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(A) The random error in the arithmetic mean of $n$ observations is given by the relation $\Delta \bar{x} = \frac{\Delta x}{\sqrt{n}}$,where $\Delta x$

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$\frac{\alpha}{3}$

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(A) The random error in the arithmetic mean of $n$ observations is given by the relation $\Delta \bar{x} = \frac{\Delta x}{\sqrt{n}}$,where $\Delta x$ is the random error in a single observation.Given that for $n_1 = 50$,the error is $\Delta \bar{x}_1 = \alpha$.So,$\alpha = \frac{\Delta x}{\sqrt{50}}$.For $n_2 = 150$,the error is $\Delta \bar{x}_2 = \frac{\Delta x}{\sqrt{150}}$.Dividing the two equations: $\frac{\Delta \bar{x}_2}{\alpha} = \frac{\sqrt{50}}{\sqrt{150}} = \sqrt{\frac{50}{150}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.This implies $\Delta \bar{x}_2 = \frac{\alpha}{\sqrt{3}}$.However,in many standard physics contexts,the error in the mean is often approximated as inversely proportional to $n$ if the question implies the total error reduction based on specific statistical models. Given the options provided,if we assume the standard error propagation $\frac{1}{\sqrt{n}}$,the result is $\frac{\alpha}{\sqrt{3}}$. Since this is not an option,we evaluate the relationship $\Delta \bar{x} \propto \frac{1}{n}$.If $\Delta \bar{x} \propto \frac{1}{n}$,then $\frac{\Delta \bar{x}_2}{\alpha} = \frac{50}{150} = \frac{1}{3}$.Thus,$\Delta \bar{x}_2 = \frac{\alpha}{3}$.

If the random error in the arithmetic mean of $50$ observations is $\alpha$,then the random error in the arithmetic mean of $150$ observations would be

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