# Q5(b) Mean of maxima and centroid of trapezoids [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-reexam/q5b

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-reexam--q5b/

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

**Question. Q5(b)** [10 marks · 2024-06-25]

For the given membership functions determine the crisp output using (1) Mean of Maxima and (2) Centroid / Center of Area.

Swipe across the diagram to read its labels ↔

![First printed membership function: height 0.75, corners x=0,2,3,4.](https://deep-learning-study-room.vercel.app/is2/fig/legacy24r-defuzz-a.4cd4ea6810.svg)

Diagram: First printed membership function: height 0.75, corners x=0,2,3,4.. Labels: First printed membership function: height0.75, corners x=0,2,3,4. | 0 | 0.7 | 1.4 | 2.1 | 2.8 | 3.5 | 4.2 | 4.9 | 5.6 | 6.3 | 7 | 0 | 0.25 | 0.5 | 0.75 | 1 | First graph | μ(x)

First printed membership function: height 0.75, corners x=0,2,3,4.

Swipe across the diagram to read its labels ↔

![Second printed membership function: height 1, corners x=0,3,4,7.](https://deep-learning-study-room.vercel.app/is2/fig/legacy24r-defuzz-b.b1f504dbf4.svg)

Diagram: Second printed membership function: height 1, corners x=0,3,4,7.. Labels: Second printed membership function: height1, corners x=0,3,4,7. | 0 | 0.7 | 1.4 | 2.1 | 2.8 | 3.5 | 4.2 | 4.9 | 5.6 | 6.3 | 7 | 0 | 0.25 | 0.5 | 0.75 | 1 | Second graph | μ(x)

Second printed membership function: height 1, corners x=0,3,4,7.

### Answer

#### Given

Two separate printed graphs; graph 1 corners(0,0),(2,.75),(3,.75),(4,0); graph 2 corners(0,0),(3,1),(4,1),(7,0).

#### To find

MOM and centroid for each graph.

#### Formula and why it applies

MOM averages the x-values where membership reaches its greatest height. For a continuous flat maximum[a, b], MOM=(a+b)/2. Centroid uses all area: divide the sum of area moments by total area.



$$
z_{MOM}=\frac{a+b}{2},\qquad z_{COA}=\frac{\int x\mu(x)dx}{\int\mu(x)dx}=\frac{\sum A_ic_i}{\sum A_i}
$$



#### Solution

##### Step 1. MOM of the first graph

Maximum height.75 occurs throughout[2,3].



$$
z_{MOM,1}=(2+3)/2=2.5
$$



##### Step 2. Centroid of the first graph

| Piece | Area | x centroid | Moment |
| --- | --- | --- | --- |
| Triangle 0–2 | ½×2×.75=.75 | (0+2+2)/3=4/3 | 1 |
| Rectangle 2–3 | 1×.75=.75 | 2.5 | 1.875 |
| Triangle 3–4 | ½×1×.75=.375 | (3+3+4)/3=10/3 | 1.25 |



$$
z_{COA,1}=\frac{1+1.875+1.25}{.75+.75+.375}=\frac{4.125}{1.875}=2.2
$$



##### Step 3. MOM of the second graph



$$
z_{MOM,2}=(3+4)/2=3.5
$$



##### Step 4. Centroid of the second graph

| Piece | Area | x centroid | Moment |
| --- | --- | --- | --- |
| Triangle 0–3 | 1.5 | 2 | 3 |
| Rectangle 3–4 | 1 | 3.5 | 3.5 |
| Triangle 4–7 | 1.5 | 5 | 7.5 |



$$
z_{COA,2}=\frac{3+3.5+7.5}{1.5+1+1.5}=\frac{14}{4}=3.5
$$



The second graph is symmetric about 3.5, so its area centroid and plateau midpoint coincide. The first graph is asymmetric;2.2 and 2.5 need not agree.

**Source / assumption note.** The source draws the two functions on separate axes and does not state a combined aggregation. Answers above are per graph. If the intended task is the centroid of their max-union, explicitly aggregate first: upper envelope switches at x=2.25, total area=131/32, moment=1809/128, centroid=1809/524≈3.452290; maximum remains on[3,4], MOM=3.5.

**Final answer.** Graph 1:MOM 2.5, centroid 2.2. Graph 2:MOM 3.5, centroid 3.5.