Find the values of
Question:

Find the values of $n$ and $\bar{X}$ in each of the following cases:

(i). $\sum_{i=1}^{n}\left(x_{i}-12\right)=-10$ and $\sum_{i=1}^{n}\left(x_{i}-3\right)=62$

(ii). $\sum_{i=1}^{n}\left(x_{i}-10\right)=30$ and $\sum_{i=1}^{n}\left(x_{i}-6\right)=150$

Solution:

(i). Given $\sum_{i=1}^{n}\left(x_{i}-12\right)=-10$

$\Rightarrow\left(x_{1}-12\right)+\left(x_{2}-12\right)+\ldots \ldots \ldots+\left(x_{n}-12\right)=-10$

$\Rightarrow\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+\cdots+x_{n}\right)-(12+12+12+12+\cdots+12)=-10$

$\Rightarrow \sum x-12 n=-10 \cdots(1)$

And $\sum_{i=1}^{n}\left(x_{i}-3\right)=62$

$\Rightarrow\left(x_{1}-3\right)+\left(x_{2}-3\right)+\cdots+\left(x_{n}-3\right)=62$

$\Rightarrow\left(x_{1}+x_{2}+\cdots+x_{n}\right)-(3+3+3+\cdots+3)=62$

$\Rightarrow 5 x-3 n=62 \cdots \cdots(2)$

By subtracting equation (1) from equation (2), we get

$\sum x-3 n-\sum x+12 n=62+10$

⇒ 9n = 72

⇒ n = 72/9 = 8

Put value of n in equation (1)

$\sum x-12 \times 8=-10$

$\Rightarrow \sum x-96=-10$

$\Rightarrow \Sigma x=96-10=86$

$\therefore \overline{\mathrm{x}}=\frac{\sum \mathrm{x}}{\mathrm{n}}=\frac{86}{8}=10.75$

(ii). $\sum_{i=1}^{n}\left(x_{i}-10\right)=30$

$\left(x_{1}-10\right)+\left(x_{2}-10\right)+\ldots \ldots \ldots \ldots+\left(x_{n}-10\right)=30$

$\Rightarrow\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+\cdots+x_{n}\right)-(10+10+10+10+\cdots+10)=30$

$\Rightarrow \sum x-10 n=30 \cdots(1)$

$\Rightarrow\left(x_{1}-6\right)+\left(x_{2}-6\right)+\cdots+\left(x_{n}-6\right)=150$

$\Rightarrow\left(x_{1}+x_{2}+\cdots+x_{n}\right)-(6+6+6+\cdots+6)=150$

$\Rightarrow \sum x-6 n=150 \cdots(2)$

By subtracting equation (1) from equation (2), we get

$\sum x-10 \times 30=30$

$\Rightarrow \sum x-300=30$

$\Rightarrow \Sigma x=30+300=330$

$\therefore \overline{\mathrm{x}}=\frac{\sum \mathrm{x}}{\mathrm{n}}=\frac{330}{30}=11$