Question

The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen's School of Business in Kingston, Ontario requires a minimum LSAT score of 157 for admission. Find the 35th percentile of the LSAT scores. Give your answer accurate to one decimal place. Use the applet. (Example: 124.7) Your Answer:

Answer 1

Answer:

108 rooms

Step-by-step explanation:

Answer 2

Answer:

$a=150.7 -0.385*10.2=146.773$

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:

$X \sim N(150.7,10.2)$  

Where $\mu=150.7$ and $\sigma=10.2$

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

$P(X>a)=0.65$   (a)

$P(X$   (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:

$P(X$  

$P(z$

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

$Z=-0.385$

And if we solve for a we got

$a=150.7 -0.385*10.2=146.773$

So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.