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
PR-4 Standard Deviation and Mean
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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.
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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How are you getting the percentage of 34%?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.
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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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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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I have attached a figure of NORMAL DISTRIBUTIONildude02 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.
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.
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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.
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Regards.
PS: - Where are Ian and Stuart when you need them
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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.
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.
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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
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.