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(Let M

') = 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) no info of k...insuff

combined..again

no info of k...insuff

Ans E