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You have probably asked too much for answers to all parts to be accommodated within the constraints of this Q&A system, including space limits (8000 characters) and weaknesses in the system which make submitting and editing and uploading difficult or virtually impossible. 

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=(b-a)/n. The summation is for r ∈[1,n-1]. (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=(b-a)/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...

 

 

by Top Rated User (825k points)

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