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PR-4 Standard Deviation and Mean

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PR-4 Standard Deviation and Mean

by crackitpal » Sun Aug 10, 2008 10:57 pm
Hello Everybody,

Could anyone explain how to solve following problem?

Two different groups of test-takers received scores on the GXYZ standardized test. Group A's scores had a normal distribution with a mean of 460 and a standard deviation of 20. Group B's scores had a normal distribution with a mean of 520 and a standard deviation of 40. If each group has the same number of test-takers, what fraction of the test-takers who scored below 440 belonged to Group B?

A. 1/9
B. 1/8
C. 1/6
D. 4/17
E. 4/21

A
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by sudhir3127 » Mon Aug 11, 2008 4:16 am
I am not sure if this is tested on GMAT ..but here it goes..

u need to know thw 68-95-99.7 rule of statisitcs( Normal distribution )

consider group A.
Mean 460 SD 20

hence 440-460 is 1 SD hence 34%
420 - 440 - 2SD 14%
400 - 420 - 3 SD 2%

16% below 440

Similarly B
Mean 520 SD 40
480-520 - 34%
440-480 - 14%
400-440 - 2%

2% below 440

hence total who scored below 440 belonged to group B is 2/(16+2)

hence 2/18 = 1/9.

Hope it helps..though u need to be conversant with the rule to follow it.
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by ildude02 » Mon Aug 11, 2008 11:23 am
sudhir3127 wrote:I am not sure if this is tested on GMAT ..but here it goes..

u need to know thw 68-95-99.7 rule of statisitcs( Normal distribution )

consider group A.
Mean 460 SD 20

hence 440-460 is 1 SD hence 34%


Hope it helps..though u need to be conversant with the rule to follow it.
How are you getting the percentage of 34%?
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by sudhir3127 » Mon Aug 11, 2008 11:36 am
i am considering only one side of the deviation ...

hence its 68/2 = 34
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by parallel_chase » Mon Aug 11, 2008 11:38 am
I did exactly like Sudhir3127. You need to to know the bell curve normal distribution 68% - 95% - 99.7% rule, which I would say easy to remember. Just take the mean as the mid point of the curve with 1sd, 2sd & 3sd for both sides.
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by ildude02 » Mon Aug 11, 2008 12:35 pm
Can you brief a little more as to how to apply it. Seems like if a value is within 1 time the SD of the mean, 68% of the the values lie within that range, and 95% for 2 times the SD of the mean and 99.75% for 3 times the SD of the meean. But I'm failing to consider how to apply that to solving a specific question without knowing the number of terms to find the percentage of it. Appreciate your response in a little elaborate way.
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by parallel_chase » Mon Aug 11, 2008 2:57 pm
ildude02 wrote:Can you brief a little more as to how to apply it. Seems like if a value is within 1 time the SD of the mean, 68% of the the values lie within that range, and 95% for 2 times the SD of the mean and 99.75% for 3 times the SD of the meean. But I'm failing to consider how to apply that to solving a specific question without knowing the number of terms to find the percentage of it. Appreciate your response in a little elaborate way.
I have attached a figure of NORMAL DISTRIBUTION

GROUP A mean = 460, SD=20

if you look at the figure mean is the mid point, therefore we only have to consider the values, which are on the left side of the graph.

460-20 = 440 1st SD , 460--440=34%
440-20 = 420 2nd SD, 440--420=14%
420-20 = 400 3rd SD, 420--400=2%

Similarly, GROUP B mean=520, SD=40
520-40=480 1st SD, 520--480=34%
480-40=440 2nd SD, 480--440=14%
440-40=400 3rd SD, 440--400=2%

The question is asking what fraction of group B is out of total students who scored below 440

Scored below 440
Group A = 14% + 2% = 16%
Group B = 2%

Fraction = 2/(16+2) = 2/18 = 1/9

Hope its clear.
Attachments
SD.jpg
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by ildude02 » Mon Aug 11, 2008 3:16 pm
thanks a bunch, parallel_chase. So I would assume the 34%, 14% and 2% distribution is always the same no matter what the mean value is, for 1SD, 2SD and 3SD respectively. Correct me if I'm wrong.
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by gabriel » Tue Aug 12, 2008 5:01 am
While I too could not think of an alternative way to solve this question without using the formula you guys discussed, the point is that if this is a GMAT question then it should be solvable without the use of the knowledge of the Chebyshev formula. So if you guys have any alternative method to solve it please post it. Also OP can you post the OA to this question.

Regards.

PS: - Where are Ian and Stuart when you need them
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by Ian Stewart » Tue Aug 12, 2008 5:39 am
Sudhir's solution above is perfect, but this is not a realistic GMAT question- it can only be answered if you know the 68-95-99.7 rule of normal distributions, something which is not tested on the GMAT. There is no possible alternative approach (incidentally, it's not Chebyshev's inequality that we use here, and if we did use it, we wouldn't get an exact answer). I could imagine the question being asked if the question told you that the distributions are symmetric, and told you what percent of data points were within one, two and three standard deviations of the mean. But as the question is presented, GMAT test-takers shouldn't worry about it.

edit: I've solved this problem using exact values (the 68/95 rule just gives approximate values) and the answer is not closest to 1/9; it's actually closest to 1/8. So while the approach used above by Sudhir is perfect, logically, the rounding off here affects the answer too much. Clearly the question designer rounded off in the same way, and therefore also got the answer wrong! There is no way one could possibly answer this on the GMAT, however- you'd need to consult a statistics table.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by fc135 » Mon Oct 18, 2010 5:54 pm
Hi Sudhir

Your explanation was really good but could you please let me know how to calculate to get the 34%, 14% and 2 %. Appreciate your help.

Thanks
Manju


sudhir3127 wrote:I am not sure if this is tested on GMAT ..but here it goes..

u need to know thw 68-95-99.7 rule of statisitcs( Normal distribution )

consider group A.
Mean 460 SD 20

hence 440-460 is 1 SD hence 34%
420 - 440 - 2SD 14%
400 - 420 - 3 SD 2%

16% below 440

Similarly B
Mean 520 SD 40
480-520 - 34%
440-480 - 14%
400-440 - 2%

2% below 440

hence total who scored below 440 belonged to group B is 2/(16+2)

hence 2/18 = 1/9.

Hope it helps..though u need to be conversant with the rule to follow it.
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