# 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:

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