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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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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
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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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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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Re: Jane scored in the 68th percentile on a test, and John score [#permalink] New post 15 Dec 2018, 22:12
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Carcass wrote:
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)



Help on this

Plz if someone can guide me on this :| :

My reasoning,

Since it is a normally distributed curve, and if I consider with the total of 100 students,

so the number of students from 32 to 68 = 50 i.e if 50 is taken as mean then 18 will be 1 SD below = 32 and 18 will 1 SD above mean = 68

i.e. 68% consist of the students whose score is in between John and jane

and the rest 32% will consist the score that are either below or above the score of John and Jane. i.e. half of 32% i.e. 16% will be below John's score and the other 1/2 i.e. 16% will be above Jane's score.

Hence the two quantities will be equal
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Re: Jane scored in the 68th percentile on a test, and John score [#permalink] New post 16 Dec 2018, 11:12
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Jane scored in the 68th percentile on a test, and John scored in the 32nd percentile.
Jane is on 68th percentile
so difference from mean to percentile is 18

John e is on 32nd percentile
so difference from mean to percentile is 18

Quantity A:The proportion of the class that received a score less than John's score
John score is 32. The class received a score less than 32 is 32

Quantity B:The proportion of the class that scored equal to or greater than Jane's score
Jane score is 68. The class received a score greater than 68 is 32

Hence C
Re: Jane scored in the 68th percentile on a test, and John score   [#permalink] 16 Dec 2018, 11:12
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Jane scored in the 68th percentile on a test, and John score

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