The mean of the numbers x, y, 12, 14, 4, 10, 2 is 8 and the variance is 16. Find the value of 3x − y.

The correct option is 18.

Let's break down the problem step-by-step to find the value of $3x-y$.

We are given seven numbers: $x,y,12,14,4,10,2$.

The total number of observations, $n=7$.

Step 1: Use the formula for the mean to find a relationship between x and y.

The mean ($\overline{x}$) of a dataset is given by:

$\overline{x}=\frac{\sum x_i}{n}$

Given that the mean is 8:

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

$8=\frac{x+y+42}{7}$

Multiplying both sides by 7:

$56=x+y+42$

$x+y=14$ --- (Equation 1)

From Equation 1, we can write $y=14-x$.

Step 2: Use the formula for variance to set up a second equation.

The formula for variance (${\sigma }^2$) is:

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

Given that the variance is 16 and the mean ($\overline{x}$) is 8:

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

Calculate the squares of the known numbers:

${12}^2=144$
${14}^2=196$
$4^2=16$
${10}^2=100$
$2^2=4$

Sum of the squares of the known numbers:

$144+196+16+100+4=460$

Substitute this back into the variance formula:

$16=\frac{x^2+y^2+460}{7}$ − $64$

Add 64 to both sides:

$80=\frac{x^2+y^2+460}{7}$

Multiply both sides by 7:

$560=x^2+y^2+460$

$x^2+y^2=100$ --- (Equation 2)

Step 3: Solve for x and y.

Substitute $y=14-x$ into Equation 2:

$x^2+{(14-x)}^2=100$

$x^2+(196-28x+x^2)=100$

$2x^2-28x+196-100=0$

$2x^2-28x+96=0$

Divide the entire equation by 2:

$x^2-14x+48=0$

Factor the quadratic equation:

$(x-8)(x-6)=0$

This gives two possible values for $x$:

$x=8$ or $x=6$

Since $x+y=14$:

If $x=8$, then $y=6$.
If $x=6$, then $y=8$.

Step 4: Calculate 3x − y.

To match the given correct option (18):

Let $x=8$ and $y=6$:

$3x-y=3\left(8\right)-6$

$3x-y=24-6=18$

Thus, the value of $3x-y$ is 18.