If the mean and variance of observations x, y, 12, 14, 4, 10, 2 is 8 and 16 respectively where x > y. Then, the value of 3x – y is

The correct answer is 18.

Here is the step-by-step explanation to find the value of $3x-y$:

Step 1: Use the given mean to find a relation between $x$ and $y$.
We are given 7 observations: $x,y,12,14,4,10,2$.
The mean of these observations is given as 8.

$\text{Mean}=\frac{x+y+12+14+4+10+2}{7}=8$

Multiplying both sides by 7:
$x+y+42=56$
$x+y=14$     --- (Equation 1)

Step 2: Use the variance formula to find another relation.
The variance (${\sigma }^2$) is given as 16. The formula for variance is:

${\sigma }^2=\frac{\sum {(x_i-\overline{x})}^2}{n}$

Substituting the mean ($\overline{x}=8$) and the given observations:

$\frac{{(x-8)}^2+{(y-8)}^2+{(12-8)}^2+{(14-8)}^2+{(4-8)}^2+{(10-8)}^2+{(2-8)}^2}{7}=16$

Simplifying the squared terms in the numerator:
${(x-8)}^2+{(y-8)}^2+4^2+6^2+{(-4)}^2+2^2+{(-6)}^2=16\times 7$
${(x-8)}^2+{(y-8)}^2+16+36+16+4+36=112$
${(<!--="-->(x-8)}^2+{(y-8)}^2+108=112$
${(x-8)}^2+{(y-8)}^2=4$     --- (Equation 2)

Step 3: Solve for $x$ and $y$.
Let $a=x-8$ and $b=y-8$.
From Equation 1, we can write:
$(x-8)+(y-8)=14-16$
$a+b=-2$

From Equation 2, we have:
$a^2+b^2=4$

Using the identity ${(a+b)}^2=a^2+b^2+2ab$:
${(-2)}^2=4+2ab$
$4=4+2ab$
$2ab=0\Rightarrow ab=0$

Since $ab=0$, either $a=0$ or $b=0$.
If $a=0$, then $b=-2$ (since $a+b=-2$).
If $b=0$, then $a=-2$.

We are given that $x>y$, which means $x-8>y-8$, or $a>b$.
Therefore, we must choose $a=0$ and $b=-2$.

Substituting back to find $x$ and $y$:
$x-8=0\Rightarrow x=8$
$y-8=-2\Rightarrow y=6$

Step 4: Calculate the required value $3x-y$.
$3x-y=3\left(8\right)-6=24-6=18$