Using centro id mean of maxima and center of sums controlled apply defuzzification.

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[1] Centro-id method.

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Computation of $x^x$

Area segment No	Area (A)	$ \bar x$	$A( \bar x )$
1	$\frac{1}{2} \times 0.3 \times 1 = 0.15$	$\frac{0 + 1 + 1}{3} = 0.67$	0.1005
2	$2.5 \times 3 = 0.78$	(1 + 3.6) 12 = 2.3	1.794
3	$0.4 \times 0.3 = 0.12$	(3.6 + 4) 12 = 3.8	0.456
4	$\frac{1}{2} \times 0.4 \times 0.2 = 0.04$	3.8667	0.1547
5	$1.5 \times 0.5 = 0.75$	4.75	3.56
6	$0.5 \times 0.5 = 0.25$	5.75	1.4375
7	$\frac{1}{2} \times 0.5 \times 0.5 = 0.125$	5.833	0.729
8	$1 \times 1 = 1$	6.5	6.5
9	$\frac{1}{2} \times 1 \times 1 = 0.5$	7.33	3.665

$x^* = \frac{\sum a ( \bar X)}{\sum A} = \frac{18.3967}{3.7} = 4.9$

[b] Weighted Average : (valid for symmetric mean)

$ x^* = \frac{2.5 * 0.3 + 5 \times 0.5 + 6.5 * 1}{0.3 + 0.5 + 1}$

= 5.41

[c] Centre of sums:

Area of $A_1$ = $\frac{1}{2} \times 0.3 \times 1 + 3 \times 0.3 + \frac{1}{2} \times 1 \times 0.3$

= 0.15 + 0.9 + 0.15

= 1.2

Area of $A_2$ = $\frac{1}{2} \times 1 \times 0.5 + 2 \times 0.5 + \frac{1}{2} \times 1 + 0.5$

= 0.25 + 1.0 + 0.25

= 1.5

Area of $A_3$ = $\frac{1}{2} \times 1 \times 1 + 1 \times 1 + \frac{1}{2} \times 1 \times 1$

$ = \frac{1}{2} + 1 + \frac{1}{2} = 2$

$x^x = \frac{1.2 * 2.5 + 1.5 * 5 + 2 * 6.5}{1.2 + 1.5 + 2}$

$= \frac{23.5}{4.7} = 5$

[d] First of maxima (last of maxima):

$A_3$ is having the largest area.

First of maxima = 6 Last of maxima 7

$x^* = \frac{6 + 7}{2} = 6.5$

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