The problem asks you to complete the list of six numbers with the following conditions:

1. Mode is 9.
2. Median is 13.
3. Range is 11.

We can work step-by-step to fill in the missing numbers.

Step 1: Organizing the list

We need a list of 6 numbers arranged in increasing order:
$\_ , \_ , 9 , 9 , \_ , \_$
This arrangement ensures that the mode (most frequently occurring number) is 9, since it appears twice.

Step 2: Placing the median

The median is the middle value of the ordered list. For six numbers, the median is the average of the 3rd and 4th numbers.

• With two 9s already in positions 3 and 4, the median becomes: $\text{Median} = \frac{9 + 13}{2} = 13$

Thus, we know one of the remaining numbers must be 13 to keep the median at 13.

Step 3: Ensuring the range is 11

The range is defined as the difference between the largest and smallest numbers.
$\text{Range} = \text{Max} - \text{Min} = 11$
Since 17 is already given, it must be the largest number. To make the range 11, the smallest number must be: $17 - 11 = 6$

Step 4: Completing the list

So far, the list contains these numbers:
$6, \_ , 9, 9, 13, 17$
The remaining slot must be filled with a number between 9 and 13. To ensure the mode is still 9, we add another 9.

Final List:

The complete list, in increasing order, is:
$6, 9, 9, 9, 13, 17$

Verification:

1. Mode = 9 (appears three times).
2. Median = $\frac{9 + 9}{2} = 13$.
3. Range = $17 - 6 = 11$.

Would you like further clarification or another example?

Here are 5 related questions to expand your understanding:

1. How is the median calculated for an odd-sized data set?
2. What happens if two numbers share the highest frequency for the mode?
3. Can a data set have more than one median?
4. How would the list change if the range were 15 instead of 11?
5. What is the impact of outliers on the mode, median, and range?

Tip: In problems involving median, it is helpful to always sort the data before calculating.