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. $a=3$, $b=4$
• B. $a=0$, $b=7$
• C. $a=5$, $b=2$
• D. $a=1$, $b=6$

Answer 1

The correct option is A

$a=3$, $b=4$

Explanation for the correct option:

Step 1: Find the relation between $a$ and $b$.

Since, it is given that the mean of $6$ observations is $6$ and the observations are$a,b,8,5,10$.

Therefore,

$$\begin{gathered} \frac{a+b+8+5+10}{5}=6 \\ \Rightarrow a+b+8+5+10=30 \\ \Rightarrow a+b=7 \\ \Rightarrow a=7-b...1 \end{gathered}$$

Also, it is given that the variance of $6$ observations is $6.80$.

$$\begin{gathered} \frac{{\left(a-6\right)}^2+{\left(b-6\right)}^2+{\left(8-6\right)}^2+{\left(5-6\right)}^2+{\left(10-6\right)}^2}{5}=6.8 \\ \Rightarrow {\left(a-6\right)}^2+{\left(b-6\right)}^2+{\left(2\right)}^2+{\left(-1\right)}^2+{\left(4\right)}^2=34 \\ \Rightarrow {\left(a-6\right)}^2+{\left(b-6\right)}^2+4+1+16=34 \\ \Rightarrow {\left(a-6\right)}^2+{\left(b-6\right)}^2=13...2 \end{gathered}$$

Step 2: Find the values of $a$ and $b$.

From equation $1$ and equation $2$.

$$\begin{gathered} {(7-b-6)}^2+{\left(b-6\right)}^2=13 \\ \Rightarrow {(1-b)}^2+{\left(b-6\right)}^2=13 \\ \Rightarrow b^2+1-2b+b^2+36-12b=13 \\ \Rightarrow 2b^2-14b+37=13 \\ \Rightarrow 2b^2-14b+24=0 \\ \Rightarrow b=\frac{14\pm \sqrt{{14}^2-4\left(2\right)\left(24\right)}}{2\left(2\right)} \\ \Rightarrow b=\frac{14\pm \sqrt{196-192}}{4} \\ \Rightarrow b=\frac{14\pm \sqrt{4}}{4} \\ \Rightarrow b=\frac{14\pm 2}{4} \\ \Rightarrow b=3,4 \end{gathered}$$

For $b=3$,

$$\begin{gathered} a=7-3 \\ \Rightarrow a=4 \end{gathered}$$

For $b=4$,

$$\begin{gathered} a=7-4 \\ \Rightarrow a=3 \end{gathered}$$

Therefore, the values of $a$ and $b$ are $a=3,b=4$ or $a=4,b=3$.

Hence, option $A$ is the correct option.