Discussion
The normality of a solution is determined by four separate titrations, the results being 0.2041, 0.2049, 0.2039 and 0.2043. The average deviation for the results is :
Answer: A
The average deviation is 0.0003.
First find the mean, then the mean of the absolute deviations from it:
\[ \begin{aligned} \bar{x} &= \frac{0.2041 + 0.2049 + 0.2039 + 0.2043}{4} \\ &= \frac{0.8172}{4} = 0.2043 \end{aligned} \]
The deviations are |0.2041 - 0.2043| = 0.0002, 0.0006, 0.0004 and 0, which sum to 0.0012:
\[ \bar{d} = \frac{0.0012}{4} = 0.0003 \]
0.2043 is the mean itself, not a deviation. 0.0006 is the largest single deviation, and 0.0009 does not follow from any standard measure of these data.
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