Verified Solution • 2025–26 Session moderate

QJEEM19Ma7a00: JEE Main 2019 Mathematics - Class 11 JEEM19Ma7a00 Solution: A data consists of n observations : x 1 , x 2 , . . . . . . .,…

A data consists of n observations : x1, x2, . . . . . . ., xn.    

If     ∑i=1n(xi+1)2=9n\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n    and

∑i=1n(xi−1)2=5n,\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,

then the standard deviation of this data is :

Multiple Choice Options

  • A 2
  • B 5\sqrt 5
  • C 5
  • D 7\sqrt 7

Step-by-Step Verified Solution

∑i=1n(xi+1)2=9n\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n

⇒∑i=1nxi2+2∑i=1nxi+n=9n     ... (1)\Rightarrow \sum\limits_{i = 1}^n {x_i^2} + 2\sum\limits_{i = 1}^n {{x_i}} + n = 9n\,\,\,\,\,...\,(1)

∑i=1n(xi−1)2=5n\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n

⇒∑i=1nxi2−2∑i=1nxi+n=5n     ... (2)\Rightarrow \sum\limits_{i = 1}^n {x_i^2} - 2\sum\limits_{i = 1}^n {{x_i}} + n = 5n\,\,\,\,\,...\,(2)

Performing (1) + (2), we get

2∑i=1nxi2+2n=14n2\sum\limits_{i = 1}^n {x_i^2} + 2n = 14n

∑i=1nxi2=6n\sum\limits_{i = 1}^n {x_i^2} = 6n

Performing (1) −- (2), we get

⇒4∑i=1nxi=4n\Rightarrow 4\sum\limits_{i = 1}^n {{x_i}} = 4n

⇒\Rightarrow ⇒∑i=1nxi=n\Rightarrow \sum\limits_{i = 1}^n {{x_i}} = n

S.D( σ\sigma ) =∑xi2n−(x‾)2= \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}}

σ\sigma =6nn−(1)= \sqrt {{{6n} \over n} - \left( 1 \right)}

σ\sigma =5= \sqrt 5

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