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Re: The graph represents the normally distributed scores on a te [#permalink]
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GreenlightTestPrep wrote:
sandy wrote:
Attachment:
Capture.PNG


The graph represents the normally distributed scores on a test. The shaded area represents approximately 68% of the scores.

Quantity A
Quantity B
The mean score on the test
\(550\)


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.


Given: The shaded area represents approximately 68% of the scores.
In other words, 68% of the area under the curve is shaded.
However, even though it LOOKS like the shaded area is SYMMETRICAL about the mean, we aren't told that. So we cannot assume that that is the case.

If the shaded area IS symmetrical about the mean,....
Image
....then the mean is 550, in which case the two quantities are EQUAL.


However, it could also be the case that the shaded area is NOT symmetrical about the mean
Image
For example, in the above diagram, 68% of the area under the curve is shaded. HOWEVER, it is not symmetrical about the mean.
In this case, the mean could be something like 530, in which case Quantity B is greater than Quantity A

Answer: D

Cheers,
Brent


Hi, thanks for the explanation. Could you also please guide me as to which chapter/section this question comes under? I am pretty weak at questions like this and could brush up on the basics for this.
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Re: The graph represents the normally distributed scores on a te [#permalink]
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This question is under statistics.
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The graph represents the normally distributed scores on a te [#permalink]
I thought graphs are drawn to scale in the GRE. Is it not true for normal distribution graphs? Please help.
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Re: The graph represents the normally distributed scores on a te [#permalink]
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The graph showed above by Brent is to explain how it works to deliver the OA which is D

However, the distribution is just a bell curve. it does not need to drawn or not the scale

regards
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Re: The graph represents the normally distributed scores on a te [#permalink]
godxyz wrote:
I thought graphs are drawn to scale in the GRE. Is it not true for normal distribution graphs? Please help.


The graphs that are drawn to scale either usually have detailed x-axis and y-axis values alongside them to pinpoint all the necessary values, or stated in the question itself. However, in this question, it is not stated and we are only given 2 y-axis values, not enough data to make sure the mean is ACTUALLY in the middle of 500 and 600.
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Re: The graph represents the normally distributed scores on a te [#permalink]
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In the GRE, graphs are not drawn to scale. Only 1 section: coordinate geometry is drawn to scale.

Also, the normal distribution curve's centre-line is always the median, but not necessarily the mean (to piggyback on Bren't explanation).
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Re: The graph represents the normally distributed scores on a te [#permalink]
Hi, I have a doubt on this, the question mentioned that it is "Normal Distribution". When its normally distributed doesn't it mean that the data is symmetric by default? Because normal distribution questions rarely say they are symmetric with mean, they just read like " normally distributed with mean of 67 and deviation of 8". So in my example does it mean, that distribution is symmetric with mean? If not what could the wording be like? because I haven't any question specifically saying symmetric with/around mean
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Re: The graph represents the normally distributed scores on a te [#permalink]
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JaxHammer wrote:
Hi, I have a doubt on this, the question mentioned that it is "Normal Distribution". When its normally distributed doesn't it mean that the data is symmetric by default? Because normal distribution questions rarely say they are symmetric with mean, they just read like " normally distributed with mean of 67 and deviation of 8". So in my example does it mean, that distribution is symmetric with mean? If not what could the wording be like? because I haven't any question specifically saying symmetric with/around mean


If we're told we have a normal distribution, then we can also conclude that the distribution is symmetric with the mean in the middle.
In other words: normal distribution = symmetric about the mean
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The graph represents the normally distributed scores on a te [#permalink]
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Hi, then as per this question why cant we say that mean/median is 550 ie., option C as question says normally distributed scores, which means LHS and RHS are the same, as per numbers 550 sets perfectly,

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Re: The graph represents the normally distributed scores on a te [#permalink]
Carcass sir, if it is given that graph is normally distributed thus total 68% will lie between 1SD of mean only ,
then how we can take "NOT symmetrical" case as taken by GreenlightTestPrep ?
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Re: The graph represents the normally distributed scores on a te [#permalink]
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In other words: normal distribution = symmetric about the mean
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Re: The graph represents the normally distributed scores on a te [#permalink]
Carcass wrote:
In other words: normal distribution = symmetric about the mean


So Carcass Sir , answer will be C then ? because as per GreenlightTestPrep it is D
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Re: The graph represents the normally distributed scores on a te [#permalink]
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OE

While the shaded area may appear to be evenly located on either side of the mean, it isn’t necessarily. For example, the 68% could be more lopsided, like so


Attachment:
GRe SD (2).jpg
GRe SD (2).jpg [ 22.5 KiB | Viewed 583 times ]



This area could still represent 68% of the scores, even if it’s not 1 standard deviation to either side of the mean. In order to determine that the mean is 550, the problem would need to state explicitly that 500 and 600 each represent 1 standard deviation from the mean (or at least that 500 and 600 are equally far from the mean). The fact that 68% of the data is located between 500 and 600 is a trick implying that 500 and 600 are –1 and +1 standard deviation from the mean, but this is not necessarily true. While it is always true that, in a normal distribution, about 68% (some people memorize the approximation as two-thirds) of the data is within 1 standard deviation of the mean, the reverse is not true: do not assume that any chunk of data that is about 68% of the whole is therefore within 1 standard deviation of the mean.
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