JEE MAIN - Mathematics (2008 - No. 1)
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following
gives possible values of a and b?
a = 0, b = 7
a = 5, b = 2
a = 1, b = 6
a = 3, b = 4
Explanation
Given that,
Mean of a, b, 8, 5, 10 = 6
$$\therefore\,\,\,$$ $${{a + b + 8 + 5 + 10} \over 5} = 6$$
$$ \Rightarrow \,\,\,$$ a + b + 23 = 30
$$ \Rightarrow \,\,\,$$ a + b = 7 ....... (1)
Variance $$ = {{\sum {{{\left( {{x_i} - A} \right)}^2}} } \over n} = 6.8$$
$$ \Rightarrow \,\,\,{{{{\left( {6 - a} \right)}^2} + {{\left( {6 - b} \right)}^2} + {{\left( {6 - 8} \right)}^2} + {{\left( {6 - 5} \right)}^2} + {{\left( {6 - 10} \right)}^2}} \over 5} = 6.5$$
$$ \Rightarrow \,\,\,{\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} + 4 + 1 + 16 = 34$$
$$ \Rightarrow {\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} = 13$$
$$ \Rightarrow \,\,\,{a^2} + {b^2} = 25$$
By using (1) we get,
$${a^2} + \left( {7 - a} \right){}^2 = 25$$
$$ \Rightarrow \,\,\,a{}^2 - 7a + 12 = 0$$
$$\therefore\,\,\,$$ a = 3, 4
then b = 4, 3.
Mean of a, b, 8, 5, 10 = 6
$$\therefore\,\,\,$$ $${{a + b + 8 + 5 + 10} \over 5} = 6$$
$$ \Rightarrow \,\,\,$$ a + b + 23 = 30
$$ \Rightarrow \,\,\,$$ a + b = 7 ....... (1)
Variance $$ = {{\sum {{{\left( {{x_i} - A} \right)}^2}} } \over n} = 6.8$$
$$ \Rightarrow \,\,\,{{{{\left( {6 - a} \right)}^2} + {{\left( {6 - b} \right)}^2} + {{\left( {6 - 8} \right)}^2} + {{\left( {6 - 5} \right)}^2} + {{\left( {6 - 10} \right)}^2}} \over 5} = 6.5$$
$$ \Rightarrow \,\,\,{\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} + 4 + 1 + 16 = 34$$
$$ \Rightarrow {\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} = 13$$
$$ \Rightarrow \,\,\,{a^2} + {b^2} = 25$$
By using (1) we get,
$${a^2} + \left( {7 - a} \right){}^2 = 25$$
$$ \Rightarrow \,\,\,a{}^2 - 7a + 12 = 0$$
$$\therefore\,\,\,$$ a = 3, 4
then b = 4, 3.
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