Jun 19 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. Jun 22 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jun 23 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.
Author 
Message 
TAGS:

Hide Tags

Director
Status: I don't stop when I'm Tired,I stop when I'm done
Joined: 11 May 2014
Posts: 530
Location: Bangladesh
Concentration: Finance, Leadership
GPA: 2.81
WE: Business Development (Real Estate)

For each student in a certain class, a teacher adjusted the student’s
[#permalink]
Show Tags
12 Jun 2017, 15:23
Question Stats:
68% (01:21) correct 32% (01:36) wrong based on 1578 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6567
Location: United States (CA)

Re: For each student in a certain class, a teacher adjusted the student’s
[#permalink]
Show Tags
19 Oct 2017, 10:29
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
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.




Retired Moderator
Joined: 13 Apr 2015
Posts: 1679
Location: India
Concentration: Strategy, General Management
GPA: 4
WE: Analyst (Retail)

Re: For each student in a certain class, a teacher adjusted the student’s
[#permalink]
Show Tags
12 Jun 2017, 20:48
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




Retired Moderator
Joined: 22 Aug 2013
Posts: 1437
Location: India

Re: For each student in a certain class, a teacher adjusted the student’s
[#permalink]
Show Tags
12 Jun 2017, 21:10
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



Intern
Joined: 21 Jun 2015
Posts: 43
Location: India
Concentration: Finance, General Management
GPA: 3.32
WE: Programming (Computer Software)

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



Intern
Joined: 17 Sep 2017
Posts: 2

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



Intern
Joined: 08 Oct 2017
Posts: 1

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



VP
Joined: 09 Mar 2016
Posts: 1275

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



Retired Moderator
Joined: 22 Aug 2013
Posts: 1437
Location: India

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



Manager
Joined: 26 Feb 2018
Posts: 56
WE: Sales (Internet and New Media)

Re: For each student in a certain class, a teacher adjusted the student’s
[#permalink]
Show Tags
16 Mar 2018, 09:34
Made a blunder , by adding 16 + 20 to the actual standard deviation . Corrected it QA : 16 (ANSWER)
_________________
" Can't stop learning and failing"



Intern
Joined: 30 Nov 2017
Posts: 23

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



Intern
Joined: 05 Mar 2015
Posts: 43
Location: Azerbaijan
GMAT 1: 530 Q42 V21 GMAT 2: 600 Q42 V31 GMAT 3: 700 Q47 V38

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



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6567
Location: United States (CA)

Re: For each student in a certain class, a teacher adjusted the student’s
[#permalink]
Show Tags
31 Jul 2018, 08:27
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!
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.




Re: For each student in a certain class, a teacher adjusted the student’s
[#permalink]
31 Jul 2018, 08:27






