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