JEE MAIN - Mathematics (2016 (Offline) - No. 4)
If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?
3$$a$$2 - 26$$a$$ + 55 = 0
3$$a$$2 - 32$$a$$ + 84 = 0
3$$a$$2 - 34$$a$$ + 91 = 0
3$$a$$2 - 23$$a$$ + 44 = 0
Explanation
The formula for standard deviation (S.D)
$$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {{{\sum {{x_i}} } \over n}} \right)}^2}} $$
Where $$\sum {x_i^2 = } $$ Sum of square of the numbers
$$ = {2^2} + {3^2} + {a^2} + {11^2}$$
$$ = 4 + 9 + {a^2} + 121$$
$$ = 134 + {a^2}$$
$$\sum {{x_i}} = $$ Sum of numbers
$$ = 2 + 3 + a + 11$$
$$ = 16 + a$$
$$\therefore\,\,\,$$ $$SD = \sqrt {{{134 + a{}^2} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} $$
$$ \Rightarrow \sqrt {{{134 + {a^2}} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} = 3.5 = {7 \over 2}$$ (given)
$$ \Rightarrow {{134 + {a^2}} \over 4} - {\left( {{{16 + a} \over 4}} \right)^2} = {{49} \over 4}$$
$$ \Rightarrow 4\left( {134 + {a^2}} \right) - \left( {256 + 32a + {a^2}} \right) = 4 \times 49$$
$$ \Rightarrow 3{a^2} - 32a + 84 = 0$$
$$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {{{\sum {{x_i}} } \over n}} \right)}^2}} $$
Where $$\sum {x_i^2 = } $$ Sum of square of the numbers
$$ = {2^2} + {3^2} + {a^2} + {11^2}$$
$$ = 4 + 9 + {a^2} + 121$$
$$ = 134 + {a^2}$$
$$\sum {{x_i}} = $$ Sum of numbers
$$ = 2 + 3 + a + 11$$
$$ = 16 + a$$
$$\therefore\,\,\,$$ $$SD = \sqrt {{{134 + a{}^2} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} $$
$$ \Rightarrow \sqrt {{{134 + {a^2}} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} = 3.5 = {7 \over 2}$$ (given)
$$ \Rightarrow {{134 + {a^2}} \over 4} - {\left( {{{16 + a} \over 4}} \right)^2} = {{49} \over 4}$$
$$ \Rightarrow 4\left( {134 + {a^2}} \right) - \left( {256 + 32a + {a^2}} \right) = 4 \times 49$$
$$ \Rightarrow 3{a^2} - 32a + 84 = 0$$
Comments (0)
