QJEEM21Ma4cac: JEE Main 2021 Mathematics - Class 11 JEEM21Ma4cac Solution: Let X 1 , X 2 , ......., X 18 be eighteen observations such…

Question

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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 ____________.

Accepted Answer

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 )

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