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

# Coded Data Proofs: Mean & Standard Deviation, y=x/k

Coded Data Poofs (2):

Say y=x/k and that: x={p, q}, y={p/k, q/k}.

This would mean that: $\frac { \Sigma y }{ n } =\frac { \frac { p }{ k } +\frac { q }{ k } }{ n } \\ \\ =\frac { \frac { 1 }{ k } \left( p+q \right) }{ n } \\ \\ =\frac { 1 }{ k } \cdot \frac { \left( p+q \right) }{ n } \\ \\ =\frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } \\$

It would also mean that: $\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( \frac { p }{ k } \right) }^{ 2 }+{ \left( \frac { q }{ k } \right) }^{ 2 } }{ n } \\ \\ =\frac { \frac { { p }^{ 2 } }{ { k }^{ 2 } } +\frac { { q }^{ 2 } }{ { k }^{ 2 } } }{ n } \\ \\ =\frac { \frac { 1 }{ { k }^{ 2 } } \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ =\frac { 1 }{ { k }^{ 2 } } \cdot \frac { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ =\frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } \\$

And if the above is true: ${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } -\frac { 1 }{ { k }^{ 2 } } \cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \left( \frac { \Sigma { x }^{ 2 } }{ n } -{ \left\{ \frac { \Sigma x }{ n } \right\} }^{ 2 } \right) } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\frac { 1 }{ k } \cdot { \sigma }_{ x }\\$

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

# Proving That Two Different Variance Formulas Are Equal To One Another

Prove that: $\frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } =\frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 }$

Well you first identify these truths: $x=\left\{ a,\quad b,\quad c \right\} \\ \\ \therefore \quad \overline { x } =\frac { \Sigma x }{ n } =\frac { a+b+c }{ n } ,\quad \\ \\ \therefore \quad n=3\\ \\ \therefore \quad n\overline { x } =a+b+c\\ \\ \therefore \quad \frac { \Sigma { x }^{ 2 } }{ n } =\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ n } .\\ \\ \therefore \quad \Sigma { x }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$

And next: The standard deviation formula looks like this: $\sigma =\sqrt { \frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } } =\sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } }$

# Obtaining The Mean From Two Sets Of Data

Prove that if set A, of size ${ n }_{ 1 }$, has mean $\overline { { x }_{ 1 } }$ and set B, of size ${ n }_{ 2 }$, has a mean $\overline { { x }_{ 2} }$ then the mean of the combined set of A and B is $\overline { x } =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } }$.

——————

Set A: ${ q }_{ 1 },{ q }_{ 2 },{ q }_{ 3 },{ q }_{ 4 }$ ${ n }_{ 1 }=4$

Therefore: $\overline { { x }_{ 1 } } =\frac { { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } }{ 4 } =\frac { { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } }{ { n }_{ 1 } }$

Therefore: ${ n }_{ 1 }\overline { { x }_{ 1 } } ={ q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 }$

——————

Set B: ${ q }_{ 5 },{ q }_{ 6 },{ q }_{ 7 },{ q }_{ 8 },{ q }_{ 9 }$ ${ n }_{ 2 }=5$

Therefore: $\overline { { x }_{ 2 } } =\frac { { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } }{ 5 } =\frac { { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } }{ { n }_{ 2 } }$

Therefore: ${ n }_{ 2 }\overline { { x }_{ 2 } } ={ q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 }$

So: $\frac { \left( { q }_{ 1 }+{ q }_{ 2 }+{ q }_{ 3 }+{ q }_{ 4 } \right) +\left( { q }_{ 5 }+{ q }_{ 6 }+{ q }_{ 7 }+{ q }_{ 8 }+{ q }_{ 9 } \right) }{ 9 } \\ \\ =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ 4+5 } \\ \\ =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } }$