https://www.perplexity.ai/search/prob-and-stata-questions-analy-zC0lDjvpT4CZQZwvtZk2pA

Perplexity

Q1. (Interview Room Assignments & Derangements)

• Fundamental Probability Concepts

  • Sample Space: All possible ways to assign 5 candidates to 5 rooms; total = 5! arrangements.
  • Event Probability: $$P(E) = \frac{\text{favorable outcomes}}{\text{total outcomes}}$$.

• Combinatorial Analysis

  • Permutations: Counting number of ways candidates can be assigned.
  • Derangements: Number of permutations where no item appears in its original position; required for "none in correct room".
    • Formula: $$ !n = n! \sum_{k=0}^n \frac{(-1)^k}{k!} $$
    • Application: Use for (b) and conditional derangement in (c).
  • Conditional Probability: Used in (c), considering one assignment is fixed.

Q2. (Balls in a Box with Replacement)

• Discrete Probability Distributions

  • Experiment & Sample Space: Drawing balls with replacement, two draws.
  • Random Variables: X = red, Y = black, Z = white in two draws.

• Probability Calculations

  • Probability Mass Function (PMF): For (X, Y), find probability for each pair using multinomial formula, considering probabilities of each color:
    • Red: 1/6, Black: 2/6, White: 3/6.
    • PMF uses binomial/multinomial counting for two draws.

• Conditional Probability

  • $$P\{X=1|Z=0\}$$
    • Compute probability of 1 red in 2 draws given 0 white was drawn.

Q3. (Joint Continuous PDF)

• Continuous Random Variables

  • Joint Probability Density Function (PDF): Function $$f(x, y)$$ provided.
  • Normalization: Find C such that $$\iint_{-\infty}^{\infty} f(x, y) dx,dy = 1$$.
    • Use Gaussian integrals; linear algebra for multivariate cases.

• Independence

  • Test if $$f(x, y) = f_X(x)f_Y(y)$$.
  • Marginals: Find $$f_X(x), f_Y(y)$$ by integrating out other variable.

• Conditional Density

  • $$f_{Y|X}(y|x)=\frac{f(x,y)}{f_X(x)}\$\$,pluginfunctionalforms.$$

• Statistical Moments

  • Variance: $$Var(aX + bY) = a^2Var(X) + b^2Var(Y) + 2abCov(X,Y)$$.
  • Covariance: $$Cov(X_i, Y)$$, with $$Y = \frac{1}{n}\sum X_i$$.

• IID Random Variables

  • Properties: Variance and covariance calculations for identical, independent variables.

Q5. (Estimation: Exponential Distribution)

• Parameter Estimation

  • Method of Moments: Match sample moments to theoretical moments.
  • Maximum Likelihood Estimation (MLE): Find parameter maximizing likelihood function.

• Estimator Properties

  • Bias: Estimator is unbiased if $$E[\hat{\theta}] = \theta$$.
  • Justification: Use expectation and variance results for exponential.

Q6. (Hypothesis Testing: Service Time)

• Statistical Inference

  • Hypothesis Formulation: $$H_0: \mu \leq 200$$, $$H_1: \mu > 200$$.
  • Significance Level: $$\alpha = 0.05$$.
  • Rejection Region: Critical value for normal distribution.
  • Type II Error ($$\beta$$): Probability test fails to reject $$H_0$$ when alternative is true.

• Normal Distribution Theory

  • Standardization: Use known $$\sigma$$, sample mean for Z-test.
  • p-value: Probability of observing a more extreme sample mean under $$H_0$$.
  • Cumulative Distribution: Involves $$\Phi()$$, the CDF of standard normal.

Q7. (Confidence Intervals: Normally Distributed Measurements)

• Estimation Theory

  • Point Estimate: Sample mean of measurements.
  • Confidence Interval: $$ \bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}} $$, using known standard deviation.
  • Lower/Upper Confidence Bound: One-sided confidence intervals.

