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Find the values of $n$ and $X$ in each of the following cases:
$\sum\limits _{i=1}^{n}( x_{i} -12) =-10$ and $\sum\limits _{i=1}^{n}( x_{i} -3) =62$.
Given:
$\sum\limits _{i=1}^{n}( x_{i} -12) =-10$ and $\sum\limits _{i=1}^{n}( x_{i} -3) =62$.
To do:
We have to find the values of $n$ and $X$.
Solution:
We know that,
Mean $\overline{X}=\frac{Sum\ of\ the\ observations}{Number\ of\ observations}$
Therefore,
$\sum\limits _{i=1}^{n}( x_{i} -12) =-10$.........(i)
$\sum_{i=1}^{n}(x_{i}-3)=62$......(ii)
From (i) and (ii), we get,
$n \bar{x}-12 n=-10$.........(iii)
$n \bar{x}-3 n=62$........(iv)
Subtracting (iv) from (iii), we get,
$-9 n=-72$
$n=\frac{-72}{-9}$
$n=8$
From (iii),
$n \bar{x}-12 \times 8=-10$
$8 \overline{\mathrm{X}}-96=-10$
$\overline{\mathrm{X}}=\frac{-10+96}{8}$
$=\frac{86}{8}$
$=10.75$
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