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Jane scored in the 68th percentile on a test, and John score

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Jane scored in the 68th percentile on a test, and John score [#permalink] New post 06 Apr 2018, 06:47
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75% (00:11) correct 25% (00:15) wrong based on 4 sessions
Jane scored in the 68th percentile on a test, and John scored in the 32nd percentile.

Quantity A
Quantity B
The proportion of the class that received a score less than John's score
The proportion of the class that scored equal to or greater than Jane's score



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

Last edited by Carcass on 06 Apr 2018, 07:53, edited 1 time in total.
Edited by Carcass
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Re: Jane scored in the 68th percentile on a test, and John score [#permalink] New post 06 Apr 2018, 07:56
Expert's post
my solving
but what I thought is that if there are 100 people, and score is distributed normally(1 to 100)
John's score is 32. so less than 32 is 31(31 people)
Jane's score is 68. so equal to or greater than 68 is 32(32 people with Jane)
than Quantity can be greater, but what's the problem of my solving?


this is answer from book.
Percentiles define the proportion of a group that scores below a particular benchmark. Since John scored in the 32nd percentile, by definition, 32% of the class scored worse than John. Quantity A is equal to 32%
Jane scored in the 68th percentile, so 68% of the class scored worse than she did. Since 100 - 68 = 32, 32% of the class scored equal to or greater than Jane. Quantity B is also equal to 32%
official answer is C (so Quantity A is equal to Quantity B)
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Joined: 18 Apr 2015
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Re: Jane scored in the 68th percentile on a test, and John score [#permalink] New post 06 Apr 2018, 08:05
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Please follow the rules for posting on the board

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Re: Jane scored in the 68th percentile on a test, and John score   [#permalink] 06 Apr 2018, 08:05
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Jane scored in the 68th percentile on a test, and John score

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