• Normal Distribution Properties

  • Critical Value: Use standard normal (z) values.
  • Interpretation: Probability true mean lies within interval, under repeated sampling.

Mindmap Structure (Outline)

Q1. Interview Assignments (Permutations & Probability)
  ├─ Counting Principles (Permutations, Sample Space)
  ├─ Derangements (No Fixed Points)
  ├─ Conditional Probability (with constraints)

Q2. Balls in a Box (Discrete RV, Conditional Probability)
  ├─ Discrete Probability Distributions (PMF)
  ├─ Multinomial/Combinatorial Calculations
  ├─ Conditional Probability

Q3. Joint Continuous PDF (Continuous RV, Independence, Conditioning)
  ├─ Joint PDF Normalization (Integration, Gaussian Integral)
  ├─ Marginal Distributions (Integration)
  ├─ Independence (Product of Marginals)
  ├─ Conditional PDF

Q4. Variance & Covariance for IID (Statistical Moments)
  ├─ Variance of Sums/Linear Combinations
  ├─ Covariance Calculation
  ├─ IID Properties

Q5. Parameter Estimation (Exponential MME/MLE, Bias)
  ├─ Method of Moments
  ├─ Maximum Likelihood Estimation
  ├─ Properties of Estimators (Bias, Unbiasedness)

Q6. Hypothesis Testing (Normal, One-Sample)
  ├─ Hypothesis Formulation
  ├─ Z-Test (Known Standard Deviation)
  ├─ Rejection Region/Critical Value
  ├─ Type II Error (\(\beta\))
  ├─ P-value

Q7. Confidence Intervals (Normal Population, Known SD)
  ├─ Point Estimate (Sample Mean)
  ├─ Confidence Interval Equation
  ├─ One-sided Intervals (Lower/Upper)
  ├─ Critical Values (z)

Each node branches into:

• The definition and purpose of the concept.
• The relevant formula or method.
• The step or reasoning required for the specific application in the question.

This structure explicitly links each question to its required concepts, progressing from fundamentals through to formulas and practical computation techniques needed for solution.

Gemini

Of course. Here is a detailed conceptual analysis of the provided exam questions, structured as a multi-level mindmap in markdown format.

Conceptual Analysis of Exam Questions

This analysis breaks down each question to its foundational concepts and the specific techniques required for a solution, following the hierarchical structure you requested.

Q1: Job Interviews

• Core Topic: Combinatorics and Classical Probability.

• Layer 1: Foundational Concepts

• Sample Space & Events: The set of all possible outcomes and subsets of interest.

• Theory: In this problem, the sample space S is the set of all possible assignments of 5 candidates to 5 rooms. An event is a specific type of assignment (e.g., all correct).

• Classical Probability: The probability of an event A is the ratio of the number of favorable outcomes to the total number of outcomes in the sample space.

• Formula: P(A) = |A| / |S|.

• Steps: 1. Define the sample space. 2. Count the total outcomes |S|. 3. Define the event of interest A. 4. Count the favorable outcomes |A|. 5. Calculate the ratio.

• Layer 2: Specific Techniques & Applications

• (a) All Correct

• Concept: Permutations. An arrangement of distinct objects in a specific order.

• Reasoning: The total number of ways to assign 5 candidates to 5 rooms is the number of permutations of 5 items. The "all correct" case is just one specific permutation.

• Formula: n! (n-factorial) for the total number of permutations of n items.

• Steps: Calculate |S| = 5!. Recognize |A| = 1. Compute P(A) = 1 / 5!.

• (b) None Correct

• Concept: Derangements. A permutation of elements where no element appears in its original position.

• Reasoning: This is the direct application of a derangement, counting the number of ways to assign candidates so that none are in their pre-assigned room.

• Formula: The number of derangements of n items, D_n = n! * Σ (from k=0 to n) [(-1)^k / k!].

• Steps: Calculate D_5 using the formula. The probability is D_5 / 5!.

