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JEE Main 202126 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2021 · 26 February · Evening Shift · Mathematics · Statistics

Let X _ 1 , X _ 2 , , X _ 18 be eighteen observations such that ∑ _...

Let X1,X2,…,X18{X}_{1}, {X}_{2}, \ldots , {X}_{18} be eighteen observations such that ∑i=118(Xi−α)=36and\mathop{∑}_{i = 1}^{18}\left({X}_{i} - \alpha\right) = 36 and\mathop{∑}_{i = 1}^{18}{\left({X}_{i} - \beta\right)}^{2} = 90,where, where\alphaandand\betaaredistinctrealnumbers.Ifthestandarddeviationoftheseobservationsisare distinct real numbers. If the standard deviation of these observations is1,thenthevalueof∣α−β∣, then the value of \left|\alpha - \beta\right| is _______.

Correct answer

0

Step-by-step solution

∑i=118(xi−α)=36,∑i=118(xi−β)2=90\mathop{∑}_{i = 1}^{18}\left({x}_{i} - \alpha\right) = 36, \mathop{∑}_{i = 1}^{18}{\left({x}_{i} - \beta\right)}^{2} = 90

⇒∑i=118xi=18(α+2),∑i=118xi2−2β∑i=118xi+18β2=90⇒ \mathop{∑}_{i = 1}^{18}{x}_{i} = 18\left(\alpha + 2\right), \mathop{∑}_{i = 1}^{18}{x}_{i}^{2} - 2\beta\mathop{∑}_{i = 1}^{18}{x}_{i} + 18{\beta}^{2} = 90

Hence ∑xi2=90−18β2+36β(α+2)∑ {x}_{i}^{2} = 90 - 18{\beta}^{2} + 36\beta\left(\alpha + 2\right)

Given ∑xi218−(∑xi18)2=1\dfrac{∑ {x}_{i}^{2}}{18} - {\left(\dfrac{∑ {x}_{i}}{18}\right)}^{2} = 1

⇒90−18β2+36β(α+2)−18(α+2)2=18⇒ 90 - 18{\beta}^{2} + 36\beta\left(\alpha + 2\right) - 18{\left(\alpha + 2\right)}^{2} = 18

⇒5−β2+2αβ+4β−α2−4α−4=1⇒ 5 - {\beta}^{2} + 2\alpha\beta + 4\beta - {\alpha}^{2} - 4\alpha - 4 = 1

⇒(α−β)2+4(α−β)=0⇒∣α−β∣=0⇒ {\left(\alpha - \beta\right)}^{2} + 4\left(\alpha - \beta\right) = 0 ⇒ \left|\alpha - \beta\right| = 0 or 44

As α\alpha and β\beta are distinct ∣α−β∣=4| \alpha - \beta | = 4

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