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Where,
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Percentage relative error (% From equation (1) and (3)
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Given that: measured value The absolute error De is given as:
Substituting the given values, we get:
The correction for this ammeter is obtained as: –
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Given that: measured value The absolute error De is given as:
Substituting the given values, we get:
Now, the relative error The percentage error % % = – 0.02 × 100 = – 2%
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(i) The arithmetic mean is calculated as follows
(ii) The deviations from each value are given by d1 = x1 – x¯ = 49.7 – 49.86 = – 0.16 d2 = x2 – x¯ = 50.1 – 49.86 = + 0.24 d3 = x3 – x¯ = 50.2 – 49.86 = + 0.34 d4 = x4 – x¯ = 49.6 – 49.86 = – 0.26 d5 = x5 – x¯ = 49.7 – 49.86 = – 0.16 (iii) The algebraic sum of the deviation is
= + 0.58 – 0.58 = 0
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