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I propose to answer as much as I can piecemeal before I encounter problems. The questions require lots of calculations and, depending on how lengthy these are, I may have to limit or summarise results, but I hope you will be able to extrapolate the methods sufficiently to work out for yourself what I omit.
(1) The trapezoidal method of approximating areas under a curve (Normal Distribution) yields 0.92125 approx when n=16. I’ll try to upload tables for n=4, 8, 12, 16, which I’m currently preparing. Using Z scores 2 and 1.6, the approximations can be compared with actual values from Normal Distribution tables or graphs. For n=12, area=0.9203; n=8, area=0.9176; n=4, area=0.9032.
P(x) is the function and we are approximating the area beneath this curve (which is the integral) using trapezoids made up of a set of a right triangle atop a rectangle. We simply add together the areas of the triangles and the areas of the rectangles. The narrower the bases (usually called ∆x or h), that is, the higher n is, the better the approximation. The true value is about 0.9225.
n 
4 


h 
0.9 
Trapezoid 

a 
2 
0.05399 (=P(a)) 

x₁ 
1.1 
0.43570 (=2P(x₁)) 

x₂ 
0.2 
0.78209 (=2P(x₂)) 

x₃ 
0.7 
0.62451 (=2P(x₃)) 

b 
1.6 
0.11092 (=P(b)) 
Area 


2.00721 
0.90324 (=(h/2)*2.00721) 
This is the table for n=4. It’s the pattern for the other tables. The sum 2.00721 is the result of applying the Trapezoid Area formula before multiplying by h/2. The Area is the final approximation of the trapezoidal area.
<unable to upload tables>
Trapezoidal Area=(h/2)(P(a)+2∑P(xᵣ)+P(b)) where the limits are [a,b]=[2,1.6] and h=(ba)/n. The summation is for r ∈[1,n1]. (xᵣ=xᵣ₋₁+h. x₁=a+h.) This formula can be easily derived using basic geometry. The tables clarify how to apply the formula. I used Excel to do the many calculations and build the tables.
(2) Simpson’s ⅓ Rule.
n=4, 8, 12, 16
h=0.9, 0.45, 0.3, 0.225
h/3=0.3, 0.15, 0.1, 0.075
Applying Simpson’s ⅓ Rule:
A₄=0.3[P(2)+4P(1.1)+2P(0.2)+4P(0.7)+P(1.6],
A₈=0.15[P(2)+
4P(1.55)+2P(1.1)+4P(0.65)+2P(0.2)+
4P(0.25)+2P(0.7)+4P(1.15)+
P(1.6)],
A₁₂=0.1[P(2)+
4P(1.7)+2P(1.4)+4P(1.1)+2P(0.8)+
4P(0.5)+2P(0.2)+4P(0.1)+2P(0.4)+
4P(0.7)+2P(1)+4P(1.3)+
2P(1.6)],
A₁₆=0.075[P(2)+
4P(1.775)+2P(1.55)+4P(1.325)+2P(1.1)+
4P(0.875)+2P(0.65)+4P(425)+2P(0.2)+
4P(0.025)+2P(0.25)+4P(0.475)+2P(0.7)+
4P(0.925)+2P(1.15)+4P(1.375)+
P(1.6)].
<unable to upload tables>
Summarising the tables:
A₄=0.92023
A₈=0.92243
A₁₂=0.92245
A₁₆=0.92245
True value is 0.92245058 approx.
(3) Simpson’s ⅜ Rule when n=3r (r is a positive integer).
General formula:
A₃ᵣ=⅜h[P(a)+3(P(x₁)+P(x₂)+P(x₄)+P(x₅)+...)+2(P(x₃)+P(x₆)+P(x₉)+...)+P(b)],
where h=(ba)/3r, x₀=a and x₃ᵣ=b; xᵢ=xᵢ₋₁+h.
<unable to upload tables>
n=6, 9, 12, 15
h=0.6, 0.4, 0.3, 0.24
⅜h=0.225, 0.15, 0.1125, 0.09
a=x₀=2, b=x₃ᵣ=1.6 in all cases.
For r=2, h=0.6, x₁=1.4, x₂=0.8, x₃=0.2, x₄=0.4, x₅=1:
A₆=0.225[P(2)+3(P(1.4)+P(0.8)+P(0.4)+P(1))+2P(0.2)+P(1.6)]
For r=3, h=0.4, x₁=1.6, x₂=1.2, x₃=0.8, x₄=0.4, x₅=0, x₆=0.4, x₇=0.8, x₈=1.2:
A₉=0.15[P(2)+3(P(1.6)+P(1.2)+P(0.4)+P(0)+P(0.8)+P(1.2))+
2(P(0.8)+P(0.4))+P(1.6)]
For r=4, h=0.3, x₁=1.7, x₂=1.4, x₃=1.1, x₄=0.8, x₅=0.5, x₆=0.2, x₇=0.1, x₈=0.4, x₉=0.7, x₁₀=1, x₁₁=1.3:
A₁₂=0.1125[P(2)+
3(P(1.7)+P(1.4)+P(0.8)+P(0.5)+P(0.1)+P(0.4)+P(1)+P(1.3))+
2(P(1.1)+P(0.2)+P(0.7))+P(1.6)]
For r=5, h=0.24, x₁=1.76, x₂=1.52, x₃=1.28, x₄=1.04, x₅=0.8, x₆=0.56, x₇=0.32, x₈=0.08, x₉=0.16, x₁₀=0.4, x₁₁=0.64, x₁₂=0.88, x₁₃=1.12, x₁₄=1.36:
A₁₅=0.09[P(2)+
3(P(1.76)+P(1.52)+P(1.04)+P(0.8)+P(0.32)+P(0.08)+P(0.4)+P(0.64)+P(1.12)+P(1.36))+
2(P(1.28)+P(0.56)+P(0.16)+P(0.88))+P(1.6)]
A₆=0.9215948
A₉=0.9224139
A₁₂=0.9224433
A₁₅=0.9224483
More to follow...