The mean and variance of a data of \(10\) observations are \(10\) and \(2\) , respectively. If an observations \(\alpha\…

Let, in first case, \(10\) numbers are \(x_1, x_2, \ldots \ldots x_9, \alpha\)

Then, \(\frac{\sum _{i=1}^{9}{x}_{i}+\alpha }{10}=10\)

\(\Rightarrow \alpha +\sum _{i=1}^{9}{x}_{i}=100\Rightarrow \sum _{i=1}^{9}{x}_{i}=100−\alpha\)

And \({\sigma }^{2}=\left(\frac{\sum {x}_{i}^{2}+{\alpha }^{2}}{n}\right)−{\left(\frac{\sum {x}_{i}+\alpha }{n}\right)}^{2}\)

\(\Rightarrow 2=\left(\frac{\sum {x}_{i}^{2}+{\alpha }^{2}}{10}\right)−{\left(10\right)}^{2}\)

\(\Rightarrow \frac{\sum {x}_{i}^{2}+{\alpha }^{2}}{n}=102\)

\(\Rightarrow {x}_{1}^{2}+{x}_{2}^{2}+\ldots \ldots {x}_{9}^{2}+{\alpha }^{2}=1020\)

\(\Rightarrow \sum {x}_{i}^{2}=1020−{\alpha }^{2}\)

In second case, let number are

\({x}_{1},{x}_{2},\ldots \ldots {x}_{9},\beta\)

Now \(\frac{\sum_{i=1}^9 x_i+\beta}{10}=\frac{100-\alpha+\beta}{10}\)

\(\Rightarrow 10.1=\frac{100-\alpha+\beta}{10}\)

\(\Rightarrow \alpha-\beta=-1\)

Now, \({\sigma }^{2}=\left(\frac{\sum {x}_{i}^{2}+{\beta }^{2}}{n}\right)−{\left(\frac{\sum {x}_{i}+\beta }{n}\right)}^{2}\)

\(\Rightarrow \frac{\sum {x}_{i}^{2}+{\beta }^{2}}{10}−{(10.1)}^{2}=1.99\)

\(\Rightarrow 1.99=\frac{1020-{\alpha }^{2}+{\beta }^{2}}{10}-102.01\\ \Rightarrow 1020-{\alpha }^{2}+{\beta }^{2}=1040\)

\(\Rightarrow {\beta }^{2}−{\alpha }^{2}=20\)

\(\Rightarrow \alpha^2-\beta^2=-20\)

\(\Rightarrow (\alpha-\beta)(\alpha+\beta)=-20 \)

\(\Rightarrow (-1)(\alpha+\beta)=-20\)

Hence, \(\alpha+\beta=20\)