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OG For each student in a certain class, a teacher adjusted

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OG For each student in a certain class, a teacher adjusted

by AbeNeedsAnswers » Sat Aug 19, 2017 3:04 pm

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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

B
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by DavidG@VeritasPrep » Tue Aug 22, 2017 11:27 am
AbeNeedsAnswers 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

B
Two important identities here.

1) If the standard deviation of some set were "d," and every element of the set were multiplied by 'a,' the standard deviation would also be multiplied by "a," giving us a new standard deviation of
"d * a."

2) If the standard deviation of some set were "d," and every element of the set were increased by a value of 'a,' the new standard deviation would be unchanged.

Here, we can think of the formula y = .8x + 20, as telling us to multiply each term by .8 and then increase that term by 20. If we multiply every term by .8, and the old standard deviation was 20, the new standard deviation would be 20*.8 = 16. If we then increased each term by 20, the standard deviation would remain unchanged. So the answer is B.
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Re: OG For each student in a certain class, a teacher adjusted

by Scott@TargetTestPrep » Sun Feb 16, 2020 7:17 am
AbeNeedsAnswers wrote: ↑
Sat Aug 19, 2017 3:04 pm
 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

B
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: OG For each student in a certain class, a teacher adjusted

by [email protected] » Thu Apr 22, 2021 2:33 pm
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: B

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