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normal distribution [#permalink] New post 07 Dec 2009, 18:08
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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-1/8
b-1/9
c-1/6
d-4/17
e-4/21
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Re: normal distribution [#permalink] New post 07 Dec 2009, 21:46
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1/9 I think. This could be kinda tricky even if you know the normal distribution, bc of the estimation. I did it this way. 1 std deviation away to the left would be all the people who got less than 440 in group A, which means 14% (1 std below)+2% (2 std below)=16% in the Group A. Now for Group B, it's two std deviations to the left. So about 2% in group B. Assume there are 100 people in each group. 16% of Group A would give you 16 people below 440, and 2% of group B would give you 2 people in Group B who scored below 440.

To find the fraction of students below 440 from B ====> Group B/(Total)

===> 2/(2+16)

Reduces to 1/9
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Re: normal distribution [#permalink] New post 09 Dec 2009, 03:30
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There is no way to answer this question without consulting a stats table. Since that's impossible during a GMAT, it's not a realistic question. I know that the source gives an OA of B for this question, but that is the wrong answer. While you don't need to know this for the GMAT, when data is normally distributed, approximately 68.3% of values are within one standard deviation of the mean, and approximately 95.4% are within two standard deviations of the mean. If you round these values off to 68% and 95%, you'll get the answer given by the source. This rounding affects the answer, however; if you use exact values, you'll find that the correct answer to this question is closer to 1/8 than to 1/9. It's a very poorly designed question, and completely out of the scope of GMAT statistics, so it's best to ignore this question (and similar questions from the same source) and move on to better materials.
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Re: normal distribution [#permalink] New post 18 May 2011, 00:39
got the concept now.

in A between 440-460 34% in b between 480-520 34%
in A between 420-440 14% in B between 440-480 14%

in A less than 420 2% in B less than 440 2%

hence (14+2) in A the percentage of people
and 2 in B the percentage of people having scores <440

thus ratio = 2/ (16+2) = 1/9 B
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Re: normal distribution   [#permalink] 18 May 2011, 00:39
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