• (c) Conditional Probability with Derangements

• Concept 1: Conditional Probability. The probability of an event occurring given that another event has already occurred.

• Formula: P(A|B) = P(A ∩ B) / P(B).

• Concept 2: Reduced Sample Space. A simpler way to handle conditional probability where the condition fixes part of the outcome.

• Reasoning: The condition "E ends up in room 5" fixes one candidate's position. The problem is now reduced to arranging the remaining 4 candidates (A,B,C,D) in the remaining 4 rooms (1,2,3,4) such that none are in their correct room.

• Steps: 1. Acknowledge the sample space is now the 4! ways to arrange the 4 remaining candidates. 2. The favorable outcomes are the number of derangements of these 4 candidates, D_4. 3. The probability is D_4 / 4!.

Q2: Balls in a Box

• Core Topic: Discrete Random Variables, Joint and Conditional Probability Distributions.

• Layer 1: Foundational Concepts

• Discrete Random Variables: Variables (X, Y, Z) that take on a countable number of values, representing the counts of different colored balls.

• Probability Mass Function (PMF): A function that gives the probability that a discrete random variable is equal to some value.

• Experiment with Replacement: Each trial is independent, and the probabilities of drawing each color remain constant.

• Probabilities: P(Red) = 1/6, P(Black) = 2/6, P(White) = 3/6.

• Layer 2: Specific Techniques & Applications

• (a) Conditional Probability P(X=1|Z=0)

• Concept: Conditional Probability for Random Variables.

• Formula: P(A|B) = P(A ∩ B) / P(B). Here, A is the event {X=1} and B is the event {Z=0}.

• Reasoning & Steps:

1. Calculate P(Z=0): This means in two draws, no white balls were selected. The probability of not drawing white in one trial is 1 - P(W) = 1 - 3/6 = 3/6. Since trials are independent, P(Z=0) = (3/6) * (3/6) = 1/4.
2. Calculate P(X=1 ∩ Z=0): This means exactly one red ball and no white balls were drawn in two trials. This implies the other ball must be black. The outcomes are (Red, Black) or (Black, Red).

• P(R,B) = P(R) * P(B) = (1/6) * (2/6) = 2/36.
• P(B,R) = P(B) * P(R) = (2/6) * (1/6) = 2/36.
• P(X=1, Z=0) = 2/36 + 2/36 = 4/36 = 1/9.

3. Compute the Ratio: P(X=1|Z=0) = (1/9) / (1/4) = 4/9.

• (b) Joint PMF of (X, Y)
• Concept: Multinomial Distribution. A generalization of the binomial distribution for k > 2 outcomes. The joint PMF of (X, Y, Z) follows a multinomial distribution.
• Formula: P(X=x, Y=y, Z=z) = (n! / (x!y!z!)) * p_R^x * p_B^y * p_W^z, where n=2 and x+y+z=2.
• Reasoning: Since Z = 2 - X - Y, the joint PMF of (X,Y) completely defines the system. We list all possible pairs (x, y) such that x+y ≤ 2 and calculate their probabilities.
• Steps: Systematically list possible (x,y) pairs and calculate P(X=x, Y=y):
• (2,0): (R,R) -> (1/6)^2
• (0,2): (B,B) -> (2/6)^2
• (1,1): (R,B) or (B,R) -> 2 * (1/6) * (2/6)
• (1,0): (R,W) or (W,R) -> 2 * (1/6) * (3/6)
• (0,1): (B,W) or (W,B) -> 2 * (2/6) * (3/6)
• (0,0): (W,W) -> (3/6)^2

Q3: Bivariate Continuous PDF

• Core Topic: Bivariate Continuous Random Variables, specifically Bivariate Normal Distribution.

• Layer 1: Foundational Concepts

• Joint Probability Density Function (PDF): A function f(x,y) for continuous variables where the volume under the surface over a region gives the probability for that region.

• Property of PDF: The integral of the PDF over its entire domain must equal 1. ∫∫ f(x,y) dx dy = 1.

