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3/11/25

Homework Review

  • Question number four from discrete distributions homework discussed.

Overview of Life Insurance Example

  • A 40-year-old man in the U.S. has a 0.242% risk of dying in the next year.

  • Insurance company charges $250 per year for a life insurance policy that pays $100,000 on death.

  • The profit for the insurance company depends on the outcome of the year, classified into two scenarios:

    • If you live: Profit = $250 (premium paid).

    • If you die: Profit = $100,000 - $250 = $99,750.

Probability Calculations

  • Calculating Probabilities:

    • Probability of dying ( x = 1) = 0.242% = 0.00242.

    • Probability of living = 1 - 0.00242 = 0.99758.

Expected Value (Mean) Calculation

  • Calculated in the context of the life insurance profits:

    • Mean = Sum of (Value * Probability).

  • Given values and their probabilities for the table:

    • 4 * 0.12 = 0.48

    • 3 * 0.45 = 1.35

    • 2 * 0.33 = 0.66

    • 1 * 0.07 = 0.07

    • 0 * 0.03 = 0.00

  • The total mean is estimated to be around 2.56.

Variance Calculation

  • To find variance, the following is computed:

    • Formula: (Value - Mean)² * Probability for each value.

    • Example for four:

      • (4 - 2.56)² * 0.12 = 0.248832

    • Continue for values of 3, 2, 1, and 0.

  • Variance result estimated at 0.8064.

    • Standard Deviation = √Variance = 0.898.

Introduction to Binomial Distribution

  • Definition of binomial distribution details:

    • Fixed number of trials (n).

    • Two outcomes (success/failure).

    • Independent trials with constant probability of success (p).

    • Focus on counting the number of successes (X).

Notation for Binomial Distribution

    • Number of trials: n.

    • Probability of success: p (e.g., 0.6 for 60%).

    • Random variable (success count): X.

  • Example of drawing cards (success = heart):

    • n = 10 trials, p = 0.25 for picking a heart, with x-values ranging from 0 to 10.

Probability Calculation Example

  • Example scenario with trials of a 3-part experiment:

    • Probability of success of p = 0.4, two successes across three trials.

    • Combinations calculated through C(n, x) = C(3, 2) = 3 ways.

    • Probability for each sequence leading to say, success, failure, success.

    • General formula for binomial probabilities identified:

    • P(X = k) = C(n, k) * p^k * (1-p)^(n-k).

Actual Problem Example

  • Multiple choice quiz scenario:

    • 10 questions, 5 choices each, chance to guess correct = 0.2.

    • To find probability of guessing 6 correct answers:

    • Use the binomial formula to evaluate:

    • C(10, 6) * 0.2^6 * 0.8^4.

    • Importance of calculator and correct inputs highlighted.

Calculator Use and Binomial CDF

  • CDF used for non-exact probabilities (e.g., less than/equal to, greater than/equal to).

  • Key calculations involve understanding setup and using correct calculator functionality:

    • Example questions were solved demonstrating the nuances of inputs required.

  • Emphasis on subtle differences like whether to use PDF (specific) or CDF (cumulative).

Summary of Topics Covered

  • Reviewed life insurance case for discrete probabilities.

  • Introduced binomial distributions and their specific criteria.

  • Provided numerous examples and probability calculations using specific formulas.

  • Emphasized importance of calculators and practice with real problems.