Examination Student Scores Analysis

Let's analyze the scores of the 8 students in the examination to determine the relationship between the Median score (N) and the Mean score (M).

We are given the following information:

• Total number of students: 8
• Scores of 2 brilliant students: 100 and 100
• Scores of the remaining 6 students: Less than 12. Since there are no negative marks, the score must be $\ge 0$. Let the scores of these six students be $s_1, s_2, s_3, s_4, s_5, s_6$. Each $s_i$ satisfies $0 \le s_i \lt 12$. Assuming integer scores (common in exams), $0 \le s_i \le 11$.

Calculating the Mean Score (M)

The Mean score (M) is the total sum of scores divided by the number of students.

Total sum of scores = (Score of student 1) + ... + (Score of student 8)

Total sum of scores = $100 + 100 + s_1 + s_2 + s_3 + s_4 + s_5 + s_6 = 200 + \sum_{i=1}^6 s_i$

The sum of the six scores $\sum_{i=1}^6 s_i$ can range from $6 \times 0 = 0$ (minimum, if all six score 0) to $6 \times 11 = 66$ (maximum, if all six score 11).

• Minimum total sum = $200 + 0 = 200$
• Maximum total sum = $200 + 66 = 266$

The Mean (M) is calculated as:

M = $\frac{\text{Total sum of scores}}{\text{Number of students}} = \frac{200 + \sum_{i=1}^6 s_i}{8}$

• Minimum Mean (M) = $\frac{200}{8} = 25$
• Maximum Mean (M) = $\frac{266}{8} = 33.25$

So, the Mean score (M) is always in the range $25 \le M \le 33.25$.

Calculating the Median Score (N)

The Median score (N) is the middle value when the scores are arranged in ascending order. For 8 students, the median is the average of the 4th and 5th scores in the sorted list.

Let the scores arranged in ascending order be $x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8$.

We know that 6 students scored less than 12 ($ \le 11$) and 2 students scored 100. When sorted, the 6 scores less than 12 will come before the scores of 100.

So, the first six scores ($x_1$ to $x_6$) are the six scores less than 12. The last two scores ($x_7$ and $x_8$) are 100.

$x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \lt 12$

$x_7 = 100$

$x_8 = 100$

The Median (N) is the average of the 4th and 5th scores:

N = $\frac{x_4 + x_5}{2}$

Since both $x_4$ and $x_5$ are among the six scores less than 12, $x_4 \lt 12$ and $x_5 \lt 12$. Therefore, their average N must also be less than 12.

Minimum Median (N): If the scores of the six students are 0, 0, 0, 0, 0, 0, the sorted list is 0, 0, 0, 0, 0, 0, 100, 100. N = $\frac{0+0}{2} = 0$.

Maximum Median (N): If the scores of the six students are 11, 11, 11, 11, 11, 11, the sorted list is 11, 11, 11, 11, 11, 11, 100, 100. N = $\frac{11+11}{2} = 11$.

In any case where the six scores are between 0 and 11 (inclusive), the 4th and 5th scores will also be between 0 and 11 (inclusive). Thus, the Median (N) will be between 0 and 11.

So, the Median score (N) is always in the range $0 \le N \le 11$. More precisely, $0 \le N \lt 12$.

Comparing Mean (M) and Median (N)

We found the following ranges:

• Mean (M): $25 \le M \le 33.25$
• Median (N): $0 \le N \le 11$ (assuming integer scores) or $0 \le N \lt 12$ (if scores can be non-integer less than 12)

Let's consider the maximum possible value for 2N based on the range of N:

• Maximum $2N = 2 \times 11 = 22$ (assuming integer scores)
• Maximum $2N$ is less than $2 \times 12 = 24$ (if scores can be non-integer less than 12)

Now let's compare the minimum possible value of M with the maximum possible value of 2N.

Minimum M = 25.

Maximum 2N is less than 24.

Since the smallest possible value for M (25) is greater than the largest possible value for 2N (which is less than 24), it is always true that M is greater than 2N.

$M \gt 2N$

Evaluating the Given Options

Let's check which option matches our finding:

• Option 1: $M \gt 2N$. This matches our conclusion.
• Option 2: $3N/2 \lt M \le 2N$. This includes $M \le 2N$, which contradicts our finding.
• Option 3: $N \le M \le 3 N/2$. This includes $M \le 1.5N$. Given $M \ge 25$ and $1.5N \le 1.5 \times 11 = 16.5$ (or $1.5 \times (\text{just below } 12) = \text{just below } 18$), this is clearly false.
• Option 4: From the above information nothing can be said. We have found a definitive relationship, so this is false.

The only true relationship based on the given information is $M \gt 2N$.