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A and B are graphical representations of normally distribute
[#permalink]
30 Jul 2018, 09:06

1

Expert Reply

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Question Stats:

Attachment:

#GREpracticequestion A and B are graphical representations of normally.PNG [ 79.62 KiB | Viewed 2530 times ]

A and B are graphical representations of normally distributed random variables X and Y, respectively, with relative positions, shapes, and sizes as shown. Which of the following must be true?

Indicate all such statements.

A. Y has a greater standard deviation than X.

B. The probability that Y falls within 2 standard deviations of its mean is greater than the probability that X falls within 2 standard deviations of its mean.

C. Y has a greater mean than X.

_________________

Sandy

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Re: A and B are graphical representations of normally distribute
[#permalink]
23 Aug 2018, 09:22

1

The dataset in B is more flat than A so S.D of B> S.D of A

Probability range for all SD in a normal distribution is fixed at 34%, 14% and 2% above and below the mean for 1st, 2nd and 3rd SD

The value directly beneath the protruded part of the curve is the mean. Clearly from the two graphs mean of y>mean of x

Posted from my mobile device

_________________

Probability range for all SD in a normal distribution is fixed at 34%, 14% and 2% above and below the mean for 1st, 2nd and 3rd SD

The value directly beneath the protruded part of the curve is the mean. Clearly from the two graphs mean of y>mean of x

Posted from my mobile device

_________________

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Re: A and B are graphical representations of normally distribute
[#permalink]
13 Jun 2019, 07:51

amorphous wrote:

The dataset in B is more flat than A so S.D of B> S.D of A

Probability range for all SD in a normal distribution is fixed at 34%, 14% and 2% above and below the mean for 1st, 2nd and 3rd SD

The value directly beneath the protruded part of the curve is the mean. Clearly from the two graphs mean of y>mean of x

Posted from my mobile device

Probability range for all SD in a normal distribution is fixed at 34%, 14% and 2% above and below the mean for 1st, 2nd and 3rd SD

The value directly beneath the protruded part of the curve is the mean. Clearly from the two graphs mean of y>mean of x

Posted from my mobile device

------------------------------

Need little more explication of the option C:

Flatter graph has highly dispersed data while the narrower one is not- meaning most of its values are centred around the mean.If the values has low dispersion, then they must have a mean closer to centralised values. But the highly dispersed values pull down the mean (?),therefore small mean.

Is the above correct ?

Re: A and B are graphical representations of normally distribute
[#permalink]
20 Jun 2019, 19:06

1

Any explanations for why option C is correct?

Do we consider the X-axis value of the protruded parts in A and B curves? If yes, then C is true.

Thanks!

Do we consider the X-axis value of the protruded parts in A and B curves? If yes, then C is true.

Thanks!

Re: A and B are graphical representations of normally distribute
[#permalink]
21 Jun 2019, 18:17

1

Expert Reply

The mean of a normal curve is the point along the horizontal axis below the “peak” of the curve. Red and blue lines, respectively

B is to the right of the highest point of curve A, so the mean of Y is larger than the mean of X.

Ask for further assistance.

#GREpracticequestion A and B are graphical representations of normally.PNG [ 77.73 KiB | Viewed 2521 times ]

Regards

_________________

B is to the right of the highest point of curve A, so the mean of Y is larger than the mean of X.

Ask for further assistance.

Attachment:

#GREpracticequestion A and B are graphical representations of normally.PNG [ 77.73 KiB | Viewed 2521 times ]

Regards

_________________

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Re: A and B are graphical representations of normally distribute
[#permalink]
16 Oct 2019, 01:28

Carcass wrote:

The mean of a normal curve is the point along the horizontal axis below the “peak” of the curve. Red and blue lines, respectively

B is to the right of the highest point of curve A, so the mean of Y is larger than the mean of X.

Ask for further assistance.

Regards

B is to the right of the highest point of curve A, so the mean of Y is larger than the mean of X.

Ask for further assistance.

Attachment:

#GREpracticequestion A and B are graphical representations of normally.PNG

Regards

do not get it, please elaborate again

_________________

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Re: A and B are graphical representations of normally distribute
[#permalink]
20 Dec 2020, 07:40

Why B is not the answer?

Re: A and B are graphical representations of normally distribute
[#permalink]
20 Dec 2020, 07:58

Expert Reply

daina1323031 wrote:

Why B is not the answer?

B is to the right of the highest point of curve A, so the mean of Y is larger than the mean of X.

See my explanation and graph above

regards

_________________

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Questions' Banks and Collection:

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Re: A and B are graphical representations of normally distribute
[#permalink]
21 Dec 2020, 16:45

1

Expert Reply

daina1323031 wrote:

Why B is not the answer?

B. The probability that Y falls within 2 standard deviations of its mean is greater than the probability that X falls within 2 standard deviations of its mean.

For ALL normal distributions, the probability that a selected value falls within two standard deviations of the mean is ALWAYS 96% (14% + 34% + 34% + 14% = 96%)

We know this because, for all normal distributions, 96% of the population is within two standard deviations of the mean of the population.

Since the probability that Y falls within 2 standard deviations of its mean EQUALS the probability that X falls within 2 standard deviations of its mean, statement B is false.

_________________

gmatclubot

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