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QJEEM16Ma7835: JEE Main 2016 Mathematics - Class 11 JEEM16Ma7835 Solution: If the standard deviation of the numbers 2, 3, a and 11 is 3.5, …

If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

Multiple Choice Options

  • A 3 aa 2 - 26 aa + 55 = 0
  • B 3 aa 2 - 32 aa + 84 = 0
  • C 3 aa 2 - 34 aa + 91 = 0
  • D 3 aa 2 - 23 aa + 44 = 0

Step-by-Step Verified Solution

The formula for standard deviation (S.D)

=∑xi2n−(∑xin)2= \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {{{\sum {{x_i}} } \over n}} \right)}^2}}

Where ∑xi2=\sum {x_i^2 = } Sum of square of the numbers

=22+32+a2+112= {2^2} + {3^2} + {a^2} + {11^2}

=4+9+a2+121= 4 + 9 + {a^2} + 121

=134+a2= 134 + {a^2}

∑xi=\sum {{x_i}} = Sum of numbers

=2+3+a+11= 2 + 3 + a + 11

=16+a= 16 + a

∴   \therefore\,\,\, SD=134+a24−(16+a4)2SD = \sqrt {{{134 + a{}^2} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}}

⇒134+a24−(16+a4)2=3.5=72\Rightarrow \sqrt {{{134 + {a^2}} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} = 3.5 = {7 \over 2} (given)

⇒134+a24−(16+a4)2=494\Rightarrow {{134 + {a^2}} \over 4} - {\left( {{{16 + a} \over 4}} \right)^2} = {{49} \over 4}

⇒4(134+a2)−(256+32a+a2)=4×49\Rightarrow 4\left( {134 + {a^2}} \right) - \left( {256 + 32a + {a^2}} \right) = 4 \times 49

⇒3a2−32a+84=0\Rightarrow 3{a^2} - 32a + 84 = 0

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