Question

The number examined , the mean weight and standard deviations in each group of examination by two medical examiners are given below . Calculate mean and standard deviation of both the groups taken together.

Medical Examiner	Number Examined	Mean Weight (lbs)	Standard deviation (lbs)
A	50	113	6.5
B	60	120	8.2

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {\overline{\mathit{X}}}_{\mathrm{A}}=113,{\sigma }_{\mathrm{A}}=6.5,N_{\mathrm{A}}=50 \\ {\overline{\mathit{X}}}_{\mathrm{B}}=120,{\sigma }_{\mathrm{B}}=8.2,N_{\mathrm{B}}=60 \\ \mathbf{Combined}\mathbf{}\mathbf{Mean} \\ {\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B}}=\frac{{\overline{\mathit{X}}}_{\mathrm{A}}{N}_{\mathrm{A}}+{\overline{\mathit{X}}}_{\mathrm{B}}{N}_{\mathrm{B}}}{N_{\mathrm{A}}+N_{\mathrm{B}}} \\ =\frac{113\times 50+120\times 60}{50+60}=\frac{5650+7200}{110} \\ =\frac{12850}{110} \\ =116.82 \end{gathered}$$

Combined Standard Deviation
$$\begin{gathered} d_{\mathrm{A}}^2={\left({\overline{\mathit{X}}}_{\mathrm{A}}-{\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B}}\right)}^2={\left(113-116.82\right)}^2 \\ ={\left(-3.82\right)}^2=14.59 \\ {d}_{\mathrm{B}}^2={\left({\overline{\mathit{X}}}_{\mathrm{B}}-{\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B}}\right)}^2={\left(120-116.82\right)}^2={\left(3.18\right)}^2 \\ =10.11 \\ {\sigma }_{\mathrm{A},\mathrm{B}}=\sqrt{\frac{N_{\mathrm{A}}{\sigma }_{\mathrm{A}}^2+N_{\mathrm{B}}{\sigma }_{\mathrm{B}}^2+N_{\mathrm{A}}{d}_{\mathrm{A}}^2+N_{\mathrm{B}}{d}_{\mathrm{B}}^2}{N_{\mathrm{A}}+N_{\mathrm{B}}}} \\ =\sqrt{\frac{50{\left(6.5\right)}^2+60{\left(8.2\right)}^2+50\left(14.59\right)+60\left(10.11\right)}{50+60}} \\ =\sqrt{\frac{2112.5+4034.4+729.5+606.6}{110}} \\ =\sqrt{\frac{7483}{110}}=\sqrt{68.02} \\ =8.25 \end{gathered}$$
Thus, mean and standard deviation of both the groups taken together are 116.82 and 8.25 respectively.