Question

If ${\overline{\mathrm{X}}}_1,{\overline{\mathrm{X}}}_2,...,{\overline{\mathrm{X}}}_k$ are the means of n group with n1,n2,...,nk number of observations respectively, then the mean $\overline{\mathrm{X}}$ of all the groups taken together is govern by

$$\begin{gathered} \left(\mathrm{a}\right)\sum _{i=1}^k{n}_i{\overline{\mathrm{X}}}_i \\ \left(\mathrm{b}\right)\frac{1}{n^2}\sum _{i=1}^k{n}_i{\overline{\mathrm{X}}}_i \\ \left(\mathrm{c}\right)\frac{{\displaystyle \sum _{i=1}^k}{n}_i{\overline{\mathrm{X}}}_i}{{\displaystyle \sum _{i=1}^k}{n}_i} \\ \left(\mathrm{d}\right)\frac{{\displaystyle \sum _{i=1}^k{n}_i{\overline{\mathrm{X}}}_i}}{{\displaystyle 2_n}} \end{gathered}$$

Answer 1

We know

Mean = $\frac{\mathrm{Sum}\mathrm{of}\mathrm{observations}}{\mathrm{Number}\mathrm{of}\mathrm{observations}}$

⇒ Sum of observations = Mean × Number of observations


Mean of n₁ observations of first group = $\overline{X_1}$

∴ Sum of n₁ observations of first group = $n_1\overline{X_1}$

Mean of n₂ observations of second group = $\overline{X_2}$

∴ Sum of n₂ observations of second group = $n_2\overline{X_2}$

. . . . . . .
. . . . . . .

Mean of n_k observations of kth group = $\overline{X_k}$

∴ Sum of n_k observations of kth group = $n_k\overline{X_k}$

Now,

Sum of all observations in the k groups

= Sum of n₁ observations of first group + Sum of n₂ observations of second group + ... + Sum of n_k observations of kth group

= $n_1\overline{X_1}+n_2\overline{X_2}+...+n_k\overline{X_k}$

= $\sum _{i=1}^k{n}_i\overline{X_i}$ .....(1)

Total number of observations in the k groups = $n_1+n_2+...+n_k=\sum _{i=1}^k{n}_i$ .....(2)

∴ Mean $\overline{X}$ of all the groups = $\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{observations}\mathrm{in}\mathrm{the}k\mathrm{groups}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{observations}\mathrm{in}\mathrm{the}k\mathrm{groups}}=\frac{{\displaystyle \sum _{i=1}^k}{n}_i\overline{X_i}}{{\displaystyle \sum _{i=1}^k}{n}_i}$ [Using (1) and (2)]

Hence, the correct answer is option (c).