Question

If `barx_1, barx_2, barx_3, ..., barx_n` are the means of n groups with n₁, n₂, ..., n_n number of observations respectively, then the mean `barx` of all the groups taken together is given by ______.

Options

• `sum_(i = 1)^n n_i barx_i`

• `(sum_(i = 1)^n n_i barx_i)/n^2`

• `(sum_(i = 1)^n n_i barx_i)/(sum_(i = 1)^n n_i)`

• `(sum_(i = 1)^n n_i barx_i)/(2n)`

Answer 1

If `barx_1, barx_2, barx_3, ..., barx_n` are the means of n groups with n₁, n₂, ..., n_n number of observations respectively, then the mean `barx` of all the groups taken together is given by `underlinebb((sum_(i = 1)^n n_i barx_i)/(sum_(i = 1)^n n_i))`.

Explanation:

Given, `barx_1, barx_2, barx_3, ..., barx_n` are the means of n groups having number of observations n₁, n₂, ..., n_n, respectively.

Then, `n_1 barx_1 = sum_(i = 1)^(n_1) x_i, n_2 barx_2` 

= `sum_(j = 1)^(n_2 ) x_j, n_3 barx_3`

 = `sum_(k = 1)^(n_3) x_k, ..., n_n barx_n`

 = `sum_(p = 1)^(n_n) x_p`

Now, the mean `barx` of all the groups taken together is given by

`barx = (sum_(i = 1)^(n_1) x_i + sum_(j = 1)^(n_2) x_j + sum_(k = 1)^(n_3) x_k + .... + sum_(p = 1)^(n_n) x_p)/(n_1 + n_2 + ... + n_n)`

= `(n_1 barx_1 + n_2 barx_2 + n_3 barx_3 + ... + n_n barx_n)/(n_1 + n_2 + ... + n_n)`

= `(sum_(i = 1)^n n_i barx_i)/(sum_(i = 1)^n n_i)`

Hence, the mean of all the groups taken together is given by `barx = (sum_(i = 1)^n n_i barx_i)/(sum_(i = 1)^n n_i)`