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

Coded Data Proofs (1):

Say y=kx, and also that: x={p, q}, y={kp, kq}.

This would mean that:

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

Therefore, when y=kx:

$\frac { \Sigma y }{ n } =k\cdot \frac { \Sigma x }{ n } \\$

What we’d also be able to conclude is that:

$\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( kp \right) }^{ 2 }+{ \left( kq \right) }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+{ k }^{ 2 }{ q }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } \\$

When considering the above, we can deduce that:

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

Therefore, when y=kx:

${ \sigma }_{ y }=k\cdot { \sigma }_{ x }\\$