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oh_nimesh
11-16-2004, 08:09 PM
1. This year the mean score on the GRE’s was 510, with a standard deviation of 102. Compute the Z score and the percentile rank for each of the following GRE scores:

X Z score percentile

270

780

500

410

565

2. What is the probability that a student would have a scored 300 or less or of 700 or more on the GRE this year?

3. What GRE score corresponds to a percentile ranking of 97.5?

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thanks

Econ
11-16-2004, 08:42 PM
Hi nimesh:

First, here is a useful link for the normal distribution: http://davidmlane.com/hyperstat/nor...stribution.html

Second, the only thing that is missing in the file I send you is the word '' way.'' . Sorry for the misconduct.

Now for the question (I guess that the question must also state that the obseravtions are taken from a normal distribution) here are my thoughts:

(1) The z-scores are the found by normalizing the variables, i.e. when x~N(m,s^2), where m is the mean and s is the s.d. then z = (x-m)/s and z~N(0,1). In our case then:
x=210 -------------> z= (210-510) / 102 = - 2.94
and construct the z-values for the rest observations in a similar way.
For the percentiles, we need the standarized table so as to convert the z-value into a percentile value. For God shake, no one will ask you to do this on the real GRE!However, there are some z values that you need to memorize (check the link above for more info), namely:
z=1.96 -----------> 97.5%
z= -1.96----------> -2.5%
z=0-------------> 50%
z=1 ------------> ~84.1%
z=-1----------> ~16%
z=1.69 ---------> 95%

{ Keep in mind that the normal dist is a symmetic one}

(2) It helps a lot if you plot the standard normal dist (again check the link). First convert the values into z-values, i.e. X1=300 ----------> Z1 = -2.06 and X2= 700 -----------> Z2 = 1.86. Now check what you are asked for:
P ( X<300 and X>700) that is equivalent to P (z<-2.06 and z>1.86).
Having in mind the figure you see that the desired probability is the sum of the two areas; the one on the left of z=-2.06 and the one on the right of z=1.86 {Again you must have the standarized tables -- however, to get a picture, if the two z values were z=-1.96 and z=1.96 then the probability would have been 2.5% + 2.5%=5%).

P.S. The equality (X<=300) is irrelevant because the distribution is continuous and the probability of a specific point occuring is always zero (because there are infinite points).

(3) Well, here you are expected to know that the 97.5% corresponds to the z-value 1.96. So solve z= (x-m) / s w.r.t x, i.e. 1.96 = (x-510) / 102 -----> x= 709.92

Hope I helped :)

777
11-17-2004, 12:57 AM
Hey econ,

I tried to open that link and I couldn't. Can you recommend anything else?

Dingus
11-17-2004, 07:46 AM
Hey 777!

Try this one:--> http://davidmlane.com/hyperstat/normal_distribution.html

The software parsed the earlier link.

You can also read this thread started by Econ:

Also, try searching for "normal distribution" in the forum, you will come up with many threads discussing questions on the topic.

krakow
11-17-2004, 11:52 AM
(1) The z-scores are the found by normalizing the variables, i.e. when x~N(m,s^2), where m is the mean and s is the s.d. then z = (x-m)/s and z~N(0,1). In our case then:
x=210 -------------> z= (210-510) / 102 = - 2.94
and construct the z-values for the rest observations in a similar way.

Hi!

why x=210?

Econ
11-17-2004, 01:07 PM
For no reason...I just miscopied the question....I meant 270....anyway, the intuition does not change

777
11-17-2004, 08:11 PM
Thanks Dingus. I'll definitely try this.

Hey 777!

Try this one:--> http://davidmlane.com/hyperstat/normal_distribution.html

The software parsed the earlier link.

You can also read this thread started by Econ: