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Results on  A.M.

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☼ Results on  A.M. ( \large \overline{x} )  :

 (i) Let  there be n observations x_{1}+x_{2}+...+x_{n}    then :

A.M. ( \overline{x} )  =   \left (\frac{x_{1}+x_{2}+...+x_{n}}{n} \right )     =        \frac{\sum x}{n}

=     A  +  \frac{\sum d_i}{n}      ;   d_{i} =  x_{i} – A

(  Where,  A  =  Assumed  Mean  )

 (ii) For Grouped – Data,

  

A.M.  ( \overline{x} )   =    \frac{\sum f_{i}x_{i}}{n}   ;    n   =  \sum f_{i}.

=      A   +   \frac{\sum f_{i} d_{i}}{n}   ;   n  = \sum f_{i}      &  d_{i} =  x_{i} – A

=      A   +   \frac{\sum f_{i} d_{i}}{n}  \times  c   ;    n  = \sum f_{i}      &  d_{i} =  x_{i} – A

 (iii) If  observations x_{1}+x_{2}+...+x_{k}  have weights  w_{1}+w_{2}+...+w_{k}  respectively, then :

Weighted  A.M.  = \frac{x_{1}w_{1} + x_{2}w_{2} + x_{3}w_{3}+...+x_{k}w_{k}}{w_{1}+w_{2}+w_{3}+...+w_{k}}

 (iv) The mean is influenced by change of origin and scale.

  i.e.   If  A.M. of  x_{1}, x_{2},..., x_{n}   be  , \overline{x}  then :

\large \mapsto   A.M. of  \left (x_{1}\pm k \right ), \left (x_{2}\pm k \right ), ... , \left (x_{n} \pm k \right )   is \left (\overline{x} \pm k \right )

\large \mapsto   A.M. of  kx_{1}, kx_{2},...,kx_{n}   is  k\overline{x} .

\large \mapsto   A.M. of  \frac{x_{1}}{k},\frac{x_{2}}{k},... , \frac{x_{n}}{k}  is  \frac{\overline{x}}{k}    ;    where,    C \ne 0.

 (v) The sum of deviations of all the observations from mean is equal to zero.

i.e.  \sum \left (x_{i}- \overline{x} \right )=0     =    0                  But           \sum_{i=1}^{n} x_{i}-\overline{x}=\left (n-1 \right ) \overline{x}–    =   (n – 1)

 

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