To construct a set of 4 positive whole numbers satisfying the condition $\text{median} < \text{mean} < \text{mode}$, let's analyze the problem step-by-step:

Conditions:

1. Median: For 4 numbers, the median is the average of the two middle numbers when the numbers are ordered.
2. Mean: The mean is the sum of all numbers divided by 4.
3. Mode: The mode is the number that appears most frequently.

Approach:

• Median < Mean: This means the central tendency of the data skews higher.
• Mean < Mode: This implies the most frequently occurring number is larger than the average.

Let's construct such a set:

Example Set:

$\{1, 2, 2, 6\}$

Step-by-Step Verification:

1. Median: The ordered set is $\{1, 2, 2, 6\}$. Median = $\frac{2 + 2}{2} = 2$.

2. Mean: Mean = $\frac{1 + 2 + 2 + 6}{4} = \frac{11}{4} = 2.75$.

3. Mode: Mode = $2$, since it appears twice, which is more frequent than other numbers.

4. Condition Check:

   • Median $= 2$,
   • Mean $= 2.75$,
   • Mode $= 2$,
   • Clearly, $2 < 2.75 < 6$.

This set satisfies $\text{median} < \text{mean} < \text{mode}$.

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Would you like more examples or explanations? Here are some questions to explore further:

1. Can you find another set of numbers meeting the condition?
2. Why is the mean typically larger in skewed distributions?
3. Can this condition be satisfied with non-integers?
4. How does increasing the range of the data affect the relationship between mean, median, and mode?
5. Can there be cases where the mean equals the mode?

Tip: Understanding how each central tendency measure behaves with skewness helps in constructing examples like these.