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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Updated on: 10-Oct-2022

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