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# The mean score in a certain test is 80. If the test scores have a norm

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Senior Manager
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The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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31 Jul 2009, 20:08
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Question Stats:

53% (01:31) correct 47% (01:01) wrong based on 90 sessions

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The mean score in a certain test is 80. If the test scores have a normal distribution and the standarad distribution is 7, what percent of examinees score between 66 and 94?

A) 50%
B) 68%
C) 82%
D) 96%
E) 98%
[Reveal] Spoiler: OA

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Senior Manager
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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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02 Aug 2009, 20:30
Thank you.

I read through the description now i get it.

The percentages 34: 14 : 2 corresponds to the 1st , 2nd and 3rd standard deviation on each side of the mean.

So our case 66 and 94 are two std deviation away from the mean.

14 + 34 + 34 + 14 = 96%

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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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02 Aug 2009, 22:41
tkarthi4u wrote:
Thank you.

I read through the description now i get it.

The percentages 34: 14 : 2 corresponds to the 1st , 2nd and 3rd standard deviation on each side of the mean.

So our case 66 and 94 are two std deviation away from the mean.

14 + 34 + 34 + 14 = 96%

Hi,
Can you please explain how did you get that?

Thanks

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Manager
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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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07 Aug 2009, 07:27
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Normal distribution says

1 sigma = 68% (34% on either side) population
2 sigma = 96% (14% on either side) population so 68% +14%x 2=96%
3 sigma = 99.99% (2% on either side) population so 96% + 2% x 2 ~ 99.99 %
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Manager
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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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16 Sep 2009, 08:23
Try wikipedia to make it clear.

http://en.wikipedia.org/wiki/Normal_distribution
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Manager
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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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19 Sep 2009, 07:06
The mean score in a certain test is 80. If the test scores have a normal distribution and the standarad distribution is 7, what percent of examinees score between 66 and 94?

A) 50%
B) 68%
C)82%
D)96%
E)98%

Soln: Here mean = 80
and standard deviation sig = 7
Since for a range of scores from 66 to 94, the scores vary between 2 * sig about mean
that is from -2 * sig to +2 * sig
for this range with normal distribution the percent of examinees is
= 13.6 + 34.1 + 34.1 + 13.6
= 95.4
Approximately 96%
Ans is D

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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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18 May 2011, 00:28
interesting concept.

73 - 80 and 80-87 = 1 sigma = 34*2

66-73 and 87- 94 = 2 sigma = 14 * 2

total 96%
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Manager
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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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24 Sep 2015, 02:25
i need more explanation from experts please. will I see this type of Qs in Gmat Test ?

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Manager
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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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07 Nov 2015, 22:41
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This question is just testing standard knowledge of a normal distribution graph.

The mean=80, mean 50% of the data falls above and below this point,

When we consider 1 standard deviation on either side of it, we are cutting the data points of interest down to 34% on each side, or 68% between them

As for the question we are being asked about the range +/- 2 SD from the mean. This tests us just knowing that this represents 96% of the data of the data points, or 48% on either side of the mean.

Hope this helps someone

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Re: The mean score in a certain test is 80. If the test scores have a norm [#permalink]

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28 Nov 2015, 10:31
normal distribution represents the bell curve, in which 50% is above mean, and 50% below mean
34% lies within 1 standard deviation or mean+SD and mean-SD
mean+-SD is 73 and 87.
68% of people have scores between 73 and 87
now, aprox. 14% is between (mean+2SD)-(Mean+SD), this is the same for mean minus 2 SD.
Mean+2SD=94
Mean-2SD=66

ok, so in total, we have 68% + 28 = aprox. 98%.

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Re: The mean score in a certain test is 80. If the test scores have a norm   [#permalink] 28 Nov 2015, 10:31
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