3  Chapter 2: Continuous Probability Distributions

3.1 Module 2: Continuous Probability Distributions

3.1.1 Topic Summary

This module explores continuous random variables through Probability Density Functions (PDFs), where the integral over the range equals 1, and cumulative probabilities are found via integration. Uniform distributions have constant density over an interval, with mean \((a+b)/2\) and variance \((b-a)^2/12\), used for equal-probability scenarios. Exponential distributions model waiting times with PDF \(\lambda e^{-\lambda x}\), mean \(1/\lambda\), and variance \(1/\lambda^2\), applied to bus waits or repair times. Normal distributions follow the bell curve, with probabilities calculated using Z-scores and standard tables for weights or other measurements. The Central Limit Theorem states that sums of independent variables approximate normality, enabling probability calculations for large samples or sums exceeding thresholds (Bertsekas and Tsitsiklis 2008).

Tips for Beginners

  • PDF vs PMF: PDFs can be >1, but integral =1. Probabilities are areas under the curve, so use integration (or tables for standard distributions).

  • Uniform: Easy to remember: mean midpoint \(\frac{a+b}{2}\), variance \(\frac{(b-a)^2}{12}\). For probs, just proportion of interval.

  • Exponential: “Memoryless” – past doesn’t affect future. Mean = \(\frac{1}{\lambda}\), variance = \(\frac{1}{\lambda^2}\). For P(X > t), use \(e^{-\lambda t}\).

  • Normal: Use \(Z = \frac{x - \mu}{\sigma}\). Look up in table for \(P(Z < z)\). CLT applies when \(n\geq 30\) or known normal.

  • Common Mistakes: Don’t integrate PDFs directly for probs – use definite integrals. For CLT, ensure independence and identical distributions.

  • Practice: Sketch PDFs to visualize. Use online calculators for integrals initially.

3.1.2 Solved Problems

  1. Continuous Constants: A continuous random variable \(X\) has PDF \(f(x) = kx\) for \(0 \le x \le 4\). Find the value of \(k\).

    • Solution: Apply the normalization property \(\int_0^4 kx \,dx = 1\). Integrating gives \([kx^2/2]_0^4 = 1 \implies 8k = 1\), so \(k = 1/8\).

  2. Uniform Distribution Limits: If \(X\) is uniformly distributed in the interval \([a, b]\) with mean 3 and variance 3, find \(a\) and \(b\).

    • Solution: The mean formula gives \((a+b)/2 = 3 \implies a+b = 6\). The variance formula gives \((b-a)^2/12 = 3 \implies b-a = 6\). Solving these simultaneous equations yields \(a=0\) and \(b=6\).

  3. Exponential Wait Times: The waiting time for a bus is exponentially distributed with a mean of 10 minutes. Find the probability that a person waits for more than 15 minutes.

    • Solution: Since the mean is \(1/\lambda = 10\), we have \(\lambda = 0.1\). \(P(X > 15) = \int_{15}^\infty 0.1e^{-0.1x} dx = e^{-1.5} \approx 0.2231\).

  4. Normal Distribution Curve: The weights of students are normally distributed with mean 65kg and standard deviation 5kg. Find the probability that a student weighs more than 70kg.

    • Solution: First, find the Z-score: \(Z = \frac{70 - 65}{5} = 1\). Using the standard normal table, \(P(X > 70) = P(Z > 1) = 0.5 - 0.3413 = 0.1587\).

  5. Central Limit Theorem: Let \(X_1, X_2, \dots, X_{100}\) be independent exponential random variables with a mean of 50. Use the Central Limit Theorem to find the probability that their sum exceeds 6000.

    • Solution: For an exponential distribution, \(\mu = 50\) and \(\sigma = 50\). For the sum of 100 variables, the approximated normal mean is \(100 \times 50 = 5000\) and standard deviation is \(\sqrt{100} \times 50 = 500\). The Z-score for 6000 is \(Z = \frac{6000 - 5000}{500} = 2\). From Normal tables, \(P(Z > 2) = 0.0228\).

  6. PDF Constant: If \(X\) is a continuous random variable with PDF \(f(x) = kx^2\) for \(0 < x < 2\) and \(0\) otherwise, find the value of \(k\).

    • Solution: The integral must equal 1: \(\int_0^2 kx^2 dx = 1 \implies k[x^3/3]_0^2 = 1 \implies k(8/3) = 1\), so \(k = 3/8\).

  7. Uniform Probabilities: If \(X\) is uniformly distributed over \((0, 10)\), find \(P(X < 3)\) and \(P(3 < X < 8)\).

    • Solution: PDF is \(f(x) = 1/10\). \(P(X < 3) = \int_0^3 1/10 dx = 0.3\). \(P(3 < X < 8) = \int_3^8 1/10 dx = 0.5\).

  8. Exponential Probability: For an exponential distribution with mean 5, calculate \(P(X > 10)\).

    • Solution: Mean is \(5 \implies \lambda = 1/5 = 0.2\). \(P(X > 10) = e^{-2} \approx 0.1353\).

  9. Exponential Repair Time: The time taken to repair a machine follows an exponential distribution with mean 20 minutes. Find the probability that the repair takes more than 30 minutes.

    • Solution: \(P(X > 30) = \int_{30}^\infty \frac{1}{20} e^{-x/20} dx = e^{-1.5} \approx 0.2231\).

3.1.3 Practice Problems

  • In a normal distribution, 31% of items are under 45 and 8% are over 64. Find the mean and standard deviation.

  • The joint PDF of X and Y is \(f(x, y) = e^{-(x+y)}\) for \(x > 0, y > 0\). Find \(P(X+Y \le 1)\).

  • A continuous random variable X is uniformly distributed with mean 1 and variance 4/3. Find P(X < 0).

  • A sample of 100 items is taken from a population with mean 50 and variance 25. Use Central Limit Theorem to find P(49 < X < 51).

Exam Preparation Tips

  • Memorize PDFs: Uniform \(\frac{1}{b-a}\), Exponential \(\lambda e^{-\lambda x}\), Normal \(\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\).

  • For probabilities, use integration for continuous variables, or Z-scores for Normal.

  • CLT: Sample mean ~ Normal(μ, σ/√n) for large n.

  • Practice: Sketch PDFs to visualize areas. Use online tools for integration if needed.

  • For integration, know basic antiderivatives: \(\int kx \,dx = \frac{kx^2}{2}\), \(\int e^{-ax} \,dx = -\frac{1}{a} e^{-ax}\).

  • CLT: Sample mean ~ Normal(μ, σ/√n) for large n.

  • Use symmetry in Normal: \(P(Z > z) = 1 - P(Z < z)\).

  • Practice: Calculate Z-scores and look up tables. Sketch integrals for visual aid.