JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 21)
Let X1, X2, ......., X18 be eighteen observations such
that $$\sum\limits_{i = 1}^{18} {({X_i} - } \alpha ) = 36$$ and $$\sum\limits_{i = 1}^{18} {({X_i} - } \beta {)^2} = 90$$, where $$\alpha$$ and $$\beta$$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of | $$\alpha$$ $$-$$ $$\beta$$ | is ____________.
that $$\sum\limits_{i = 1}^{18} {({X_i} - } \alpha ) = 36$$ and $$\sum\limits_{i = 1}^{18} {({X_i} - } \beta {)^2} = 90$$, where $$\alpha$$ and $$\beta$$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of | $$\alpha$$ $$-$$ $$\beta$$ | is ____________.
Answer
4
Explanation
Given, $$\sum\limits_{i = 1}^{18} {({x_1} - \alpha ) = 36} $$
$$ \Rightarrow \sum {{x_i} - 18\alpha = 36} $$
$$ \Rightarrow \sum {{x_i} = 18(\alpha + 2)} $$ .... (1)
Also, $$\sum\limits_{i = 1}^{18} {{{({x_1} - \beta )}^2} = 90} $$
$$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} - 2\beta \sum {{x_i} = 90} } $$
$$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} + 2\beta \times 18(\alpha + 2) = 90} $$ (using equation (1))
$$ \Rightarrow \sum {x_i^2 = 90} - 18{\beta ^2} + 36\beta (\alpha + 2)$$
Given, $${\sigma ^2} = 1 \Rightarrow {1 \over {18}}{\sum {x_i^2 - \left( {{{\sum {{x_i}} } \over {18}}} \right)} ^2} = 1$$
$$ = {1 \over {18}}(90 - 18{\beta ^2} + 36\alpha \beta + 72\beta ) - {\left( {{{18(\alpha + 2)} \over {18}}} \right)^2} = 1$$
$$ \Rightarrow 90 - 18{\beta ^2} + 36\alpha \beta + 72\beta - 18{(\alpha + 2)^2} = 18$$
$$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {(\alpha + 2)^2} = 1$$
$$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {\alpha ^2} - 4 - 4\alpha = 1$$
$$ \Rightarrow {\alpha ^2} - {\beta ^2} + 2\alpha \beta + 4\beta - 4\alpha = 0$$
$$ \Rightarrow (\alpha - \beta )(\alpha - \beta + 4) = 0$$
$$ \Rightarrow \alpha - \beta = - 4$$
$$ \therefore $$ $$|\alpha - \beta |\, = 4$$ $$(\alpha \ne \beta )$$
$$ \Rightarrow \sum {{x_i} - 18\alpha = 36} $$
$$ \Rightarrow \sum {{x_i} = 18(\alpha + 2)} $$ .... (1)
Also, $$\sum\limits_{i = 1}^{18} {{{({x_1} - \beta )}^2} = 90} $$
$$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} - 2\beta \sum {{x_i} = 90} } $$
$$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} + 2\beta \times 18(\alpha + 2) = 90} $$ (using equation (1))
$$ \Rightarrow \sum {x_i^2 = 90} - 18{\beta ^2} + 36\beta (\alpha + 2)$$
Given, $${\sigma ^2} = 1 \Rightarrow {1 \over {18}}{\sum {x_i^2 - \left( {{{\sum {{x_i}} } \over {18}}} \right)} ^2} = 1$$
$$ = {1 \over {18}}(90 - 18{\beta ^2} + 36\alpha \beta + 72\beta ) - {\left( {{{18(\alpha + 2)} \over {18}}} \right)^2} = 1$$
$$ \Rightarrow 90 - 18{\beta ^2} + 36\alpha \beta + 72\beta - 18{(\alpha + 2)^2} = 18$$
$$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {(\alpha + 2)^2} = 1$$
$$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {\alpha ^2} - 4 - 4\alpha = 1$$
$$ \Rightarrow {\alpha ^2} - {\beta ^2} + 2\alpha \beta + 4\beta - 4\alpha = 0$$
$$ \Rightarrow (\alpha - \beta )(\alpha - \beta + 4) = 0$$
$$ \Rightarrow \alpha - \beta = - 4$$
$$ \therefore $$ $$|\alpha - \beta |\, = 4$$ $$(\alpha \ne \beta )$$
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