• Layer 2: Specific Techniques & Applications

• (a) Find the constant C

• Concept: Normalization of a PDF.

• Technique 1: Completing the Square. The exponent -2x² + 2xy - y² is a quadratic form. This is the key step to simplify the integral.

• Steps: Rewrite the exponent: -2x² + 2xy - y² = -(y² - 2xy + 2x²) = -( (y-x)² - x² + 2x² ) = -(y-x)² - x².

• Technique 2: Gaussian Integral.

• Formula: ∫ e^(-au²) du = √(π/a).

• Steps:

1. Set up the normalization integral: ∫[-∞,∞] ∫[-∞,∞] C * e^(-(y-x)² - x²) dy dx = 1.
2. Integrate with respect to y first: ∫ C * e^(-x²) * [∫ e^(-(y-x)²) dy] dx. The inner integral is a Gaussian integral with a=1 and evaluates to √π.
3. The integral becomes ∫ C * e^(-x²) * √π dx = C√π ∫ e^(-x²) dx.
4. The remaining integral is another Gaussian with a=1, evaluating to √π.
5. The equation is C√π * √π = Cπ = 1. So, C = 1/π.

• (b) Independence of X and Y
• Concept: Independence of Continuous Random Variables.
• Test: X and Y are independent if and only if their joint PDF f(x,y) can be factored into the product of their marginal PDFs, f(x,y) = g(x)h(y).
• Reasoning: The presence of the cross-term 2xy in the exponent e^(-2x²+2xy-y²) prevents the function from being separated into a product of a function of x only and a function of y only.
• Conclusion: X and Y are not independent.
• (c) Conditional PDF f_{Y|X}(y|x)
• Concept: Conditional Density Function.
• Formula: f_{Y|X}(y|x) = f(x,y) / f_X(x).
• Steps:

1. Find the Marginal PDF f_X(x): Integrate the joint PDF with respect to y.

• f_X(x) = ∫ f(x,y) dy = ∫ (1/π) * e^(-(y-x)² - x²) dy.
• f_X(x) = (1/π) * e^(-x²) ∫ e^(-(y-x)²) dy = (1/π) * e^(-x²) * √π = (1/√π) * e^(-x²). (This is a Normal distribution).

2. Apply the formula:

• f_{Y|X}(y|x) = [ (1/π) * e^(-(y-x)² - x²) ] / [ (1/√π) * e^(-x²) ].
• f_{Y|X}(y|x) = (1/√π) * e^(-(y-x)²).
• Interpretation: Given X=x, Y is a Normal random variable with mean x and variance 1/2.

Q4: Variance and Covariance of Sample Means

• Core Topic: Properties of Expectation, Variance, and Covariance for linear combinations of random variables.

• Layer 1: Foundational Concepts

• IID Random Variables: X_i are Independent and Identically Distributed.

• Implication: E[X_i] is the same for all i. Var(X_i) = σ² is the same for all i. Cov(X_i, X_j) = 0 for i ≠ j.

• Variance & Covariance Properties:

• Var(aX) = a²Var(X).

• Cov(X, Y) = E[XY] - E[X]E[Y].

• Cov(X, X) = Var(X).

• Cov(aX, bY) = ab*Cov(X, Y).

• Bilnearity of Covariance: Cov(Σa_i X_i, Σb_j Y_j) = ΣΣ a_i b_j Cov(X_i, Y_j).

• Variance of a Sum: Var(A+B) = Var(A) + Var(B) + 2*Cov(A,B).

• Layer 2: Specific Techniques & Applications

• (b) Find Cov(X_1, Y) (It's easier to find this first)

• Steps:

1. Substitute Y: Cov(X_1, Y) = Cov(X_1, (1/n)ΣX_i).
2. Use linearity of covariance: (1/n) * Cov(X_1, ΣX_i) = (1/n) * Cov(X_1, X_1 + X_2 + ... + X_n).
3. Expand the covariance: (1/n) * [ Cov(X_1,X_1) + Cov(X_1,X_2) + ... + Cov(X_1,X_n) ].
4. Apply independence: Cov(X_1, X_j) = 0 for j > 1.
5. Simplify: The only non-zero term is Cov(X_1, X_1) = Var(X_1) = σ².
6. Result: Cov(X_1, Y) = (1/n) * σ².

