Notes, summaries, assignments, exams, and problems for Mathematics

Understanding the distribution of sample means is crucial in statistics. Let's analyze two distinct scenarios.

MLB Player Salaries in 2012

In 2012, there were 855 major league baseball players. The mean salary was \(\mu = 3.44\) million dollars, with a standard deviation of \(\sigma = 4.70\) million dollars. We will examine random samples of size \(n = 50\) players to understand the distribution of their mean salaries.

Distribution of Sample Means

To describe the shape, center, and spread of the distribution of sample means, we apply the Central Limit Theorem (CLT). The CLT states that for sufficiently large sample sizes (typically \(n \ge 30\)), the sampling distribution of the sample mean will be approximately normal, irrespective of the population... Continue reading "MLB Player Salaries & Dark Chocolate's Vascular Health Impact" »

Matrices (Question 6a)

Verify that A · (\text{adj } A) = (\text{adj } A) · A = |A| · I_3 for
A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 0 & 1 \\ 2 & 1 & 5 \end{bmatrix}.

Tasks:

• Find the determinant |A|:
• Find the Adjoint (\text{adj } A): This involves finding the cofactor of each element and then transposing the resulting matrix.
• Cofactors: C₁₁ = -1, C₁₂ = -13, C₁₃ = 3, C₂₁ = -11, C₂₂ = 2, C₂₃ = 4, C₃₁ = 3, C₃₂ = 10, C₃₃ = -9
• Multiply A · (\text{adj } A)

4. Financial Arithmetic (Question 2g)

Find the compound interest on Rs. 8,000 for 1 1/2 years at 10% per annum, compounded annually.

• Amount for the first year:
• Interest for the next half year: Use simple interest on the new principal.
• Total Compound Interest:

Answers use standard... Continue reading "Matrix Determinant and Adjoint Verification with AP/GP and CI" »

Calculating the Future Value of an Annuity Due

Step 1: Determine the Variables

The problem provides the following details:

• Annual payment: Rs. 200. Therefore, the half-yearly payment (Pmt) is:
  
  Rs. 200 / 2 = Rs. 100
  Rs. 200 / 2 = Rs. 100
• Annual interest rate (r): 4% or 0.04. Since the interest is compounded half-yearly, the interest rate per period (i) is:
  
  0.04 / 2 = 0.02
  0.04 / 2 = 0.02
• Term: 20 years. Payments are made half-yearly, so the total number of periods (n) is:
  
  20 × 2 = 40
  20 × 2 = 40
• The annuity type is an annuity due, meaning payments are made at the beginning of each period.

Step 2: Apply the Future Value Formula

The formula for the Future Value (FV) of an annuity due is given by:

FV = Pmt × [((1 + i)^n - 1) / i] × (1 + i)

FV = Pmt × [((1

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*

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" »

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(

Combine Independent Unbiased Estimators

Let d₁ and d₂ be independent unbiased estimators of θ with variances σ₁² and σ₂², respectively:

• E[d_i] = θ for i = 1,2.
• Var(d_i) = σ_i².

Any estimator of the form d = λ d₁ + (1 - λ) d₂ is also unbiased for any constant λ.

The variance (mean square error for an unbiased estimator) is
Var(d) = λ²σ₁² + (1 - λ)²σ₂².

To minimize Var(d) with respect to λ, differentiate and set to zero:

d/dλ Var(d) = 2λσ₁² - 2(1 - λ)σ₂² = 0.

Solving gives the optimal weight

λ_* = σ₂² / (σ₁² + σ₂²).

Question 1: Posterior PMF for a Third Dice Roll

Assume there are five dice with sides {4, 6, 8, 12, 20}. One of these five dice is selected uniformly at random (probability 1/5) and rolled twice. The two observed results are... Continue reading "Optimal Estimators, Dice Posterior & Statistical Problems" »