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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # The scores for the 500 students who took Ms. Johnson’s fina  Question banks Downloads My Bookmarks Reviews Important topics
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The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
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Question Stats: 57% (02:24) correct 42% (01:46) wrong based on 216 sessions

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.
[Reveal] Spoiler: OA

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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? Founder  Joined: 18 Apr 2015
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
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OE

Attachment: deviation.jpg [ 23.65 KiB | Viewed 12991 times ]

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.

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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 Intern Joined: 29 Jun 2018
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
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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:

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.

Intern Joined: 15 May 2018
Posts: 38
Followers: 0

Kudos [?]: 6 , given: 1

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

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. Intern Joined: 27 Nov 2014
Posts: 30
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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.
Intern Joined: 25 May 2019
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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%)? GRE Instructor Joined: 10 Apr 2015
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
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Expert's post
Carcass wrote:

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

-------ASIDE-----------------------------
A little extra background on standard deviations above and below the mean

If, for example, a set has a standard deviation of 4, then:
1 standard deviation = 4
2 standard deviations = 8
3 standard deviations = 12
1.5 standard deviations = 6
0.25 standard deviations = 1
etc

So, if the mean of a set is 9, and the standard deviation is 4, then:
2 standard deviations ABOVE the mean = 17 [since 9 + 2(4) = 17]
1.5 standard deviations BELOW the mean = 3 [since 9 - 1.5(4) = 3]
3 standard deviations ABOVE the mean = 21 [since 9 + 3(4) = 21]
etc.
-------------------------------------------------

We're told that 16% scored at least 92 points (80/500 = 10%)
When we examine the normal distribution curve, we see that 16% of all values are more than 1 standard deviation ABOVE the mean. So, we know that a test score of 92 points is 1 standard deviation ABOVE the mean.
So, if m = the mean of all the scores
and if d = the standard deviation of all the scores...
We can write: m + d = 92

We're also told that 2% scored less than 56 points (10/500 = 2%)
When we examine the normal distribution curve, we see that 2% of all values are more than 2 standard deviations BELOW the mean. So, we know that a test score of 56 points is 2 standard deviations BELOW the mean.
So, if m = the mean of all the scores
and if d = the standard deviation of all the scores...
We can write: m - 2d = 56

So, we now know two things:
m + d = 92
m - 2d = 56

Take the TOP equation, and multiply both sides by 2 to get:
2m + 2d = 184
m - 2d = 56

ADD the two equations to get: 3m = 240
Solve: m = 80

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course. Founder  Joined: 18 Apr 2015
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Re: The scores for the 500 students who took Ms. Johnson’s fina [#permalink]
Expert's post
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? Re: The scores for the 500 students who took Ms. Johnson’s fina   [#permalink] 06 Apr 2020, 13:38
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