Question
Download Solution PDFIf the probable error in single observation is ± 0.04 m and that of the mean is ± 0.01 m, then the number of observations are
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Most probable error: If the error for which there are equal chances that the true error will be less than the probable error and equal chances that the true error will be more than the probable error i.e. each is having the probability of 50%
\(Most\;probable\;error\;\left( E \right) = \; \pm \sqrt {\frac{{{\rm{\Sigma }}{v^2}}}{{n - 1}}} \)
Most Probable Error of mean:
\({E_m} = \frac{E}{{\sqrt n }}\)
Where,
n = Number of observation
Calculation:
Given,
E = ± 0.04
Em = ± 0.01
∵ we know, Em = \(\frac{E}{{\sqrt n }}\)
This relation can be written as,
\(n = {\left( {\frac{E}{{{E_m}}}} \right)^2}\)
\(n = {\left( {\frac{{0.04}}{{0.01}}} \right)^2} = 16\)
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