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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
How should I proceed with this kind of qustions?
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
Ummm does that mean 2% of students scored (92+12)= 104 out of 100? ... I think there is something wrong with the question
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
1
Let sd be the standard deviation, so 3 sd = (92 - 10) (Think why ?)

sd = 82 / 3

The average would be 92 -sd = 92 - (82/3) which would be less than 87
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project




The scores for the 500 students who took Ms. Johnson’s final exam have a normal distribution. There are 80 students who scored at least 92 points out of a possible 100 total points and 10 students who scored at or below 56.

Quantity A
Quantity B
The average (arithmetic mean) score on the final exam
87


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.

answer Is B
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project




The scores for the 500 students who took Ms. Johnson’s final exam have a normal distribution. There are 80 students who scored at least 92 points out of a possible 100 total points and 10 students who scored at or below 56.

Quantity A
Quantity B
The average (arithmetic mean) score on the final exam
87


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.

answer Is B
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
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Good One. M + SD = 92 and M - 2SD = 56, we need to find out M from here.
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
Could you please elaborate the ans again? the given answer seems much abstruse.
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
Can someone please explain how 16% (92 or more) is displayed in the bell curve at M+SD position? And why not at M-SD position (because this marks 16%)?
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
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Thank you Sir for the stunning explanation.

Regards
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
Carcass wrote:
OE

Attachment:
deviation.jpg


Quote:
Remember that a normal distribution curve has divisions of 34 percent, 14 percent, and 2 percent on each side of the mean. 80 out of 50 is 16 percent, or 14 percent + 2 percent, and 10 out of 500 is 2 percent. Draw a normal distribution curve and label it. There are three standard deviations between 92 and 56, so 92 - 56 = 36, and 36 ÷ 3 = 12. The mean is 92 - 12 = 80, which is smaller than Quantity B.




Since this is a normal distribution, After 92 the next option would be 92+12 = 104(which is not possible)

Am I missing something?
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
It's better to memorize the normal distribution percentage: 2%, 14% & 34% then 34%,14%,2%
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
this one took me 4 mintues but I got it right, should I skip it on the actual test?
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Re: The scores for the 500 students who took Ms. Johnsons fina [#permalink]
Why 12 is added to each value? I got confused
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Re: The scores for the 500 students who took Ms. Johnsons fina [#permalink]
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Pforpeshal wrote:
Why 12 is added to each value? I got confused


I suggest you solve it this way: https://gre.myprepclub.com/forum/the-score ... tml#p33847
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Re: The scores for the 500 students who took Ms. Johnsons fina [#permalink]
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Re: The scores for the 500 students who took Ms. Johnsons fina [#permalink]
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