Chapter 5 - Continuous Random Variables and the Continuous Uniform Distribution

Course and Curricular Context

• Course identification: STAT 243 TBCC, Spring 2026 term, instructed by Dr. Jaxon Shumaker, dated April 27, 2026.

• Subject scope: Chapter 5 examines continuous random variables, focusing specifically on Section 5.1 (Continuous Probability Distributions) and Section 5.2 (The Continuous Uniform Distribution).

• Curricular progression: Builds directly upon discrete probability distributions covered in Chapter 4 and establishes mathematical frameworks for advanced continuous models in Chapter 6.

Fundamentals of Continuous Random Variables

• Definition: A continuous random variable is a variable that takes on continuous numerical values resulting from measurement rather than counting.

• Contrast with discrete variables: Discrete random variables take on distinct, separated, countable values (for example, the numeric outcome of rolling a die). Continuous random variables assume any value within an unbroken interval or continuum.

• Physical and real-world examples:

  • The length of operational lifespan of a computer chip ($X = \text{length of time a computer chip lasts}$.

  • The weight of newborn calves ($X = \text{weight of newborn calves}$.

  • The total precipitation or rainfall measured in a specific geographical area over a given month ($X = \text{rainfall per month}$.

• Primary measurement property: Outcomes for continuous random variables are continuously measured across a scale, never counted as discrete entities.

Continuous Probability Distributions and the Probability Density Function (PDF)

• Graphical representation: The graphical model of a continuous probability distribution is represented by a continuous curve drawn above the horizontal axis.

• Area as probability: Probability corresponds strictly to the geometric area situated beneath the curve and above the horizontal axis.

• Probability Density Function (PDF) definition: The function describing the curve is termed the probability density function, abbreviated as PDF, and denoted mathematically by $f(x)$.

• Mathematical role of $f(x)$: The density function $f(x)$ defines the height of the curve at each value $x$ and serves as the mathematical formula used to graph the distribution.

• Total area axiom: The total area bounded beneath the density function curve $f(x)$ and above the $x$-axis over the entire domain of the distribution is identically equal to $1$.

• Probability at a single exact point: The probability of a continuous random variable assuming any exact, singular numeric value is identically zero:   $P(X = c) = 0$

  • Geometric justification: A single numeric point on the horizontal axis has an infinitesimal width of zero ($\text{width} = 0$). Because geometric area is calculated as height multiplied by width, the area of a vertical line above a single point is zero ($\text{area} = 0$).

• Measurement across intervals: Continuous probabilities are calculated exclusively over intervals of $x$ values using inequalities (such as $P(a < X < b)$ or $P(a \le X \le b)$).

• Equivalence of strict and non-strict inequalities: Because the probability at any isolated boundary point is zero ($P(X = a) = 0$ and $P(X = b) = 0$), including or excluding boundary endpoints does not change the probability:   $P(a < X < b) = P(a \le X \le b) = P(a \le X < b) = P(a < X \le b)$

The Cumulative Distribution Function (CDF) and Computational Approaches

• Definition of Cumulative Distribution Function (CDF): The CDF, abbreviated as cdf, is a mathematical function that evaluates the accumulated area under the probability density curve up to a specified value of $x$, representing $P(X \le x)$.

• Purpose: Serves as the primary operational tool to calculate continuous probabilities as cumulative areas.

• Primary methodologies for evaluating area under a continuous curve:

  • Geometric computation: Applying classical geometric area formulas (such as areas of rectangles or trapezoids) when the PDF forms standard planar geometric shapes.

  • Tabular lookup and computational software: For complex functional curves that cannot be integrated using elementary geometry, probabilities are retrieved from standardized reference tables or computed using statistical software programs.

Introductory Geometric Density Function Examples

• Example 1: Uniform continuous function over interval $[0, 20]$

  • Function definition:     $f(x) = \frac{1}{20}$     for $0 \le x \le 20$.

  • Geometric profile: A horizontal line at constant height $f(x) = \frac{1}{20}$ stretching from $x = 0$ to $x = 20$. Total area equals $\frac{1}{20} \times (20 - 0) = 1$.

