A) MD
B) SD
C) 0
D) relative frequency
E) none of these
Correct Answer: C
Solution :
\[\because \]\[\mu \]is mean of distribution\[\{{{y}_{i}},{{f}_{i}}\}\]. \[\therefore \]\[\mu =\frac{{{y}_{1}}{{f}_{1}}+{{y}_{2}}{{f}_{2}}+....+{{y}_{n}}{{f}_{n}}}{{{f}_{1}}+{{f}_{2}}+.....+{{f}_{n}}}\] \[\Rightarrow \]\[({{f}_{1}}+{{f}_{2}}+.....+{{f}_{n}})\mu ={{y}_{1}}{{f}_{1}}+{{y}_{2}}{{f}_{2}}+...\] \[+{{y}_{n}}{{f}_{n}}\] Now, \[\Sigma {{f}_{i}}({{y}_{i}}-\mu )=({{f}_{1}}{{y}_{1}}+{{f}_{2}}{{y}_{2}}+.....+{{f}_{n}}{{y}_{n}})\] \[-({{f}_{1}}+{{f}_{2}}+....{{f}_{n}})\mu \] \[=({{f}_{1}}+{{f}_{2}}+...+{{f}_{n}})\mu -({{f}_{1}}+{{f}_{2}}+....+{{f}_{n}})\mu \] \[=0\]
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