The random error in the arithmatic mean of $10$ observations is k. Then random error in arithmatic mean of $25$ observations would be

0.1 k

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0.4 k

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2.5 k

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1.25 k

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Solution

The correct option is B $0.4 k$
we have,

$m_{1} E_{1} = m_{2} E_{2}$

$25 E_{1} = 10 k$

$\Rightarrow E_{1} = \frac{10 k}{25}$

$∴ E_{1} = 0.4 k$

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