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555-605 Level|   Statistics and Sets Problems|                        
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For each student in a certain class, a teacher adjusted the student’s 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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B
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For each student in a certain class, a teacher adjusted the student’s 19 Oct 2017, 10:29
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AbdurRakib
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
Show more

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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For each student in a certain class, a teacher adjusted the student’s 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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For each student in a certain class, a teacher adjusted the student’s 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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x is the student's original score. So how can we define directly that X= std deviation ?




amanvermagmat
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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For each student in a certain class, a teacher adjusted the student’s 16 Oct 2017, 17:54
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gmat75016
x is the student's original score. So how can we define directly that X= std deviation ?




amanvermagmat
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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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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For each student in a certain class, a teacher adjusted the student’s 02 Dec 2017, 06:32
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AbdurRakib
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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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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For each student in a certain class, a teacher adjusted the student’s 02 Dec 2017, 11:11
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AbdurRakib
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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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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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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For each student in a certain class, a teacher adjusted the student’s Updated on: 13 Apr 2021, 22:47
Originally posted by GMATinsight on 03 Feb 2020, 22:26.
Last edited by GMATinsight on 13 Apr 2021, 22:47, edited 1 time in total.
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AbdurRakib
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
Show more

Rules of Standard Deviation



1) Addition/Subtraction of the constant in each term does NOT change the standard deviation of the set

2) Multiplication by a constant, changes the standard deviation by the same factor multiplies

i.e. +20 doesn't affect the SD

multiplication of 0.8 changes the standard deviation by same factor

i.e new Standard deviation = 0.8*20 = 16

Answer: Option B
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For each student in a certain class, a teacher adjusted the student’s 28 Apr 2020, 09:11
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AbdurRakib
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
Show more

Given: 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.

Asked: 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?

standard deviation of y = .8 standard deviation of x = .8 *20 = 16

IMO B
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For each student in a certain class, a teacher adjusted the student’s 03 Jan 2021, 16:14
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AbdurRakib
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
Show more

All the items of the score will be multiplied by 0.80 to get the adjusted score (y). So, the standard deviation (20) will also be multiplied by 0.80.

Thus the adjusted standard deviation is =20*0.80=16

The answer is B.
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Hi all,
I tried a different way to solve it, please let me know if this is acceptable.

Here, y = 0.8x + 20, where x is student’s original test score and y is student’s adjusted test score.

Using x=20,
Y = 0.8*20+20 =16+20 =36
Now, subtract the standard deviation of original test score to find the adjusted standard deviation i.e.
36-20=16.
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boncool
Hi all,
I tried a different way to solve it, please let me know if this is acceptable.

Here, y = 0.8x + 20, where x is student’s original test score and y is student’s adjusted test score.

Using x=20,
Y = 0.8*20+20 =16+20 =36
Now, subtract the standard deviation of original test score to find the adjusted standard deviation i.e.
36-20=16.
Show more

Actually: 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.
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For each student in a certain class, a teacher adjusted the student’s 17 Apr 2021, 13:40
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Just read the OG explanation for this question and thought, wow :shh:

Anyways, ScottTargetTestPrep 's explanation is perfect.

Standard Deviation is the amount by which a range of data is skewed around the mean. Adding a constant to any value is not going to change it.

For example, if the lowest score was 20, and the highest 40; S.D. is 20.

Just add 20 to each score, and the lowest would become 40, highest 60. S.D. would still be 20.

But when you multiply the scores by something, then stuff starts to change.

Take the same example as above, multiple by 0.8.

Lowest score would be 16, highest score would become 32; S.D. would become 16 i.e. 80% of 20.

And that's why the answer is B.
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For each student in a certain class, a teacher adjusted the student’s 22 Apr 2021, 14:15
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Hi All,

We’re told that a teacher adjusts each of the student’s test scores according to the formula Y = (0.8)(X) + 20 where X is the ORIGINAL score and Y is the ADJUSTED score. We’re told that the ORIGINAL Standard Deviation of the scores was 20. We’re asked for the Standard Deviation of the new (adjusted) scores.

While this question certainly looks complicated, it’s actually a great ‘concept question’, meaning that you do not need to do much math to answer it if you recognize the concepts involved. It’s also worth noting that the GMAT will NEVER actually require that you calculate the S.D. of a group of numbers, but you do have to understand that basic concepts of S.D. (re: the ‘closer’ together a group of numbers is, the smaller the S.D; the more ‘spread out’ a group of numbers is, the larger the S.D.).

We can break all of this information down into pieces to define how the ‘parts’ of the formula impact the S.D. of the class scores. As an example, if the original scores were 10, 20 and 30, then adding 20 to each of those values would have NO impact on the S.D. (since the numbers would then be 30, 40 and 50 – and the ‘spread’ of the numbers would NOT have changed). However, by first multiplying each score by 0.8, the group of numbers gets CLOSER together (the three numbers would then be 8, 16 and 24; notice how the range is now 16 instead of 20 and differences between the consecutive terms is now 8 instead of 10). This ultimately lowers the S.D., so we’re looking for an answer that’s LESS than 20… but not that much less than 20, since multiplying by 0.8 in this context isn’t that much different from multiplying by 1. Based on how the answers are written, there’s only one answer that makes sense…

Final Answer:
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B

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For each student in a certain class, a teacher adjusted the student’s 30 May 2021, 07:21
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AbdurRakib
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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Answer: Option B

Video solution by GMATinsight

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For each student in a certain class, a teacher adjusted the student’s 10 Sep 2021, 00:53
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BrentGMATPrepNow ScottTargetTestPrep

Why has everyone in the solutions multiplied 0.8 with the original SD?

What I have understood:
Per the formula in the question, As 0.8 has been multiplied with the original test score x, we have to multiply again 0.8 with the original SD in order to find the SD of the adjusted test scores. Adjusted SD is missing in the formula given.

Am I correct? Thanks for any response in advance?
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BrentGMATPrepNow ScottTargetTestPrep

Why has everyone in the solutions multiplied 0.8 with the original SD?

What I have understood:
Per the formula in the question, As 0.8 has been multiplied with the original test score x, we have to multiply again 0.8 with the original SD in order to find the SD of the adjusted test scores. Adjusted SD is missing in the formula given.

Am I correct? Thanks for any response in advance?
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Sorry, I'm not sure what you mean when you say Adjusted SD is missing in the formula given.

Here's an example that might clarify things:
Let set A = {1, 2, 3, 4, 5}
Let's create set B by taking each number in set A and multiplying it by 0.1, to get: set B = {0.1, 0.2, 0.3, 0.4, 0.5}

As you can see, the numbers in set B are much closer together.
In other words, the numbers deviate less in set B than they do in set A.

In fact, if D = the deviation of set A, then it will be true that 0.1D = the standard deviation of set B

Does that help?
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For each student in a certain class, a teacher adjusted the student’s 15 Jan 2022, 08:46
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AbdurRakib
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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Addition or Subtraction has no effect on the standard deviation.
Multiplying/Division the original score changes the standard deviation by same factor.

Standard deviation (adjusted) = 0.8 * 20 = 16
Answer: B
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