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The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results :
Number of observations $=25$, mean $=18.2$ seconds, standard deviation $=3.25$ seconds.
Further, another set of 15 observations $\mathrm{x}_1, \mathrm{x}_2, \ldots . ., \mathrm{x}_{15}$, also in seconds, is now available and we have $\sum_{i=1}^{15} x_i=279$ and $\sum_{i=1}^{15} x_i^2=5524$. Calculate the standard deviation based on all 40 observations.
Number of observations $=25$, mean $=18.2$ seconds, standard deviation $=3.25$ seconds.
Further, another set of 15 observations $\mathrm{x}_1, \mathrm{x}_2, \ldots . ., \mathrm{x}_{15}$, also in seconds, is now available and we have $\sum_{i=1}^{15} x_i=279$ and $\sum_{i=1}^{15} x_i^2=5524$. Calculate the standard deviation based on all 40 observations.
Solution:
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Verified Answer
$$
\begin{aligned}
&\mathrm{n}_1=25, \overline{\mathrm{x}}_1=18.2, \sigma_1=3.25 \\
&\mathrm{n}_2=15, \overline{\mathrm{x}}_2=\frac{279}{15}=18.6 \\
&\Sigma \mathrm{x}_2=279, \Sigma \mathrm{x}_2^2=5524 \\
&\overline{\mathrm{x}}_{12}=\frac{\mathrm{n}_1 \overline{\mathrm{x}}_1+\mathrm{n}_2 \overline{\mathrm{x}}_2}{\mathrm{n}_1+\mathrm{n}_2} \\
&=\frac{25(18.2)+15(18.6)}{40}=\frac{455+279}{40} \\
&=\frac{734}{40}=18.35 \\
&\sigma_2=\sqrt{\frac{5524}{15}-(18.6)^2}
\end{aligned}
$$
$$
\begin{aligned}
&=\sqrt{368.266-345.960}=\sqrt{22.306}\\
&=4.7229=4.723\\
&\text { Let } \mathrm{d}_1=\overline{\mathrm{x}}_1-\overline{\mathrm{x}}_{12}=18.2-18.35=(-0.15)\\
&\mathrm{d}_2=\overline{\mathrm{x}}_2-\overline{\mathrm{x}}_{12}=18.6-18.35=(0.25)\\
&\therefore \sigma_{12}=\sqrt{\frac{\mathrm{n}_1\left(\sigma_1^2+\mathrm{d}_1^2\right)+\mathrm{n}_2\left(\sigma_2^2+\mathrm{d}_2^2\right)}{\mathrm{n}_1+\mathrm{n}_2}}\\
&=\sqrt{\frac{25\left\{(3.25)^2+(-0.15)^2\right\}+15\left\{(4.723)^2+(0.25)^2\right\}}{40}}\\
&=\sqrt{\frac{25(10.5625+0.0225)+15(22.3067+0.0625)}{40}}\\
&=\sqrt{\frac{264.625+335.538}{40}}=\sqrt{\frac{600.163}{40}}\\
&=\sqrt{15.004}=3.873
\end{aligned}
$$
\begin{aligned}
&\mathrm{n}_1=25, \overline{\mathrm{x}}_1=18.2, \sigma_1=3.25 \\
&\mathrm{n}_2=15, \overline{\mathrm{x}}_2=\frac{279}{15}=18.6 \\
&\Sigma \mathrm{x}_2=279, \Sigma \mathrm{x}_2^2=5524 \\
&\overline{\mathrm{x}}_{12}=\frac{\mathrm{n}_1 \overline{\mathrm{x}}_1+\mathrm{n}_2 \overline{\mathrm{x}}_2}{\mathrm{n}_1+\mathrm{n}_2} \\
&=\frac{25(18.2)+15(18.6)}{40}=\frac{455+279}{40} \\
&=\frac{734}{40}=18.35 \\
&\sigma_2=\sqrt{\frac{5524}{15}-(18.6)^2}
\end{aligned}
$$
$$
\begin{aligned}
&=\sqrt{368.266-345.960}=\sqrt{22.306}\\
&=4.7229=4.723\\
&\text { Let } \mathrm{d}_1=\overline{\mathrm{x}}_1-\overline{\mathrm{x}}_{12}=18.2-18.35=(-0.15)\\
&\mathrm{d}_2=\overline{\mathrm{x}}_2-\overline{\mathrm{x}}_{12}=18.6-18.35=(0.25)\\
&\therefore \sigma_{12}=\sqrt{\frac{\mathrm{n}_1\left(\sigma_1^2+\mathrm{d}_1^2\right)+\mathrm{n}_2\left(\sigma_2^2+\mathrm{d}_2^2\right)}{\mathrm{n}_1+\mathrm{n}_2}}\\
&=\sqrt{\frac{25\left\{(3.25)^2+(-0.15)^2\right\}+15\left\{(4.723)^2+(0.25)^2\right\}}{40}}\\
&=\sqrt{\frac{25(10.5625+0.0225)+15(22.3067+0.0625)}{40}}\\
&=\sqrt{\frac{264.625+335.538}{40}}=\sqrt{\frac{600.163}{40}}\\
&=\sqrt{15.004}=3.873
\end{aligned}
$$
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