Study Notes on Discrete Random Variables and Their Probability Distributions

MATH 2303: Introduction to Probability & Statistics

Overview

  • Instructor: Ms. Galloway, MA Mathematics Education

  • Institution: University of the Incarnate WordⓇ

Lecture 8 Outline

  • Last Class:

    • Review of Exam 1 scheduled to be discussed on Tuesday.

  • Today's Focus:

    • Chapter 5: Discrete Random Variables and Their Probability Distributions

      • 5.1 – Random Variables

      • 5.2 – Probability Distribution of a Discrete Random Variable

      • 5.3 – Mean and Standard Deviation of a Discrete Random Variable

Chapter 5: Discrete Random Variables and Their Probability Distributions

5.1 Random Variables

Definitions
  • Random Variables:

    • A random variable is defined as a variable whose value is determined by the outcome of a random experiment.

  • Discrete Random Variable:

    • A random variable that assumes countable values.

  • Continuous Random Variable:

    • A random variable that can assume any value contained in one or more intervals.

Examples of Discrete Random Variables
  1. The number of cars sold at a dealership during a given month.

  2. The number of houses on a certain block.

  3. The number of fish caught on a fishing trip.

  4. The number of complaints received at an airline’s office on a given day.

  5. The number of customers visiting a bank during any given hour.

  6. The number of heads obtained in three tosses of a coin.

Examples of Continuous Random Variables
  1. The length of a room.

  2. The time taken to commute from home to work.

  3. The amount of milk in a gallon (not necessarily exact, may slightly exceed or be less than one gallon).

  4. The weight of a letter.

  5. The price of a house (often treated as a continuous random variable due to a large number of unique values).

5.2 Probability Distribution of a Discrete Random Variable

Definition
  • The probability distribution of a discrete random variable lists all possible values that the random variable can assume along with their corresponding probabilities.

Example 5-1
  • Scenario: Frequency and relative frequency distributions of the number of vehicles owned by families sampled from 2000 families.

Number of Vehicles Owned

Frequency

Relative Frequency

0

30

0.015

1

470

0.235

2

850

0.425

3

490

0.245

4

160

0.080

N = 2000

Sum = 1.000

Probability Distribution for Example 5-1
Table:

Number of Vehicles Owned (X)

Probability (P(x))

0

0.015

1

0.235

2

0.425

3

0.245

4

0.080

ΣP(x) = 1.000

Characteristics of a Probability Distribution
  1. For each value of x, it holds that 0P(x)10 \leq P(x) \leq 1.

  2. The sum of the probabilities must equal 1: ΣP(x)=1Σ P(x) = 1.

Example 5-2: Calculating Probabilities
  • Using the probability distribution:

    • (a) The probability that a randomly selected family owns two vehicles:

      • P(selected family owns 2 vehicles)=P(2)=0.425P(selected \ family \ owns \ 2 \ vehicles) = P(2) = 0.425.

    • (b) The probability that a randomly selected family owns at least two vehicles:

      • P(at least two vehicles)=P(2 or 3 or 4)=P(2)+P(3)+P(4)=0.425+0.245+0.080=0.750P(at \ least \ two \ vehicles) = P(2 \ or \ 3 \ or \ 4) = P(2) + P(3) + P(4) = 0.425 + 0.245 + 0.080 = 0.750.

    • (c) The probability that a randomly selected family owns at most one vehicle:

      • P(at most 1)=P(0 or 1)=P(0)+P(1)=0.015+0.235=0.250P(at \ most \ 1) = P(0 \ or \ 1) = P(0) + P(1) = 0.015 + 0.235 = 0.250.

    • (d) The probability that a randomly selected family owns three or more vehicles:

      • P(3 or 4)=P(3)+P(4)=0.245+0.080=0.325P(3 \ or \ 4) = P(3) + P(4) = 0.245 + 0.080 = 0.325.

Example 5-3: Validating Probability Distributions
  • Given tables of probabilities, you must determine:

    • (a) Invalid: Sum of probabilities is not equal to 1.

    • (b) Valid: Probabilities sum to 1.

    • (c) Invalid: Contains a negative probability.

Example 5-4: Probability Distribution for Machine Breakdowns
  • Breakdown Probability Distribution of machine:

  • Question: What is the probability of breakdowns of 0, 1, 2, etc?

  • Solution: Detailed analysis similar to prior examples will follow.

5.3 Mean and Standard Deviation of a Discrete Random Variable

Mean
  • The mean (expected value) of a discrete random variable xx, denoted as µµ, is calculated as:
    µ=ΣxP(x)µ = Σ x P(x).

Example 5-6: Calculation of Mean
  • Given the breakdown probability data, calculate as follows:

    • µ=ΣxP(x)=1.80µ = Σx P(x) = 1.80.

Standard Deviation
  • The standard deviation of a discrete random variable xx, denoted as ss, is a measure of the spread of its probability distribution, computed as:
    s=Σ(xµ)2P(x)s = \sqrt{Σ (x - µ)^2 P(x)}.

Example 5-7: Standard Deviation Calculation for Defective Parts
  • Given a probability distribution of defective parts, compute:

  • Example: s1.204.s ≈ 1.204.

Example 5-8: Company Profit Probability Distribution
  1. Define annual profits based on sales performance with respective probabilities:

    • High Sales: Profit of $4.5 million with probability 0.32

    • Medium Sales: Profit of $1.2 million with probability 0.51

    • Low Sales: Loss of $2.3 million with probability 0.17

  2. Calculate mean and standard deviation:

    • Distribution of xx shows profitable and loss scenarios clearly.

    • Implications on business strategy considered based on computed values.


Review Notes

  • Last Class:

    • Review of Exam 1 (scheduled for Tuesday).

  • Today's Focus:

    • Chapter 5 covering:

      • Random Variables (5.1)

      • Probability Distribution of a Discrete Random Variable (5.2)

      • Mean and Standard Deviation of a Discrete Random Variable (5.3)