To solve this problem, we need to find the maximum and minimum possible overall average weight of the five students and then compute the difference between them.

Step 1: Define the Variables

Let the weights of the five students be $w_1, w_2, w_3, w_4, w_5$ arranged in increasing order, so $w_1 \leq w_2 \leq w_3 \leq w_4 \leq w_5$.

Step 2: Use the Given Information

• The average weight of the 4 lightest students is 40 kg: $\frac{w_1 + w_2 + w_3 + w_4}{4} = 40$ This implies: $w_1 + w_2 + w_3 + w_4 = 160 \quad \text{(Equation 1)}$

• The average weight of the 4 heaviest students is 45 kg: $\frac{w_2 + w_3 + w_4 + w_5}{4} = 45$ This implies: $w_2 + w_3 + w_4 + w_5 = 180 \quad \text{(Equation 2)}$

Step 3: Calculate the Weight of the Fifth Student

Subtract Equation 1 from Equation 2: $(w_2 + w_3 + w_4 + w_5) - (w_1 + w_2 + w_3 + w_4) = 180 - 160$ $w_5 - w_1 = 20 \quad \text{(Equation 3)}$ This means that the difference between the heaviest and the lightest student is 20 kg.

Step 4: Calculate the Overall Average Weight

The total weight of the five students is: $w_1 + w_2 + w_3 + w_4 + w_5 = 160 + w_5 = 180 + w_1$

Thus, the overall average weight of the five students is: $\text{Overall average weight} = \frac{w_1 + w_2 + w_3 + w_4 + w_5}{5} = \frac{160 + w_5}{5} = \frac{180 + w_1}{5}$

Step 5: Determine Maximum and Minimum Possible Overall Average Weights

• Maximum Overall Average: When $w_1$ is minimized (i.e., $w_1 = w_5 - 20$): $\text{Max overall average} = \frac{180 + (w_5 - 20)}{5} = \frac{w_5 + 160}{5} = \frac{180 + 180 - 20}{5} = 44 \text{ kg}$

• Minimum Overall Average: When $w_5$ is minimized (i.e., $w_5 = w_1 + 20$): $\text{Min overall average} = \frac{160 + (w_1 + 20)}{5} = \frac{180 + w_1}{5} = 40 \text{ kg}$

Step 6: Compute the Difference

The difference between the maximum and minimum possible overall average weights is: $\text{Difference} = 44 \text{ kg} - 40 \text{ kg} = 4 \text{ kg}$

Final Answer:

The difference between the maximum and minimum possible overall average weight is 4 kg.

Would you like further details or have any questions about this problem? Here are some related questions:

1. How do average and sum calculations differ in various problems?
2. How does the arrangement of data affect the range of possible averages?
3. What are other methods to calculate the range of possible averages?
4. Can this approach be applied to other weighted averages?
5. What are the implications of outliers on average calculations?

Tip: Always check the extreme cases when determining the range of possible values in problems involving averages.