Z-score question involving LSAT scores

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ericcy

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Approximately 40% of those who take the LSAT score from 145 to 155. About 70% score from 140 to 160.
a) Determine the mean score and standard deviation for the LSAT.
b) Harvard University also uses other methods to choose students for its law school, but the minimum LSAT score that is required is about 172, What percent of people who take the LSAT would be considered by Harvard for admission?

I'm stuck on part a) of the question, and keep getting an answer different to the textbook (their answer is 9.6).

I found the mean correctly, from finding the average of the four numbers (145, 155, 160, 140) and got 150.

For the standard deviation, I know that its equation is the square root of the total of values minus the mean squared, over n.

I found the total (sigma) and got 250, divided by 4 because there are four numbers, and then square rooted the answer to get 7.9.

What am I doing wrong? Please explain in baby steps, haha.
 
Hello, and welcome to FMH! :)

Consulting a \(z\)-score chart, we find that an area of 0.20 corresponds to a \(z\)-score of about 0.52. So this means:

[MATH]\sigma=\frac{x-\mu}{z}=\frac{155-150}{0.52}\approx9.6[/MATH]
 
MarkFL said:
Hello, and welcome to FMH! :)

Consulting a \(z\)-score chart, we find that an area of 0.20 corresponds to a \(z\)-score of about 0.52. So this means:

[MATH]\sigma=\frac{x-\mu}{z}=\frac{155-150}{0.52}\approx9.6[/MATH]
Click to expand...

May I ask why exactly you chose that specific value?
 
ericcy said:
May I ask why exactly you chose that specific value?
Click to expand...

If 40% are from 145-150, then 20% is from the mean of 150 to 155. Recall the normal curve is symmetric about the mean. And so the area under the normal curve is 0.20, from the mean to the \(x\)-value of 155.
 
MarkFL said:
If 40% are from 145-150, then 20% is from the mean of 150 to 155. Recall the normal curve is symmetric about the mean. And so the area under the normal curve is 0.20, from the mean to the \(x\)-value of 155.
Click to expand...

So we use the first area from the mean to find the standard deviation?
 
ericcy said:
So we use the first area from the mean to find the standard deviation?
Click to expand...

We can use the area under the curve from the mean to a known \(x\)-value to find the standard deviation \(\sigma\), using the \(z\)-score. The \(z\)-score we get from the chart (using the given area) tells us how many standard deviations from the mean the \(x\)-value is.
 
MarkFL said:
We can use the area under the curve from the mean to a known \(x\)-value to find the standard deviation \(\sigma\), using the \(z\)-score. The \(z\)-score we get from the chart (using the given area) tells us how many standard deviations from the mean the \(x\)-value is.
Click to expand...

So hypothetically, we could use any number in the question, for example 140, but then we'd have to use a p value of 0.35 to find the z-score. Correct? Also, was my method of finding the mean correct, just finding the average?
 
I have a couple concerns about the problem.

First, I have to guess that you have been told to assume a normal distribution.

Second, I see no reason to believe that the intervals they gave you happen to be centered around the mean; you are only assuming that. (They could have given you the probabilities for any two intervals!)

Third, you will get slightly different standard deviations based on the 40% interval and the 70% interval (assuming the mean is 150); the data are inconsistent under your assumption. If I make the 70% probability correct, the 40% is not.

In principle, I suppose you could solve for the two values given the two probabilities; but that is not possible algebraically. I took some time to experiment with this (using Goal Seek in Excel) and found that no combination of mean and sd yields both probabilities exactly. I can, however, get other approximations; for example, mean of 155 and sd 8.065 gives probabilities 0.392 and 0.700 (only a little farther off than 150 and 9.65).

As an aside, MarkFL is using one type of normal probability chart, which gives the probability between the mean and a given z; many (including Excel) give the probability less than a given z. You may want to tell us what you are using for your normal distribution.
 
ericcy said:
So hypothetically, we could use any number in the question, for example 140, but then we'd have to use a p value of 0.35 to find the z-score. Correct? Also, was my method of finding the mean correct, just finding the average?
Click to expand...

As Dr. Peterson pointed out, we had to make some assumptions here, one of which is that the given intervals are centered abut the mean. Of course, the other is that the scores are normally distributed.
 
Dr.Peterson said:
As an aside, MarkFL is using one type of normal probability chart, which gives the probability between the mean and a given z; many (including Excel) give the probability less than a given z. You may want to tell us what you are using for your normal distribution.
Click to expand...

I'm using a z-score chart.
 
MarkFL said:
Hello, and welcome to FMH! :)

Consulting a \(z\)-score chart, we find that an area of 0.20 corresponds to a \(z\)-score of about 0.52. So this means:

[MATH]\sigma=\frac{x-\mu}{z}=\frac{155-150}{0.52}\approx9.6[/MATH]
Click to expand...

I'm actually just trying this equation now, from my chart I got a z-score of -0.84 which gives me a standard deviation of 5.9
 
I think you probably are in fact using the kind of table that shows P(Z < z), that is, the probability of any z score less than a given value, not MarkFL's kind that shows that shows P(|Z| < z), that is the probability of z being between 0 (the mean) and a given value. So what you found is that the probability that z is less than -0.84 is 0.2. That isn't what you want.

If you use that kind of table, then the probability you have to look up is not 0.20, but that plus 0.50. Can you see why?
 
Dr.Peterson said:
I think you probably are in fact using the kind of table that shows P(Z < z), that is, the probability of any z score less than a given value, not MarkFL's kind that shows that shows P(|Z| < z), that is the probability of z being between 0 (the mean) and a given value. So what you found is that the probability that z is less than -0.84 is 0.2. That isn't what you want.

If you use that kind of table, then the probability you have to look up is not 0.20, but that plus 0.50. Can you see why?
Click to expand...

Because it's 70 percent, not 40?
 
If 20% is between the mean and 155, then 70% will be to the left of 155 because 50% is to the left of the mean.
 
MarkFL said:
If 20% is between the mean and 155, then 70% will be to the left of 155 because 50% is to the left of the mean.
Click to expand...
So for my table, I'd have to use the percentile of 0.70, not 0.20? Also, do we always have to choose the number to the right of the mean, why couldn't I use 145, or even 160?
 
Does your table give areas that are less than 0.5?
 
Okay, then you would subtract half the given area from 0.5 if you want to use the lower bound.
 
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