The mean of the numbers $a, b, 8,5$ and 10 is 6 and the variance is 6.80 , then the possible values of $a$…
The mean of the numbers $a, b, 8,5$ and 10 is 6 and the variance is 6.80 , then the possible values of $a$ and $b$ are
- $a=2, b=3$
- $a=4, b=5$
- $a=1, b=3$
- $a=3, b=4$
Solution
Given data, $a, b, 8,5,10$
$\therefore \quad$ Mean $=\frac{a+b+8+5+10}{5}$
$6=\frac{a+b+23}{5}$
$\Rightarrow \quad a+b=7$ ...(i)
Variance
$=\frac{(a-6)^2+(b-6)^2+(8-6)^2+(5-6)^2+(10-6)^2}{5}$
$\Rightarrow 6.80=\frac{a^2+36-12 a+b^2+36-12 b+4+1+16}{5}$
$\begin{array}{ll}\Rightarrow & 34=a^2+b^2-12 a-12 b+93 \\ \Rightarrow & a^2+b^2-12(a+b)=-59\end{array}$
$\begin{array}{rrr}\Rightarrow & a^2+b^2-12(7) & =-59 \\ \Rightarrow & a^2+b^2 =25\end{array}$
$\Rightarrow \quad a^2+(7-a)^2=25$ [from Eq. (i), $b=7-a$ ]
$\begin{array}{ll}\Rightarrow & 2 a^2+49-14 a=25 \\ \Rightarrow & 2 a^2-14 a+24=0\end{array}$
$\Rightarrow \quad a^2-7 a+12=0$
$(a-4)(a-3)=0$
$\Rightarrow \quad a=4$ or $a=3$
If $a=3 \Rightarrow b=7-3=4$
Asked in: AP EAMCET 2021 (23 Aug Shift 2)
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