If `barx_1,barx_2,barx_3,...barx_n` are the means of n groups with `n_1,n_2,n_3,....n_n` numbers of observations respectively. Than the mean `barx` of all group together is given by :

Options

A. `sum_(i=1)^(n)n_(i)bar(x)_(i)`
B. `(sum_(i=1)^(n)n_(i)bar(x)_(i))/(n^(2))`
C. `(sum_(i=1)^(n)n_(i)bar(x)_(i))/(sum_(i=1)^(n)n_(i))`
D. `(sum_(i=1)^(n)n_(i)bar(x)_(i))/(2n)`

Answer

C

Explanation

If `bar(x)_(1),bar(x)_(2), bar(x)_(3),……, bar(x)_(n)` are the means of n groups with `n_(1),n_(2),……, n_(n)` numbers of observations respectively, then the mean `bar(x)`
`bar(x) = (n_(1)bar(x)_(1)+n_(2)bar(x)_(2)+n_(3)bar(x)_(3)+.......+n_(n)bar(x)_(n))/(n_(1)+n_(2)+n_(3)+.....+n_(n))=(sum_(i=1)^(n)n_(i)bar(x)_(i))/(sum_(i=1)^(n)n_(i))`
Hence, the mean `bar(x)` of all group taken together is given by `(sum_(i=1)^(n)n_(i)bar(x)_(i))/(sum_(i=1)^(n)n_(i))`
So, (c) is the correct answer.