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For each student in a certain class, a teacher adjusted the student’s [#permalink]

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12 Jun 2017, 15:23

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For each student in a certain class, a teacher adjusted the student’s test score using the formula y = 0.8x + 20, where x is the student’s original test score and y is the student’s adjusted test score. If the standard deviation of the original test scores of the students in the class was 20, what was the standard deviation of the adjusted test scores of the students in the class?

Re: For each student in a certain class, a teacher adjusted the student’s [#permalink]

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12 Jun 2017, 21:10

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When all the terms of a set are increased/decreased by a certain percent, then the Std Deviation also increases/decreases by the same percent. If all the terms of a set are increased/decreased by a certain constant, the Std Deviation does not change.

Here y = 0.8x + 20

So all the original scores are being multiplied by 0.8 (thus Std Dev will also be multiplied by 0.8), and then are increased by 20 (which will have NO further effect on Std Dev).

Re: For each student in a certain class, a teac [#permalink]

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18 Sep 2017, 13:22

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amanvermagmat wrote:

When all the terms of a set are increased/decreased by a certain percent, then the Std Deviation also increases/decreases by the same percent. If all the terms of a set are increased/decreased by a certain constant, the Std Deviation does not change.

Here y = 0.8x + 20

So all the original scores are being multiplied by 0.8 (thus Std Dev will also be multiplied by 0.8), and then are increased by 20 (which will have NO further effect on Std Dev).

So the new Std Deviation = 0.8*20 = 16.

Hence B answer

Standard deviation of adjusted score = 0.8 * 20 = 16

Re: For each student in a certain class, a teacher adjusted the student’s [#permalink]

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19 Sep 2017, 15:14

x is the student's original score. So how can we define directly that X= std deviation ?

amanvermagmat wrote:

When all the terms of a set are increased/decreased by a certain percent, then the Std Deviation also increases/decreases by the same percent. If all the terms of a set are increased/decreased by a certain constant, the Std Deviation does not change.

Here y = 0.8x + 20

So all the original scores are being multiplied by 0.8 (thus Std Dev will also be multiplied by 0.8), and then are increased by 20 (which will have NO further effect on Std Dev).

Re: For each student in a certain class, a teacher adjusted the student’s [#permalink]

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16 Oct 2017, 17:54

gmat75016 wrote:

x is the student's original score. So how can we define directly that X= std deviation ?

amanvermagmat wrote:

When all the terms of a set are increased/decreased by a certain percent, then the Std Deviation also increases/decreases by the same percent. If all the terms of a set are increased/decreased by a certain constant, the Std Deviation does not change.

Here y = 0.8x + 20

So all the original scores are being multiplied by 0.8 (thus Std Dev will also be multiplied by 0.8), and then are increased by 20 (which will have NO further effect on Std Dev).

So the new Std Deviation = 0.8*20 = 16.

Hence B answer

I actually took a way longer route... maybe it will help

If the teacher would change every single grade, he would use the same formula for every student and would get the standard deviation at the end. So every score would change after going through the same formula.

for instance, let's say there were 3 students in the class and the grades were 20, 40 and 60. if we substitute every x for the grade we will get:

1) grade 20 y = 0.8 x 20 + 20 = 36

2) grade 40 y = 0.8 x 40 + 20 = 52

3) grade 60 y = 0.8 x 60 + 20 = 68

The adjusted grades are then 36, 52, 68 , with new a standard deviation of 16

Before we add the "20" we will get 16, 32 and 48; with the same standard deviation - 16

So I guess that since the standard deviation is a way to measure the relation between the numbers, by multiplying it by the same variable (0.8) we can adjust right away for this new relation.

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