Question

Find mean and the standard deviation of the following two groups taken together:

Group	Number	Mean	S.D.
A	113	159	22.4
B	121	149	20.0

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ {\overline{X}}_{\mathrm{A}}=159,\mathit{}{\sigma }_{\mathrm{A}}=22.4,N_{\mathrm{A}}=113 \\ {\overline{X}}_{\mathrm{B}}=149,{\sigma }_{\mathrm{B}}=20,\mathit{}{N}_{\mathrm{B}}=121 \\ \mathbf{Combined}\mathbf{}\mathbf{Mean} \\ {\overline{X}}_{\mathrm{A},\mathrm{B}}=\frac{{\overline{X}}_{\mathrm{A}}{N}_{\mathrm{A}}+{\overline{X}}_{\mathrm{B}}{N}_{\mathrm{B}}}{N_{\mathrm{A}}+N_{\mathrm{B}}} \\ =\frac{159\times 113+149\times 121}{113+121} \\ =\frac{17967+18029}{234} \\ =\frac{35996}{234} \\ =153.83 \end{gathered}$$
Now, to calculate combined standard deviation, we need to find,

$$\begin{gathered} d_{\mathrm{A}}^2={\left({\overline{\mathit{X}}}_{\mathrm{A}}-{\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B}}\right)}^2={\left(159-153.83\right)}^2 \\ ={\left(5.17\right)}^2=26.73 \\ {d}_{\mathrm{B}}^2={\left({\overline{\mathit{X}}}_{\mathrm{B}}-{\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B}}\right)}^2={\left(149-153.83\right)}^2 \\ ={\left(-4.83\right)}^2=23.33 \end{gathered}$$

Combined Standard Deviation
$$\begin{gathered} {\sigma }_{\mathrm{A},\mathrm{B}}=\sqrt{\frac{N_{\mathrm{A}}{\sigma }_{\mathrm{A}}^2+N_{\mathrm{B}}{\sigma }_{\mathrm{B}}^2+N_{\mathrm{A}}{d}_{\mathrm{A}}^2+N_{\mathrm{B}}{d}_{\mathrm{B}}^2}{N_{\mathrm{A}}+N_{\mathrm{B}}}} \\ =\sqrt{\frac{113{\left(22.4\right)}^2+121{\left(20\right)}^2+113\left(26.73\right)+121\left(23.33\right)}{113+121}} \\ =\sqrt{\frac{56698.88+48400+3020.49+2822.93}{234}} \\ =\sqrt{\frac{110942.3}{234}} \\ =\sqrt{474.112} \\ =21.774\mathrm{or}21.8\mathrm{approx}. \end{gathered}$$
Thus, mean and standard deviation of the two groups-A and B taken together are 23.33 and 21.8 respectively.