Question
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.
$$a = 0,\,b = 7$$
B.
$$a = 5,\,b = 2$$
C.
$$a = 1,\,b = 6$$
D.
$$a = 3,\,b = 4$$
Answer :
$$a = 3,\,b = 4$$
Solution :
$$\eqalign{
& {\text{Mean of }}a,\,b,\,8,\,5,\,10{\text{ is }}6 \cr
& \Rightarrow \frac{{a + b + 8 + 5 + 10}}{5} = 6 \cr
& \Rightarrow a + b = 7......\left( {\text{i}} \right) \cr
& {\text{Variance of }}a,\,b,\,8,\,5,\,10{\text{ is }}6.80\, \cr
& \Rightarrow \frac{{{{\left( {a - 6} \right)}^2} + {{\left( {b - 6} \right)}^2} + {{\left( {8 - 6} \right)}^2} + {{\left( {5 - 6} \right)}^2} + {{\left( {10 - 6} \right)}^2}}}{5} = 6.80 \cr
& \Rightarrow {a^2} - 12a + 36 + {\left( {1 - a} \right)^2} + 21 = 34\,\,\,\left[ {{\text{using equation }}\left( {\text{i}} \right)} \right] \cr
& \Rightarrow 2{a^2} - 14a + 24 = 0 \cr
& \Rightarrow {a^2} - 7a + 12 = 0 \cr
& \Rightarrow a = 3{\text{ or }}4 \cr
& \Rightarrow b = 4{\text{ or }}3 \cr} $$
$$\therefore $$ The possible values of $$a$$ and $$b$$ are
$$a = 3{\text{ and }}b = 4{\text{ or }}a = 4{\text{ and }}b = 3$$