Let the original list of 6 numbers be $\(L=\left\{n_1, n_2, n_3, n_4, n_5, n_6\right\}\)$.
The range of a list of numbers is the difference between the highest and lowest values in the list.
Let $\(n_{\text {max }}\)$ be the highest number in the original list and $\(n_{\text {min }}\)$ be the lowest number in the original list.
Given, the range of the original list is 27.4 .
So, $\(n_{\max }-n_{\min }=27.4\)$.
Now, two operations are performed on each number in the list:
Step 1: 10 is added to each number in the list.
The new list, let's call it $\(L^{\prime}$, will be $\left\{n_1+10, n_2+10, \ldots, n_6+10\right\}\)$.
The highest number in $\(L^{\prime}$ will be $n_{\max }+10\)$.
The lowest number in $\(L^{\prime}$ will be $n_{\text {min }}+10\)$.
The range of $\(L^{\prime}$ will be $\left(n_{\max }+10\right)-\left(n_{\min }+10\right)=n_{\max }+10-n_{\min }-10=n_{\max }-$ $n_{\text {min }}\)$.
So, adding a constant to each number in a list does not change the range. The range of $\(L^{\prime}$\) is still 27.4.

Step 2: Each resulting number is divided by 2.
The final list, let's call it $\(L^{\prime \prime}$, will be $\left\{\frac{n_1+10}{2}, \frac{n_2+10}{2}, \ldots, \frac{n_6+10}{2}\right\}\)$.
The highest number in $\(L^{\prime \prime}$ will be $\frac{n_{\max }+10}{2}\)$.
The lowest number in $\(L^{\prime \prime}$ will be $\frac{n_{\min }+10}{2}\)$.
The range of $\(L^{\prime \prime}$ will be $\frac{n_{\max }+10}{2}-\frac{n_{\min }+10}{2}\)$.
This can be written as $\(\frac{\left(n_{\max }+10\right)-\left(n_{\min }+10\right)}{2}=\frac{n_{\max }+10-n_{\min }-10}{2}=\frac{n_{\max }-n_{\min }}{2}\)$.
So, dividing each number in a list by a constant (in this case, 2 ) divides the range by that same constant.

We know that $\(n_{\max }-n_{\min }=27.4\)$.
Therefore, the range of the final list of numbers will be $\(\frac{27.4}{2}\)$.

$$
\(\frac{27.4}{2}=13.7\)
$$


The final answer is 13.7 .