Verified Solution • 2025–26 Session moderate

QJEEM20Maaa93: JEE Main 2020 Mathematics - Class 11 JEEM20Maaa93 Solution: Let x i (1 i 10) be ten observations of a random variable X. If …

Let xi (1 ≤\le i ≤\le 10) be ten observations of a random variable X. If
∑i=110(xi−p)=3\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3 and ∑i=110(xi−p)2=9\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9
where 0 ≠\ne p ∈\in R, then the standard deviation of these observations is :

Multiple Choice Options

  • A 710{7 \over {10}}
  • B 910{9 \over {10}}
  • C 45{4 \over 5}
  • D 35\sqrt {{3 \over 5}}

Step-by-Step Verified Solution

Standard deviation = Variance\sqrt {Variance}

=∑x12n−(x‾)2= \sqrt {{{\sum {x_1^2} } \over n} - {{(\overline x )}^2}}

=∑i=110(xi−p)210−(∑i=110(xi−p)10)2= \sqrt {{{\sum\limits_{i = 1}^{10} {{{({x_i} - p)}^2}} } \over {10}} - {{\left( {{{\sum\limits_{i = 1}^{10} {({x_i} - p)} } \over {10}}} \right)}^2}}

[ Standard deviation is free from shifting of origin.]

=910−(310)2= \sqrt {{9 \over {10}} - {{\left( {{3 \over {10}}} \right)}^2}}

=910−9100= \sqrt {{9 \over {10}} - {9 \over {100}}}

=90−9100= \sqrt {{{90 - 9} \over {100}}}

=81100= \sqrt {{{81} \over {100}}}

=910= {9 \over {10}}

Related Learning Resources & Exam Preparation