How Does a Linear Transformation Affect Statistical Measures?

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A linear transformation, represented by the equation y = -2x + 1, affects statistical measures by applying specific formulas to the original values. The mean of y can be calculated as <y> = -2<x> + 1, resulting in <y> = -9. The standard deviation of y is determined by multiplying the original standard deviation by the absolute value of the multiplicative constant, yielding a standard deviation of 4. Other measures, such as the median and quartiles, can also be derived by substituting the transformed values into the respective formulas. Understanding these transformations is crucial for accurately interpreting the effects on statistical measures.
Ezekiel20
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Help please??

To anybody that can offer a hand.


<x>=5, Standard Deviation = 2, median Mx=4.5, Quartile1=4, Quartile 2=6, xmin=0, xmax=9.

After a linear tranform: y= -2x+1. What are <y>, Standard deviation y, median y, Q1y, Q2y, Ymin, Ymax.

I was given this equation for homework and I am totally lost.
 
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Except for the standard deviation, all the items you want can be botained by simply applying the linear transformation to an x quantity to get the corresponding y quantity. For example <y>=-2<x>+1. To get the standard deviation, you multiply by the absolute value of the multiplicative constant, therefore y standard deviation is 4.
 


Originally posted by Ezekiel20
To anybody that can offer a hand.


<x>=5, Standard Deviation = 2, median Mx=4.5, Quartile1=4, Quartile 2=6, xmin=0, xmax=9.

After a linear tranform: y= -2x+1. What are <y>, Standard deviation y, median y, Q1y, Q2y, Ymin, Ymax.

I was given this equation for homework and I am totally lost.

you know how you've got all the formulae for workingout those quantities? we,, instead of putting an x in the, try putting ax+b and seeing what the answer is and how it relates to the untransformed quantity:

mean: E:=\sum_nx_i/N where N is the number of measurements (or whatever) transform by ax+b \sum_n(ax_i+b)/N=a\sum_nx_i/N +\sum_nb/N = b+ a\sum_nx_i/N = b+aE
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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