Finding constant related to random variable

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Homework Help Overview

The discussion revolves around the relationship between the variance of a random variable Y and that of another variable X, specifically focusing on the constant a in the equation Var(Y) = a² . Var(X). Participants are questioning the validity of a negative value for a in the context of scaling and adjusting scores.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are exploring why the negative value of a (specifically -0.8) is rejected, with some suggesting that it could still yield valid results. Questions about the implications of scaling and the assumptions behind the term "scaled" are raised.

Discussion Status

The discussion is active, with participants providing various perspectives on the implications of using a negative scaling factor. Some express frustration over the potential outcomes of negative scores, while others attempt to clarify the assumptions involved in the scaling process. There is no explicit consensus, but multiple interpretations are being explored.

Contextual Notes

Participants are considering the implications of the scaling factor on the adjusted scores, particularly in relation to maintaining positive values. The discussion also touches on the assumptions made about the nature of scaling in the context of variance and mean adjustments.

songoku
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Homework Statement
Exam marks, X, have mean 70 and standard deviation 8.7. The marks need to be scaled using the formula Y = aX + b so that the scaled marks, Y, have mean 55 and standard deviation 6.96. Find the values of a and b
Relevant Equations
E(aX + b) = a.E(X) + b

Var(aX + b) = a^2 Var (X)
Var (Y) = a2 . Var (X)
(6.96)2 = a2 . (8.7)2
a = ± 0.8

But the answer key states that the value of a is only 0.8

Why a = -0.8 is rejected? Thanks
 
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songoku said:
Why a = -0.8 is rejected?
Why do you think?
 
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PeroK said:
Why do you think?
In my opinion, there are two possible pairs of a and b so a = -0.8 is possible
 
songoku said:
In my opinion, there are two possible pairs of a and b so a = -0.8 is possible
What do you want the mean to be?
 
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You are correct in saying that the -0.8 would give the correct mean and variance, but the term "scaled" is assumed to imply that the values of X are just shifted and stretched/shrunk -- not reversed.
 
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songoku said:
Why a = -0.8 is rejected?
I would be very angry if my raw score (X) on the exam was 70, but my adjusted score wound up being -56.
 
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Mark44 said:
I would be very angry if my raw score (X) on the exam was 70, but my adjusted score wound up being -56.
The proposed formula is ##Y = -0.8X + 111##. Your score of##70## would, indeed, scale to the average of ##55##. But a score of ##80## would scale to ##47##.
 
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PeroK said:
What do you want the mean to be?
For a = -0.8 and b = 111, the mean for Y is 55

Mark44 said:
I would be very angry if my raw score (X) on the exam was 70, but my adjusted score wound up being -56.
Well, I won't because my peers' scores will also be negative :biggrin:

Besides, there is still value of b to make the scaled mark positive

FactChecker said:
You are correct in saying that the -0.8 would give the correct mean and variance, but the term "scaled" is assumed to imply that the values of X are just shifted and stretched/shrunk -- not reversed.
I understand the assumption

Edit: I just saw post #7. That is indeed does not make sense

Thank you for all the help and explanation PeroK, FactChecker, Mark44
 
And the first one now will later be last ...
 
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