duahsolo wrote:
If the arithmetic mean of a set of n measurements was found to be m, and then it was discovered that there was an error of +k% in each measurement. What is the percentage error in the mean?
(1) m = 6
(2) n = 7
avg. /mean(M) = Sum(s) / No. of samples(s)
Actual mean without error= M = s/ n
Mean with error(LetM' ) = s(1+k%) / n
%age error =M' - M /M
= s(1+k%)/n - s/n / s/n
= k/100
so for knowing the percentage error in the mean we must know k...
(1) no info of k...insuff
(2)
...
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