Coded Data Proofs: Mean & Standard Deviation, y=kx+C

Coded Data Proofs (3):

Say y=kx+C and also that:

x={p, q} and y={kp+C, kq+C}

This would mean that:

\frac { \Sigma y }{ n } =\frac { kp+C+\left\{ kq+C \right\} }{ n } \\ \\ =\frac { kp+kq+nC }{ n } \\ \\ =\frac { k\left( p+q \right) +nC }{ n } \\ \\ =\frac { k\left( p+q \right) }{ n } +\frac { nC }{ n } \\ \\ =k\cdot \frac { \Sigma x }{ n } +C

And if the above is true:

\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( kp+C \right) }^{ 2 }+{ \left( kq+C \right) }^{ 2 } }{ n } \\ \\ =\frac { \left( kp+C \right) \left( kp+C \right) +\left( kq+C \right) \left( kq+C \right) }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+nkpC+{ C }^{ 2 }+\left\{ { k }^{ 2 }{ q }^{ 2 }+nkqC+{ C }^{ 2 } \right\} }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+{ k }^{ 2 }{ q }^{ 2 }+nkC\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) +nkC\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } +\frac { nkC\left( p+q \right) }{ n } +\frac { n{ C }^{ 2 } }{ n } \\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +kC\cdot \Sigma x+{ C }^{ 2 }\\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right)

Therefore:

{ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ \left( k\cdot \frac { \Sigma x }{ n } +C \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -\left( k\cdot \frac { \Sigma x }{ n } +C \right) \left( k\cdot \frac { \Sigma x }{ n } +C \right) } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -\left\{ { k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }+nkC\cdot \frac { \Sigma x }{ n } +{ C }^{ 2 } \right\} } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-nkC\cdot \frac { \Sigma x }{ n } -{ C }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-kC\cdot \Sigma x-{ C }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-C\left( k\cdot \Sigma x+C \right) } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\left\{ \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } \right\} } \\ \\ =\sqrt { { k }^{ 2 } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =k\cdot { \sigma }_{ x }

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