Verified Solution • 2025–26 Session moderate

QJEEM18Ma2db5: JEE Main 2018 Mathematics - Class 11 JEEM18Ma2db5 Solution: If and , then the standard deviation of the 9 items is

If ∑i=19(xi−5)=9\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9 and

∑i=19(xi−5)2=45\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45 , then the standard deviation of the 9 items
x1,x2,.......,x9{x_1},{x_2},.......,{x_9} is

Multiple Choice Options

  • A 3
  • B 9
  • C 4
  • D 2

Step-by-Step Verified Solution

IMPORTANT POINT :-

When every number is added or subtracted by a fixed number then the standard Deviation remain unchanged.

so let xi−5=yi{x_i} - 5 = {y_i}

So, new equation is ∑i=19yi=9\sum\limits_{i = 1}^9 {{y_i}} = 9

and ∑i=19yi2=45\sum\limits_{i = 1}^9 {y_i^2} = 45

As, we know. Standard Deviation (S.D)

=∑i=19yi29−(∑i=19yi9)2= \sqrt {{{\sum\limits_{i = 1}^9 {y_i^2} } \over 9} - {{\left( {{{\sum\limits_{i = 1}^9 {yi} } \over 9}} \right)}^2}}

=459−(99)2= \sqrt {{{45} \over 9} - {{\left( {{9 \over 9}} \right)}^2}}

=5−1= \sqrt {5 - 1}

== 4\sqrt 4

=2=2

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