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

The correct option is 4.

Let the 10 students' scores be ordered from lowest to highest, represented as:
$$s_1\le s_2\le \dots \le s_9\le s_{10}$$

We are given the mean of the lowest 9 scores is 42. Since there are 9 scores, their sum is:
$$s_1+s_2+\dots +s_9=9\times 42=378$$

We are also given that the mean of the highest 9 scores is 47. Their sum is:
$$s_2+s_3+\dots +s_{10}=9\times 47=423$$

Subtracting the first equation from the second equation gives the difference between the highest and lowest score:
$$(s_2+\dots +s_{10})-(s_1+\dots +s_9)=423-378$$
$$s_{10}-s_1=45$$

Let the total sum of all 10 scores be represented as S. We can express S in two ways:
$$S=(s_1+\dots +s_9)+s_{10}=378+s_{10}$$
$$S=s_1+(s_2+\dots +s_{10})=s_1+423$$

To find the range of the mean for the 10 students, we need to find the maximum and minimum possible values of S, which depend on the bounds of $s_1$ and $s_{10}$.
Since the scores are ordered, we know that:
$$s_1\le s_2$$

The average of $s_2,\dots ,s_9$ must lie between $s_2$ and $s_9$.
Let $A=s_2+\dots +s_9$.
Then we have:
$$s_1+A=378$$
$$A+s_{10}=423$$

Since $s_1\le s_2$ and $s_9\le s_{10}$, the average of the middle 8 terms, $\frac{A}{8}$, must satisfy:
$$s_1\le \frac{A}{8}\le s_{10}$$

Let us find the minimum possible value of $s_{10}$ to minimize the sum S, and the maximum possible value of $s_{10}$ to maximize S.
From $s_1\le \frac{A}{8}$, we substitute $s_1=s_{10}-45$ and $A=423-s_{10}$:
$$s_{10}-45\le \frac{423-s_{10}}{8}$$
$$8s_{10}-360\le 423-s_{10}$$
$$9s_{10}\le 783\Rightarrow s_{10}\le 87$$

Similarly, from $\frac{A}{8}\le s_{10}$, we substitute $A=423-s_{10}$:
$$\frac{423-s_{10}}{8}\le s_{10}$$
$$423-s_{10}\le 8s_{10}$$
$$423\le 9s_{10}\Rightarrow s_{10}\ge 47$$

Thus, the bounds for $s_{10}$ are:
$$47\le s_{10}\le 87$$

Since $S=378+s_{10}$, the minimum sum is:
$$S_{\text{min}}=378+47=425$$

The maximum sum is:
$$S_{\text{max}}=378+87=465$$

The minimum possible mean for the 10 students is:
$${\mu }_{\text{min}}=\frac{425}{10}=42.5$$

The maximum possible mean for the 10 students is:
$${\mu }_{\text{max}}=\frac{465}{10}=46.5$$

The difference between the maximum possible mean and the minimum possible mean is:
$${\mu }_{\text{max}}-{\mu }_{\text{min}}=46.5-42.5=4$$