1. Tardigrade
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  4. If μ is the mean distribution of yi,fi then Sigma fi yi-μ is equal to:

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Q. If $ \mu $ is the mean distribution of $ \{{{y}_{i}},{{f}_{i}}\}, $ then $ \Sigma {{f}_{i}}\{{{y}_{i}}-\mu \} $ is equal to:

KEAMKEAM 2001

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 $