  • Evaluation of $P(0 < X < 2)$:

  • Height of rectangular region: $\frac{1}{20}$.

  • Width of interval: $2 - 0 = 2$.

  • Probability calculation:       $P(0 < X < 2) = \frac{1}{20} \times 2 = \frac{2}{20} = \frac{1}{10}$

  • Evaluation of $P(2 < X < 15)$:

  • Height of rectangular region: $\frac{1}{20}$.

  • Width of interval: $15 - 2 = 13$.

  • Probability calculation:       $P(2 < X < 15) = \frac{1}{20} \times (15 - 2) = \frac{1}{20} \times 13 = \frac{13}{20}$

• Example 2: Uniform continuous function over interval $[0, 8]$

  • Function definition:     $f(x) = \frac{1}{8}$     for $0 < x < 8$.

  • Geometric profile: A rectangle of uniform height $\frac{1}{8}$ over the base interval from $0$ to $8$.

  • Evaluation of $P(2.5 < X < 7.5)$:

  • Height of rectangular region: $\frac{1}{8}$.

  • Width of interval: $7.5 - 2.5 = 5$.

  • Probability calculation:       $P(2.5 < X < 7.5) = \frac{1}{8} \times (7.5 - 2.5) = \frac{1}{8} \times 5 = \frac{5}{8}$

The Continuous Uniform Distribution Formalism

• Conceptual definition: The continuous uniform distribution describes continuous random variables where all outcomes within a bounded continuous range are equally likely to occur.

• Support interval: The continuous random variable $X$ is defined across a closed bounded interval $[a, b]$, where $a$ denotes the minimum endpoint and $b$ denotes the maximum endpoint ($a < b$).

• Standard statistical notation:   $X \sim U(a, b)$

• Probability Density Function (PDF) formula:   $f(x) = \frac{1}{b - a}$   for $a \le x \le b$.

  • The height of the function is identically zero ($f(x) = 0$) for all values where $x < a$ or $x > b$.

• Mean ($\mu$) formula: Represents the central expected value and exact midpoint of the interval:   $\mu = \frac{a + b}{2}$

• Standard Deviation ($\sigma$) formula: Quantifies the dispersion across the continuous interval:   $\sigma = \sqrt{\frac{(b - a)^2}{12}}$

  • Equivalently stated due to squaring:     $\sigma = \sqrt{\frac{(a - b)^2}{12}}$

• Core distribution parameters: Complete characterization of any uniform distribution requires identifying its PDF, mean ($\mu$), and standard deviation ($\sigma$).

• Conceptual models and physical intuition:

  • Physical box model: Dropping a ball into a two-dimensional box of length $20$ units (or shaking the box) where the ball has an equal probability of coming to rest at any linear coordinate across the base.

  • Transit wait time model: Arriving at a bus stop where a bus operates on regular, repeating intervals. Assuming a passenger does not know the specific arrival timetable, the passenger is equally likely to arrive at any moment within the headway interval between buses.

Interval Probability and Percentile Calculations

• Example distribution: Let a continuous random variable follow:   $X \sim U(0, 20)$

  • Interval endpoints: $a = 0$, $b = 20$.

  • PDF height calculation:     $f(x) = \frac{1}{20 - 0} = \frac{1}{20}$     for $0 \le x \le 20$.

• Evaluating interval probability $P(2 < X < 18)$:

  • Base width of region: $18 - 2 = 16$.

  • Height of distribution: $\frac{1}{20}$.

  • Area calculation:     $P(2 < X < 18) = \frac{1}{20} \times (18 - 2) = \frac{16}{20} = \frac{4}{5}$

• Definition and determination of percentiles:

  • Conceptual definition: The $k$-th percentile is the specific variable value $x_k$ such that $k\%$ of the distribution area (or data points) lies below $x_k$.

  • Target percentile: The 90th percentile, denoted as $x_{90}$.

