In an experiment there are $20$ observation for which following results are given
$\sum x^{2} = 8100, \sum x = 214$
One observation that was $42$ was found to be wrong and it was replaced by the correct value $28$ . Then the corrected standard deviation is

15.8

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16

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12

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21.35

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Solution

The correct option is B $16$
$\sum x^{'} = 214 - 42 + 28 = 200 \sum x^{' 2} = 8100 - 42^{2} + 28^{2} = 7120 variance (σ^{' 2}) = \frac{\sum x^{' 2}}{n} - {(\frac{\sum x^{'}}{n})}^{2} = \frac{7120}{20} - {(\frac{200}{20})}^{2} = 256 s.d = \sqrt{σ^{' 2}} = \sqrt{256} = 16.0$

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In an experiment there are 20 observation for which following results are given ∑x2=8100,∑x=214 One observation that was 42 was found to be wrong and it was replaced by the correct value 28. Then the corrected standard deviation is

The correct option is B 16∑x′=214−42+28=200∑x′2=8100−422+282=7120variance(σ′2)=∑x′2n−(∑x′n)2=712020−(20020)2=256s.d=√σ′2=√256=16.0

In an experiment there are $20$ observation for which following results are given
$\sum x^{2} = 8100, \sum x = 214$
One observation that was $42$ was found to be wrong and it was replaced by the correct value $28$ . Then the corrected standard deviation is

The correct option is B $16$
$\sum x^{'} = 214 - 42 + 28 = 200 \sum x^{' 2} = 8100 - 42^{2} + 28^{2} = 7120 variance (σ^{' 2}) = \frac{\sum x^{' 2}}{n} - {(\frac{\sum x^{'}}{n})}^{2} = \frac{7120}{20} - {(\frac{200}{20})}^{2} = 256 s.d = \sqrt{σ^{' 2}} = \sqrt{256} = 16.0$