• (a) Find Var(X_1 + Y)
• Steps:

1. Use the variance of a sum formula: Var(X_1 + Y) = Var(X_1) + Var(Y) + 2*Cov(X_1, Y).
2. Calculate each term:

• Var(X_1) = σ² (given).
• Var(Y) = Var((1/n)ΣX_i) = (1/n²) * Var(ΣX_i) = (1/n²) * (nσ²) = σ²/n (since X_i are independent).
• Cov(X_1, Y) = σ²/n (from part b).

3. Substitute and simplify: Var(X_1 + Y) = σ² + σ²/n + 2*(σ²/n) = σ² + 3σ²/n = σ²(1 + 3/n).

Q5: Parameter Estimation

• Core Topic: Statistical Inference - Point Estimation.

• Layer 1: Foundational Concepts

• Parameter Estimation: Using sample data to estimate an unknown parameter (λ) of a population distribution.

• Exponential Distribution: A continuous distribution modeling time until an event.

• PDF: f(x;λ) = λe^(-λx).

• Expectation: E[X] = 1/λ.

• Layer 2: Specific Estimation Methods

• (a) Find MME and MLE

• Method of Moments Estimation (MME):

• Principle: Equate the first population moment to the first sample moment and solve for the parameter.

• Steps:

1. Population Moment: E[X] = 1/λ.
2. Sample Moment: m_1 = (1/n)ΣX_i = X̄ (the sample mean).
3. Equate: 1/λ = X̄.
4. Solve for λ: λ_MME = 1/X̄.

• Maximum Likelihood Estimation (MLE):
• Principle: Find the parameter value that maximizes the likelihood (joint probability) of observing the given data sample.
• Steps:

1. Likelihood Function: L(λ) = Π f(x_i; λ) = Π (λe^(-λx_i)) = λ^n * e^(-λΣx_i).
2. Log-Likelihood: ln(L(λ)) = n*ln(λ) - λΣx_i. (Easier to differentiate).
3. Differentiate and set to 0: d/dλ [ln(L(λ))] = n/λ - Σx_i = 0.
4. Solve for λ: n/λ = Σx_i => λ_MLE = n / Σx_i = 1/X̄.

• (b) Unbiasedness of Estimates
• Concept: Unbiased Estimator. An estimator θ̂ is unbiased for a parameter θ if its expected value is equal to the true parameter value, i.e., E[θ̂] = θ.
• Reasoning:

1. The estimator is λ̂ = 1/X̄. We need to check if E[1/X̄] = λ.
2. We know E[X̄] = E[X] = 1/λ.
3. However, for a non-linear function g, E[g(X)] ≠ g(E[X]) in general. This is known as Jensen's Inequality.
4. Since g(u) = 1/u is a convex function, E[1/X̄] > 1/E[X̄] = 1/(1/λ) = λ.
5. Conclusion: Because E[λ̂] > λ, the estimator is biased.

Q6: Hypothesis Testing

• Core Topic: Statistical Inference - Hypothesis Testing for a single mean with known variance.

• Layer 1: Foundational Concepts

• Hypothesis Testing: A formal procedure for deciding between two competing statements (hypotheses) about a population parameter.

• Z-test for a Mean (σ known): The appropriate test when the population standard deviation is known and the sample is from a normal population or the sample size is large (Central Limit Theorem).

• Key Components:

• Null Hypothesis (H0): The statement of no effect or default state (e.g., μ = 200).

• Alternative Hypothesis (H1): The statement we are trying to find evidence for (e.g., μ > 200). This defines the test as one-tailed (right-tailed).

• Test Statistic: Z = (X̄ - μ_0) / (σ/√n).

