Let the observations x_i (1 $\le$ i $\le$ 10) satisfy the
equations, $\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)}$ = 10 and $\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}}$ = 40.
If $\mu$ and $\lambda$ are the mean and the variance of the
observations, x₁ – 3, x₂ – 3, ...., x₁₀ – 3, then
the ordered pair ($\mu$, $\lambda$) is equal to :

$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)}$ = 10 <br><br>$\Rightarrow$ x₁ + x₂ + .... + x₁₀ = 60 ....(1) <br><br>$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}}$ = 40 <br><br>$\Rightarrow$ ($x_1^2 + x_2^2 + ... + x_{10}^2$) + 25 $\times$ 10 - <br>10( x₁ + x₂ + .... + x₁₀) = 40 <br><br>$\Rightarrow$ $x_1^2 + x_2^2 + ... + x_{10}^2$ = 390 .....(2) <br><br>From question, <br> $\mu$ = $${{\left( {{x_1} - 3} \right) + \left( {{x_2} - 3} \right) + ... + \left( {{x_{10}} - 3} \right)} \over {10}}$$ <br><br>= ${{60 - 3 \times 10} \over {10}}$ = 3 <br><br>And $\lambda$ = variance = ${{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - 3} \right)}^2}} } \over {10}}$ - $\mu$² <br><br>= $${{\left( {x_1^2 + x_2^2 + ... + x_{10}^2} \right) + 90 - 6\left( {\sum {{x_i}} } \right)} \over {10}}$$ - 9 <br><br>= ${{390 + 90 - 360} \over {10}}$ - 9 <br><br>= 12 - 9 = 3