The mean and variance of a data of 10 observations are 10 and 2 , respectively.

Question

The mean and variance of a data of 10 observations are 10 and 2 , respectively. If an observations $\alpha$ in this data is replaced by $\beta$, then the mean and variance become 10.1 and 1.99 , respectively. Then $\alpha+\beta$ equals

Accepted Answer

Given, total observation $n=10$

Initial mean ${\vec{x}}_{1}=10$

$\Rightarrow \frac{\sum _{i=1}^{10}{x}_{i}}{10}=10$

Sum of 10 observation $\sum _{10}^{i=1}{x}_{i}=10\times 10=100$

Initial variance ${\sigma }_{2}^{1}$$=2$

$\begin{gathered} \Rightarrow \frac{\sum _{i=1}^{10}{x}_{2}^{i}}{n}-{({\vec{x}}_{1})}^{2}=2 \ \Rightarrow \sum _{i=1}^{10}{x}_{2}^{i}=n[2+{({\vec{x}}_{1})}^{2}]=10[2+({10}^{2})]=1020 \end{gathered}$

It is given that if $\alpha $ is replace by $\beta $

then, new mean $({\vec{x}}_{2})=10.1$

$\begin{gathered} {\vec{x}}_{2}=\frac{100-\alpha +\beta }{10}=10.1 \ \Rightarrow 100-\alpha +\beta =101 \ \Rightarrow \beta -\alpha =1......(1) \end{gathered}$

Also new variance ${\sigma }_{2}^{2}=1.99$ (given)

$\Rightarrow$ ${\sigma }_{2}^{2}=\frac{1020-{\alpha }^{2}+{\beta }^{2}}{10}-(10.1{)}^{2}=1.99$

$\Rightarrow$ $1020-{\alpha }^{2}+{\beta }^{2}=10[1.99+102.01]$$=10\times 104=1040$

$\begin{gathered} \Rightarrow {\beta }^{2}-{\alpha }^{2}=20 \ \Rightarrow (\beta +\alpha )(\beta -\alpha )=20 \end{gathered}$

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

$\Rightarrow$ $\alpha +\beta =20$

Solutions · 1