• Significance Level (α): The probability of making a Type I error (rejecting a true H0).

• p-value: The probability of observing a test statistic as extreme or more extreme than the one calculated, assuming H0 is true.

• Type II Error (β): The probability of failing to reject a false H0.

• Layer 2: Specific Techniques & Applications

• (a) Rejection Region Method

• Steps:

1. State Hypotheses: H0: μ = 200, H1: μ > 200.
2. Define Rejection Region: For a right-tailed test with α=0.05, the critical value is z_α = z_0.05 = 1.64. We reject H0 if the calculated Z_calc > 1.64.
3. Calculate Test Statistic: Z_calc = (250 - 200) / (150/√9) = 50 / 50 = 1.
4. Conclusion: Since 1 < 1.64, the test statistic does not fall in the rejection region. We fail to reject H0. There is not compelling evidence.

• (b) State the p-value
• Formula: For a right-tailed Z-test, p-value = P(Z > Z_calc).
• Steps: Using Z_calc = 1, the p-value = P(Z > 1) = 1 - P(Z ≤ 1) = 1 - Φ(1).
• (c) Type II Error Probability (β)
• Concept: β = P(Fail to reject H0 | H1 is true).
• Steps:

1. Find the "Fail to Reject" Region in terms of X̄: We fail to reject H0 when Z_calc ≤ 1.64.

• (X̄ - 200) / (150/√9) ≤ 1.64 => X̄ ≤ 200 + 1.64 * 50 = 282.

2. Calculate the probability of this event given the true mean is μ_true = 320:

• β(320) = P(X̄ ≤ 282 | μ = 320).

3. Standardize using the true mean:

• P( (X̄ - 320)/(150/√9) ≤ (282 - 320)/(150/√9) ).
• P( Z ≤ -38 / 50 ) = P(Z ≤ -0.76) = Φ(-0.76).

Q7: Confidence Intervals

• Core Topic: Statistical Inference - Interval Estimation for a single mean with known variance.

• Layer 1: Foundational Concepts

• Confidence Interval (CI): A range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter.

• Confidence Level: The probability (e.g., 95%) that a CI produced by the method will contain the true parameter value.

• Z-Interval for a Mean (σ known): The appropriate interval when the population standard deviation is known.

• Layer 2: Specific Techniques & Applications

• Preliminary Step: Calculate the sample mean X̄ from the data:

• X̄ = (11.2 + 12.4 + 10.8 + 11.6 + 12.5 + 10.1 + 11.0 + 12.2) / 8 = 91.8 / 8 = 11.475.

• (a) 95% (Two-Sided) Confidence Interval

• Formula: X̄ ± z_(α/2) * (σ/√n).

• Reasoning: The interval is centered around the sample mean. z_(α/2) defines the width based on the desired confidence level.

• Steps:

1. α = 1 - 0.95 = 0.05. α/2 = 0.025.
2. Critical value: z_0.025 = 1.96.
3. Calculate: 11.475 ± 1.96 * (0.08 / √8).

• (b) 95% Lower Confidence Interval (Lower Bound)
• Formula: [ X̄ - z_α * (σ/√n), ∞ ).
• Reasoning: We want to be 95% confident that the true mean is above a certain value. All the uncertainty (α) is placed in the left tail.
• Steps:

1. α = 1 - 0.95 = 0.05.
2. Critical value: z_α = z_0.05 = 1.64.
3. Calculate the lower bound: 11.475 - 1.64 * (0.08 / √8).

• (c) 95% Upper Confidence Interval (Upper Bound)
• Formula: ( -∞, X̄ + z_α * (σ/√n) ].
• Reasoning: We want to be 95% confident that the true mean is below a certain value. All the uncertainty (α) is placed in the right tail.
• Steps:

1. α = 1 - 0.95 = 0.05.
2. Critical value: z_α = z_0.05 = 1.64.
3. Calculate the upper bound: 11.475 + 1.64 * (0.08 / √8).