Question

The mean of the number 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=3,b=4
• B. a=0,b=7
• C. a=5,b=2
• D. a=1,b=6

Answer 1

Answer: (A)

a=3,b=4

$\Rightarrow$ Mean of $a,b,8,5,10$ is $6$

$\therefore$ $\frac{a+b+8+5+10}{5}=6$

$\Rightarrow$ $a+b+23=30$

$\therefore$ $a+b=7$ ----- ( 1 )

$\Rightarrow$ $b=7-a$ ----- ( 2 )

Given variance is $6.8$

$\Rightarrow$ $Variance=\frac{\sum (x_i-A{)}^2}{n}$

$\Rightarrow$ $6.8=\frac{(a-6{)}^2+(b-6{)}^2+(8-6{)}^2+(5-6{)}^2+(10-6{)}^2}{5}$

$\Rightarrow$ $34=(a-6{)}^2+(b-6{)}^2+4+1+16$

$\Rightarrow$ $34=a^2-12a+36+b^2-12b+36+21$

$\Rightarrow$ $34=a^2+b^2-12(a+b)+93$

$\Rightarrow$ $34=a^2+b^2-12\left(7\right)+93$ [ From ( 1 )]

$\Rightarrow$ $34=a^2+b^2-84+93$

$\therefore$ $a^2+b^2=25$

$\Rightarrow$ $a^2+(7-a{)}^2=25$

$\Rightarrow$ $a^2+49-14a+a^2=25$

$\Rightarrow$ $a^2-7a+12=0$

$\Rightarrow$ $a^2-4a-3a+12=0$

$\Rightarrow$ $a(a-4)-3(a-4)=0$

$\Rightarrow$ $(a-4)(a-3)=0$

$\therefore$ $a=4$ or $b=3$

$\therefore$ $b=4$ and $a=3$