The correct option is (A): 11.5
Explanation: Let the 11 numbers be \( N_1, N_2, N_3, \ldots, N_{11} \).
The average of these 11 numbers is 10.9, so their total sum is:
\[\text{Total sum} = 11 \times 10.9 = 119.9\]
The average of the first six numbers \( N_1, N_2, N_3, N_4, N_5, N_6 \) is 10.5, so their total sum is:
\[\text{Sum of first six} = 6 \times 10.5 = 63\]
The average of the last six numbers \( N_6, N_7, N_8, N_9, N_{10}, N_{11} \) is 11.4, so their total sum is:
\[\text{Sum of last six} = 6 \times 11.4 = 68.4\]
Now, \( N_6 \) is counted in both the sum of the first six and the last six. Therefore, the total sum can be expressed as:
\[\text{Total sum} = \text{Sum of first six} + \text{Sum of last six} - N_6\]
Substituting the known sums:
\[119.9 = 63 + 68.4 - N_6\]
This simplifies to:
\[119.9 = 131.4 - N_6\]
Solving for \( N_6 \):
\[N_6 = 131.4 - 119.9 = 11.5\]
Thus, the middle number \( N_6 \) is Option A: 11.5.