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For each student in a certain class, a teacher adjusted the student’s

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

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New post 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?

A. 12
B. 16
C. 28
D. 36
E. 40

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

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New post 19 Oct 2017, 10:29
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AbdurRakib wrote:
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?

A. 12
B. 16
C. 28
D. 36
E. 40


Let’s review two rules for the standard deviation: 1. Adding a constant to each value in a data set does not change the value of the standard deviation; 2. Multiplying each value in a data set by a constant also multiplies the standard deviation by that constant.

Thus, for y = 0.8x + 20, we see that adding 20 does not affect the value of the standard deviation. But multiplying each score by 0.8 means that the standard deviation of the entire data set is also multiplied by 0.8; thus, the standard deviation of the adjusted test scores will be 20 x 0.8 = 16.

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

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New post 12 Jun 2017, 20:48
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Addition of 20 doesn't change the standard deviation. However, multiplying the original score by 0.8 does change the standard deviation.

Standard deviation of original score = 20

Standard deviation of adjusted score = 0.8 * 20 = 16

Answer: B
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New post 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).

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

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

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New post 18 Sep 2017, 13:22
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
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Re: For each student in a certain class, a teacher adjusted the student’s  [#permalink]

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New post 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).

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

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

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New post 16 Oct 2017, 17:54
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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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New post 02 Dec 2017, 06:32
ScottTargetTestPrep wrote:
AbdurRakib wrote:
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?

A. 12
B. 16
C. 28
D. 36
E. 40


Let’s review two rules for the standard deviation: 1. Adding a constant to each value in a data set does not change the value of the standard deviation; 2. Multiplying each value in a data set by a constant also multiplies the standard deviation by that constant.

Thus, for y = 0.8x + 20, we see that adding 20 does not affect the value of the standard deviation. But multiplying each score by 0.8 means that the standard deviation of the entire data set is also multiplied by 0.8; thus, the standard deviation of the adjusted test scores will be 20 x 0.8 = 16.

Answer: B


why answer is B and not D ? id STD deviation is 20 hence by plugging it in the given formula y = 0.8 (20) + 20 ---> y = 16 +20 --> y = 36 no? :? :)
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Re: For each student in a certain class, a teacher adjusted the student’s  [#permalink]

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New post 02 Dec 2017, 11:11
dave13 wrote:
ScottTargetTestPrep wrote:
AbdurRakib wrote:
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?

A. 12
B. 16
C. 28
D. 36
E. 40


Let’s review two rules for the standard deviation: 1. Adding a constant to each value in a data set does not change the value of the standard deviation; 2. Multiplying each value in a data set by a constant also multiplies the standard deviation by that constant.

Thus, for y = 0.8x + 20, we see that adding 20 does not affect the value of the standard deviation. But multiplying each score by 0.8 means that the standard deviation of the entire data set is also multiplied by 0.8; thus, the standard deviation of the adjusted test scores will be 20 x 0.8 = 16.

Answer: B


why answer is B and not D ? id STD deviation is 20 hence by plugging it in the given formula y = 0.8 (20) + 20 ---> y = 16 +20 --> y = 36 no? :? :)


Hi

y is not the formula for standard deviation. 'y' here refers to the adjusted value or the changed value of 'x', where 'x' was the original value. So basically teacher is multiplying each original value by 0.8, and then adding 20 so as to get a new value. We have to find standard deviation of this new set of values, where we are given the standard deviation of old set of values was 20.
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New post 16 Mar 2018, 09:34
Made a blunder , by adding 16 + 20 to the actual standard deviation . Corrected it QA : 16 (ANSWER)
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Re: For each student in a certain class, a teacher adjusted the student’s  [#permalink]

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New post 24 Jul 2018, 22:42
ScottTargetTestPrep wrote:
AbdurRakib wrote:
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?

A. 12
B. 16
C. 28
D. 36
E. 40


Let’s review two rules for the standard deviation: 1. Adding a constant to each value in a data set does not change the value of the standard deviation; 2. Multiplying each value in a data set by a constant also multiplies the standard deviation by that constant.

Thus, for y = 0.8x + 20, we see that adding 20 does not affect the value of the standard deviation. But multiplying each score by 0.8 means that the standard deviation of the entire data set is also multiplied by 0.8; thus, the standard deviation of the adjusted test scores will be 20 x 0.8 = 16.

Answer: B



HI ScottTargetTestPrep,

Could you please explain a shortcut method for calculating the S.D. of numbers which are all in the form of y=0.8x+20. How is S.D. equal to 20 ?

I am sure you must have not calculated the S.D. by the long method of taking the square root of mean differences.
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New post 31 Jul 2018, 04:05
ScottTargetTestPrep explanation is perfect.

There is another way as well. If we wanted to graph original scores on coordinate plane, we will have a function y=x slope of which is 1.

y= 0.8x + 20 has a slope of 0.8

If all the of the original scores were same then the line of this scores would have slope of 0 (it would pass for example y=100), and SD of 0. There is direct relationship between slope of a line and standard deviation. when slope is 0, SD is 0. When the slope is 1 SD is 20. Reducing slope by 20% will reduce the SD by 20%
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New post 31 Jul 2018, 08:27
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kablayi wrote:
ScottTargetTestPrep explanation is perfect.

There is another way as well. If we wanted to graph original scores on coordinate plane, we will have a function y=x slope of which is 1.

y= 0.8x + 20 has a slope of 0.8

If all the of the original scores were same then the line of this scores would have slope of 0 (it would pass for example y=100), and SD of 0. There is direct relationship between slope of a line and standard deviation. when slope is 0, SD is 0. When the slope is 1 SD is 20. Reducing slope by 20% will reduce the SD by 20%


Very cool. Thank you!
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