3.36 +0.816*0.18=3.507$

The value of height that separates the bottom 90% of data from the top 10% is 3.507.  

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 grades of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.36,0.18)$  

Where $\mu=3.36$ and $\sigma=0.18$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.10$   (a)

$P(X<a)=0.90$   (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 0.90 of the area on the left and 0.10 of the area on the right it's z=0.816. On this case P(Z<0.816)=0.90 and P(Z>0.816)=0.1

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90$  

$P(z<\frac{a-\mu}{\sigma})=0.90$

But we know which value of z satisfy the previous equation so then we can do this:

$z=0.816=\frac{a-3.36}{0.18}$

And if we solve for a we got

$a=3.36 +0.816*0.18=3.507$

So the value of height that separates the bottom 90% of data from the top 10% is 3.507.