In a group of $10$ students, the mean of the lowest $9$ scores is $42$ while the mean of the highest $9$ scores is $47$. For the entire group of $10$ students, the maximum possible mean exceeds the minimum possible mean by

We have 10 students with test scores. We arrange these scores from lowest to highest: $s_1 \leq s_2 \leq s_3 \leq ... \leq s_{10}$

Given information:

Mean of lowest 9 scores = 42

Mean of highest 9 scores = 47

Key insight: The "lowest 9 scores" are $s_1, s_2, s_3, ..., s_9$, and the "highest 9 scores" are $s_2, s_3, s_4, ..., s_{10}$.

Notice that 8 scores ($s_2$ through $s_9$) appear in both groups!

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We convert the means to sums:

Sum of lowest 9 scores = $9 \times 42 = 378$

Sum of highest 9 scores = $9 \times 47 = 423$

The difference between these sums:

Sum of highest 9 - Sum of lowest 9 = $423 - 378 = 45$

But what does this difference represent?

Sum of highest 9 = $s_2 + s_3 + s_4 + ... + s_{10}$

Sum of lowest 9 = $s_1 + s_2 + s_3 + ... + s_9$

When we subtract: $(s_2 + s_3 + ... + s_{10}) - (s_1 + s_2 + ... + s_9) = s_{10} - s_1$

Therefore: $s_{10} - s_1 = 45$

This means the highest score exceeds the lowest score by exactly 45 points.

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To maximize the total sum (and thus the mean), we want to make the individual scores as large as possible while satisfying our constraints.

Since $s_1 + s_2 + ... + s_9 = 378$ and we want to maximize $s_{10}$, we should minimize $s_1$.

The maximum $s_{10}$ occurs when these 9 scores are as small as possible while maintaining $s_1 \leq s_2 \leq ... \leq s_9$.

This happens when $s_1 = s_2 = ... = s_9 = \tfrac{378}{9} = 42$.

Therefore: $s_{10} = 42 + 45 = 87$

Maximum possible sum = $378 + 87 = 465$

Maximum possible mean = $\tfrac{465}{10} = 46.5$

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For minimum mean, we want to minimize the total sum.

Since $s_2 + s_3 + ... + s_{10} = 423$ and we want to minimize $s_1$, we should make $s_2$ through $s_{10}$ as large as possible while maintaining $s_2 \leq s_3 \leq ... \leq s_{10}$.

This happens when $s_2 = s_3 = ... = s_{10} = \tfrac{423}{9} = 47$.

Therefore: $s_1 = 47 - 45 = 2$

Minimum possible sum = $2 + 423 = 425$

Minimum possible mean = $\tfrac{425}{10} = 42.5$

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Difference = Maximum mean - Minimum mean = $46.5 - 42.5 = 4$

The key insight in this problem is recognizing that the difference between the sums of the highest and lowest 9 scores directly gives us the difference between the highest and lowest individual scores, which then allows us to find the extreme possible means.