Customer Value: (Quality, Time, Flexibility, Customer Experience, Innovation) / Price

Sustainability Paradigms: Economic (Viable, Equitable), Environment (Viable, Bearable), Social (Bearable, Equitable)

- Efficiency (Optimization), Differentiator (Innovation), Driver (Motivation) | - Audits (Assess Sustainability Performance)

Table 7.3: Factors for Calculating Three-Sigma Limits for the X (Bar) Chart and R-Chart

*A sample is out of control if its value falls below the LCL or above the UCL*

Corporate Distribution Layers

Distributions are applied in the following order: Accumulated E&P + Current E&P, then basis, then the remainder is treated as capital gain.

• No Dividend: If there is no E&P, reduce current E&P first. Negative AEP still distributes E&P.
• Offset: Half of current E&P is lost to offset AEP before distribution.

Property Distributions

Recognize gains, but not losses.

• Dividend: FMV - Liabilities
• Basis: FMV
• Gains: FMV - Tax Basis

Distribution Table

The amount from CEP is calculated as: (Period distribution / Total distribution) * CEP. AEP is applied to the remainder, then to stock basis.

Stock Distributions and Redemptions

• Stock Distribution: Proportionate distributions are nontaxable and do not reduce E&

lim x->0 sinx/x = 1 | H.A.: compare degrees | V.A.: denom = 0 | Continuous if: f(a), lim, equal

DERIVATIVES: (x^n)'=nx^(n-1), (e^x)'=e^x, (a^x)'=a^x ln a, (lnx)'=1/x | (uv)'=u'v+uv', (u/v)'=(u'v-uv')/v^2 |chain: (f(g(x)))'=f'(g(x))g'(x)

TRIG DERIVATIVES: (sin)'=cos, (cos)'=-sin, (tan)'=sec^2 | (sec)'=sec·tan, (csc)'=-csc·cot, (cot)'=-csc^2

CRITICAL POINTS:f'=0 or DNE ⇒ crit pt | f'>0 inc | f'<0 dec | f''>0 conc up | f''<0 conc down | inflec = f'' signchange

INTEGRATION: ∫x^n dx = x^(n+1)/(n+1)+C | ∫e^x dx = e^x+C | ∫a^x dx = a^x/ln a+C | ∫1/x dx = ln|x|+C | ∫sin x dx

= -cos x+C | ∫cos x dx = sin x+C

FTC: Part 1: d/dx ∫_a^x f(t) dt = f(x) | Part 2: ∫_a^b f(x) dx = F(b)-F(a)

AREA & VOLUME: A = ∫_a^b (top - bot)... Continue reading "5 hr 20 min 20 sec corresponds to a longitude difference of" »

Gradient: \nabla f = \langle f_x,\ f_y,\ f_z\rangle Directional derivative (unit vector u): D_uf = \nabla f\cdot u Unit direction from A→B: u = \frac{B-A}{\|B-A\|} Tangent plane to z=f(x,y): z - z_0 = f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) Critical points: Solve f_x=0, f_y=0. D = f_{xx}f_{yy} - (f_{xy})^2 Min if D>0,f_{xx}>0. Max if D>0,f_{xx}<0. Saddle if D<0.

General double integral: Type I (y inner): \int_a^b \int_{g_1(x)}^{g_2(x)} f\, dy\, dx

Type II (x inner):

\int_c^d \int_{h_1(y)}^{h_2(y)} f\, dx\, dy

Polar:

x = r\cos\theta,\ y=r\sin\theta,\ dA=r\,dr\,d\theta

Circle: 0\le r\le R.

Line y=mx → \theta=\arctan m.

Region must be continuous in θ.

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3. TRIPLE INTEGRALS

Cylindrical:

x=r\cos\theta,\ y=r\sin\theta,\ dV=r\,dz\,dr\,... Continue reading "Multivariable Calculus and Vector Analysis Formulas" »

Section A: R Basics and Data Types (Weeks 1-4)

Model Questions and Answers

Q: Create a vector v (3, NA, Inf, -Inf). Explain adding 5 to v.

A: The operation is element-wise. Missing values (NA) propagate, resulting in NA. Infinite values (Inf, -Inf) remain infinite.

v <- c(3, NA, Inf, -Inf)
print(v + 5)
# Output: [1]  8 NA Inf -Inf

Q: Given vector a, write code to count and replace NAs.

A: Assuming a <- c(10, 15, NA, 20).

• Count NAs: sum(is.na(a)) → 1.
• Replace NAs (e.g., with 0): a[is.na(a)] <- 0.

Q: Explain the difference between a[a > 12] and a[which(a > 12)].

A: Both select elements greater than 12 (15, 20). However:

• a[a > 12] → [1] 15 NA (Uses logical indexing; preserves the position of NA in the original vector as NA).
• a[which(

Quick revision table for Mathematics and Statistics exams.

Unit I: Integration and Data Presentation

Unit II: Central Tendency and

Two-State Actuarial Model

The Two-State Model (also known as the Dead-Alive or Binary Model) is a fundamental actuarial framework used to represent processes that exist in one of two possible states, such as Alive/Dead or Working/Retired. It is widely utilized in life insurance and pension modeling to estimate the probability of transition between states. The model assumes that at any given time, an individual occupies only one state, allowing actuaries to calculate premiums, reserves, and expected present values by simplifying complex uncertainties into binary outcomes.

Core Assumptions

• Binary States: The system exists in only one of two states at any time.
• Markov Property: Transitions depend solely on the current state.
• Constant Probabilities: