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

Calculating Mean and Standard Deviation

The marks obtained in a statistics examination are normally distributed. If 15% of students scored ≥ 60 marks and 40% scored < 30 marks, find the mean (μ) and standard deviation (σ).

Given

• P(X ≥ 60) = 0.15
• P(X < 30) = 0.40

We need to find the values of μ and σ.

Step 1: Convert to Z-values

(i) P(X ≥ 60) = 0.15

Since P(X ≥ 60) = 0.15, then P(X < 60) = 0.85. From the standard normal table, P(Z < 1.04) = 0.85. Therefore:

(60 − μ) / σ = 1.04
μ = 60 − 1.04σ (Equation 1)

(ii) P(X < 30) = 0.40

From the standard normal table, P(Z < -0.25) = 0.40. Therefore:

(30 − μ) / σ = -0.25
μ = 30 + 0.25σ (Equation 2)

Step 3: Solve the Equations

Equating (1) and (2):

60 − 1.04σ = 30 + 0.25σ
30

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(

Problem 9: Z-Test for Two Means

A researcher wants to know the intelligence of students in a school. He selected two groups of students. In the first group, there are 150 students having a mean IQ of 75 with a S.D. of 15. In the second group, there are 250 students having a mean IQ of 70 with a S.D. of 20. Is there a significant difference between the means of the two groups?

Solution

Given

For Group I:

• Sample size, n₁ = 150
• Sample mean, x̄₁ = 75
• Standard deviation, s₁ = 15

For Group II:

• Sample size, n₂ = 250
• Sample mean, x̄₂ = 70
• Standard deviation, s₂ = 20

Level of significance: α = 0.05

Since both sample sizes are large (n₁, n₂ > 30), use the Z-test for the difference between two means.

Step 1: State the Hypotheses

• Null Hypothesis (H₀): μ₁ = μ₂

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:

Taylor Polynomials and Derivatives

Start by finding the derivatives of f(x) as needed. In the example question, it wants the degree 3, so we calculate up to the 3rd derivative, f'''(x), and then use the formula.

Solving Differential Equations

The equation ds/dt = sin(t) shows us that it is a pure time differential, meaning taking the integral of ds/dt will get us s(t). Since we know s(0) = 0, we use this to find the constant C. Once found, substitute it into s(3), and you have solved the problem.

Euler's Method for Numerical Solutions

To use Euler's Method, create your table with x moving at the rate given as Δ. Use the initial points and the formula provided. If given a function f(x), find the derivative. If already given a derivative or dy/dx,... Continue reading "Calculus Problem Solving Techniques and Formulas" »

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" »

Introduction to Statistical Learning

Key Notation

• Y: Response (dependent/output) variable - quantitative or qualitative
• X₁,...,Xₚ: Predictors (features/covariates/independent variables)
• p: Number of predictors
• n: Number of observations
• i: Subject index (i = 1,...,n)
• j: Variable index (j = 1,...,p)
• Data: (Yᵢ, Xᵢ), i = 1,...,n

Types of Learning

• Supervised learning: Have both Y and X (prediction or inference) - PRIMARY FOCUS
• Unsupervised learning: Have X but no Y

Model for Data

Yᵢ = f(Xᵢ) + εᵢ, i = 1,...,n
Assumptions: E(εᵢ) = 0, var(εᵢ) = σ², εᵢ are mutually independent
Properties: E(Yᵢ|Xᵢ) = f(Xᵢ), var(Yᵢ|Xᵢ) = σ²

Prediction vs Inference