Question

Mean of $6$ observations is $10$ and their variance is $\frac{20}{3}$. If their observations are $15,11,10,7,a,b$ then $\left|a-b\right|$ is equal to:

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$1$

Explanation of correct option:

Determination of mean deviation:

The observations are $15,11,10,7,a,b$, which is a total of $6$observations and their mean is $10$.

Let the mean of given observations is $\overline{x}$

$\overline{x}$ $=$$\frac{\text{sum of observations}}{\text{number of observations}}$

$\Rightarrow 10$$=$$\frac{15+11+10+7+a+b}{6}$

$\Rightarrow 60$$=$ $15+11+10+7+a+b$

$\Rightarrow 60$$=$$43+a+b$

$\Rightarrow a+b$$=$$60-43$

$\Rightarrow a+b$$=$$17\dots \dots \left(1\right)$

The variance is given by

$\begin{array}{rcl}\text{Variance}& =& \frac{\text{sum of the square of observations}}{\text{number of observations}}-{\left(\text{mean}\right)}^2\\ & \Rightarrow & \frac{\text{20}}{3}=\frac{{\left(15\right)}^2+{\left(11\right)}^2+{\left(10\right)}^2+{\left(7\right)}^2+a^2+b^2}{6}-{\left(10\right)}^2\\ & \Rightarrow & \frac{\text{20}}{3}=\frac{225+121+100+49+a^2+b^2}{6}-\left(100\right)\\ & \Rightarrow & \frac{\text{20}}{3}=\frac{225+121+100+49+a^2+b^2-600}{6}\\ & \Rightarrow & \text{120}=3\left(225+121+100+49+a^2+b^2-600\right)\\ & \Rightarrow & \text{120}=3\left(495+a^2+b^2-600\right)\\ & \Rightarrow & \text{40}=495+a^2+b^2-600\\ & \Rightarrow & a^2+b^2=600+40-495\\ & \Rightarrow & a^2+b^2=640-495\\ & \Rightarrow & a^2+b^2=145\dots \dots \left(2\right)\end{array}$

Using ${\left(a+b\right)}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{b}$ formula

Substituting the values from equations $\left(1\right)\text{and}\left(2\right)$

$\begin{array}{rcl}{\left(a+b\right)}^2& =& a^2+b^2+2ab\\ & \Rightarrow & {\left(17\right)}^2=145+2ab\\ & \Rightarrow & 289=145+2ab\\ & \Rightarrow & 2ab=289-145\\ & \Rightarrow & ab=\frac{144}{2}\\ & \Rightarrow & ab=72\end{array}$

Now, using ${\left(a-b\right)}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{a}\mathit{b}$ formula

Simplifying the above formula we get

$\begin{array}{rcl}{\left(a-b\right)}^2& =& a^2+b^2-4ab+2ab\left[-4ab+2ab=-2ab\right]\\ & =& \left(a^2+b^2+2ab\right)-4ab\end{array}$

Substituting the values, we get

$$\begin{gathered} =\left(145+2\left(72\right)\right)-4\left(72\right) \\ =289-288 \\ =1 \end{gathered}$$

Thus $\left(a-b\right)=\pm 1$

Therefore, $\left|a-b\right|=1$

Hence, the correct answer is option (B).