Consider a data consisting of 10 observations $x_1,x_2,\dots ,x_{10}$, whose mean is 5 and variance is 7. If the mean and the variance of the first 8 observations $x_1,x_2,\dots ,x_8$ are 4 and 3.5, respectively, and x₉ < x₁₀, then the value of $3x_9+2x_{10}$ is .

The correct answer is 44.

We are given a dataset of 10 observations $x_1,x_2,\dots ,x_{10}$.

Let's summarize the given information:
1. Total number of observations, N = 10.
2. Mean of all 10 observations, $\overline{x}=5$.
3. Variance of all 10 observations, ${\sigma }^2=7$.
4. Mean of the first 8 observations, ${\overline{x}}_8=4$.
5. Variance of the first 8 observations, ${\sigma }_8^2=3.5$.
6. $x_9<x_{10}$.

Step 1: Find the sum of the observations.

The total sum of all 10 observations is:

$\sum _{i=1}^{10}{x}_i=10\times 5=50$

The sum of the first 8 observations is:

$\sum _{i=1}^8{x}_i=8\times 4=32$

Therefore, the sum of the remaining two observations, $x_9$ and $x_{10}$, is:

$x_9+x_{10}=50-32=18$

So, we have our first equation:

$x_9+x_{10}=18$ --- (Equation 1)

Step 2: Find the sum of squares of the observations.

The formula for variance is given by:

${\sigma }^2=\frac{1}{n}\sum _{i=1}^n{x}_i^2-{\left(\overline{x}\right)}^2$

For the first 8 observations:

$3.5=\frac{1}{8}\sum _{i=1}^8{x}_i^2-4^2$

$3.5+16=\frac{1}{8}\sum _{i=1}^8{x}_i^2$

$19.5=\frac{1}{8}\sum _{i=1}^8{x}_i^2$

$\sum _{i=1}^8{x}_i^2=19.5\times 8=156$

For all 10 observations:

$7=\frac{1}{10}\sum _{i=1}^{10}{x}_i^2-5^2$

$7+25=\frac{1}{10}\sum _{i=1}^{10}{x}_i^2$

$32=\frac{1}{10}\sum _{i=1}^{10}{x}_i^2$

$\sum _{i=1}^{10}{x}_i^2=32\times 10=320$

Subtracting the sum of squares of the first 8 observations from the total sum of squares gives:

$x_9^2+x_{10}^2=320-156=164$ --- (Equation 2)

Step 3: Solve for $x_9$ and $x_{10}$.

Using the algebraic identity ${\left(a+b\right)}^2=a^2+b^2+2ab$:

${\left(x_9+x_{10}\right)}^2=x_9^2+x_{10}^2+2x_9{x}_{10}$

${18}^2=164+2x_9{x}_{10}$

$324=164+2x_9{x}_{10}$

$2x_9{x}_{10}=324-164=160$

$x_9{x}_{10}=80$

Now, using the identity ${\left(a-b\right)}^2={\left(a+b\right)}^2-4ab$:

${\left(x_{10}-x_9\right)}^2={18}^2-4\left(80\right)=324-320=4$

Taking square root on both sides (since $x_9<x_{10}$, $x_{10}-x_9>0$):

$x_{10}-x_9=2$ --- (Equation 3)

Adding Equation 1 and Equation 3:

$\left(x_9+x_{10}\right)+\left(x_{10}-x_9\right)=18+2$

$2x_{10}=20\Rightarrow x_{10}=10$

Substituting $x_{10}=10$ into Equation 1:

$x_9+10=18\Rightarrow x_9=8$

Step 4: Compute the value of $3x_9+2x_{10}$.

$3x_9+2x_{10}=3\left(8\right)+2\left(10\right)=24+20=44$