  • Mathematical condition:     $P(X < x_{90}) = 0.90$

  • Geometric setup using rectangular area formula:     $\text{height} \times (x_{90} - \text{start}) = 0.90$     $\frac{1}{20} \times (x_{90} - 0) = 0.90$

  • Algebraic solution:     $\frac{x_{90}}{20} = 0.90$     $x_{90} = 0.90 \times 20$     $x_{90} = 18$

  • Practical interpretation: There is a $90\%$ probability ($0.90$) that an outcome of $X$ will fall below $18$.

Conditional Probability in Uniform Distributions

• Scenario: Time required for students to complete an academic quiz.

• Variable definition: Let $X$ denote the completion time in minutes taken by a student to complete the quiz.

• Distribution specification: The variable $X$ is uniformly distributed between $6$ and $15$ minutes, inclusive:   $X \sim U(6, 15)$

  • Boundary parameters: $a = 6$, $b = 15$.

  • Total interval length: $b - a = 15 - 6 = 9$.

  • PDF height:     $f(x) = \frac{1}{15 - 6} = \frac{1}{9}$     for $6 \le x \le 15$.

• Problem 1: Unconditional probability that a randomly selected student needs at least $8$ minutes to complete the quiz:

  • Target expression:     $P(X \ge 8)$

  • Interval of interest: From $x = 8$ to the upper boundary $x = 15$.

  • Interval width: $15 - 8 = 7$.

  • Probability evaluation:     $P(X \ge 8) = \frac{1}{9} \times (15 - 8) = \frac{1}{9} \times 7 = \frac{7}{9}$

• Problem 2: Conditional probability that a student needs at least $8$ minutes given that she has already taken more than $7$ minutes:

  • Target expression:     $P(X \ge 8 \mid X > 7)$

  • General conditional probability theorem:     $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

  • Structural application:     $P(X \ge 8 \mid X > 7) = \frac{P(X \ge 8 \cap X > 7)}{P(X > 7)}$

  • Intersection simplification: Because the event $X \ge 8$ is a complete subset of $X > 7$, their intersection is simply $X \ge 8$:     $P(X \ge 8 \cap X > 7) = P(X \ge 8)$

  • Evaluation of conditioning event probability $P(X > 7)$:

  • Sub-interval width from $7$ to $15$: $15 - 7 = 8$.

  • Probability calculation:       $P(X > 7) = \frac{1}{9} \times (15 - 7) = \frac{8}{9}$

  • Computation of the conditional ratio:     $P(X \ge 8 \mid X > 7) = \frac{\frac{7}{9}}{\frac{8}{9}} = \frac{7}{8}$

• Sample space re-normalization interpretation: The given condition ($X > 7$) truncates the active sample space from its original width of $9$ down to a new effective width of $15 - 7 = 8$. This value ($8$) serves as the new denominator, while the favorable interval from $8$ to $15$ has a width of $7$, producing the ratio $\frac{7}{8}$.

Mistake on Homework Clarifications

• Why is the following statement false?: The graph of a continuous probability distribution cannot have y-values larger than 1.

  • The statement is false because, in a continuous probability distribution, the area under the curve must equal 1 rather than the y-values themselves. While individual y-values can exceed 1, their overall contribution across the range must still satisfy the total probability of 1, meaning that the function can be stretched over a sufficiently wide interval.

• Why is the statement If X ~ U(3,7), then P(X = 4) = P(X = 9) true?

  • This statement is true because for continuous random variables, specifically those following a uniform distribution, the probability of any specific point taking on a value is always zero. Therefore, both $P(X = 4)$ and $P(X = 9)$ are equal to zero, making the statement logically sound.

• How do you solve this problem?: The probability that wave will crash onto the beach exactly 4.8 seconds after the person arrives is P(x = 4.8) =

  • 0 because, similar to the uniform distribution, the probability of a continuous random variable equating to any specific value is zero.

WORD PROBLEM: Today, the waves are crashing onto the beach every 5.5 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 5.5 seconds. Round to 4 decimal places where possible.

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  • How to solve this problem? The probability that the wave will crash onto the beach between 0.7 and 5.3 seconds after the person arrives is P(0.7 < x < 5.3) =

    • 1: Identify the desired interval, which is 0.7 to 5.3

    • 2: Find the total interval length, which is b - a of the original uniform, or 5.5 - 0

    • 3: Plug the answers into the following equation:

      • $P\left(0.7<X<5.3\right)=$ Desired interval length / total interval length $\frac{\left(5.3-0.7\right)}{b-a}$

        • Or: $\frac{\left(5.3-0.7\right)}{5.5-0}=\frac{4.6}{5.5}=0.8364$

  
  

  • How to solve this problem?: The probability that it will take longer than 3 seconds for the wave to crash onto the beach after the person arrives is P(x > 3) =

    • 1: Find the interval length: 5.5 - 3 = 2.5

    • 2: Find the total interval length: 5.5 - 0 = 5.5

    • 3: use the same equation as above: $\frac{\left(5.5-3\right)}{\left(5.5-0\right)}=\frac{2.5}{5.5}=0.4545$

  
  

  • How to solve this problem?: Suppose that the person has already been standing at the shoreline for 0.5 seconds without a wave crashing in. Find the probability that it will take between 2.7 and 4.9 seconds for the wave to crash onto the shoreline.

    • IMPORTANT NOTE: When a problem says already begun, this means it’s a conditional probability.

      • So it turns into: P(2.7< X < 4.9 | X 0.5)

    • Step 1: Adjust the interval range, so instead of 5.5 - 0

      • It is now: 5.5 - 0.5

    • Step 2: Find the desired interval length.

      • So, 4.9 - 2.7 = 2.2

    • Step 3: Use the same equation as above.

      • So: $\frac{\left(4.9-2.7\right)}{\left(5.5-0.5\right)}=\frac{2.2}{5}=0.44$

  
  

  • How to solve this problem?: 63% of the time a person will wait at least how long before the wave crashes in?

    • This problem is having you work backwards from a probability, which is 0.63

      • The phrase “wait at least how long” indicates that the probability is written as: $P(X\geq x)=0.63$

    • Step 1: Think about the interval.

      • If someone waits at least x seconds, the interval goes from x to the maximum, so:

        • 5.5 - x

    • Step 2: Divide by the total interval.

    • The total range is 5.5 - 0 = 5.5

      • You set up is = $\frac{\left(5.5-x\right)}{5.5}=0.63$

        • Using some basic algebra: $\frac{\left(5.5-x\right)}{\left(5.5\cdot5.5\right)}=0.63\cdot5.5$ (Cancel out the denominator)

          • $5.5-x=3.465$

          • $-x=3.465-5.5$ (Subtract 5.5 from both sides)

          • $-x=-2.035$ (Multiply both sides by -1)

            • $x=2.035$

  • How to solve this problem?: Find the minimum for the upper quartile.

    • Remember that the upper quartile is Q3, which is 75% of the data.

      • Thus: $P\left(X\leq x)=0.75\right)$

      • Just simply multiply $0.75\left(5.5\right)=4.125$

WORD PROBLEM: Births are approximately Uniformly distributed between the 52 weeks of the year. They can be said to follow a Uniform distribution from 1 to 53 (a spread of 52 weeks). Round answers to 4 decimal places when possible.

• How to solve this problem?: P(x > 6 | x < 30) =

  • Step 1: Find your new range:

    • Because x > 30, that means we ignore everything from 30 - 53, thus 30 -1

  • Step 2: Find that portion that satisfies x > 6

    • Because you’re interested in 30 being the max, you must do 30 - 6.

  • Step 3: Calculate using: $\frac{30-6}{30-1}=\frac{24}{29}=0.8276$

• How to solve this problem?: Find the maximum for the lower quartile.

  • Step 1: Set up the percentile

    • $P(X\leq x)=0.25$

    • $\frac{\left(x-1\right)}{53-1}=0.25$ =

      • $\frac{\left(x-1\right)}{52}=0.25(52)$

        • $x-1=13$ =

          • x = 14

• How to solve this problem?: Find the 85th percentile.

  • Follow the example of finding the maximum for the lower quartile.

Probability Density Formula Problem: Consider the probability function for continuous random variable, X , given by $f\left(x\right)=0.02x$ for x between 0 and 10

• How to solve this problem?: $P(X\geq0)=$ ?

  • Step 1: Look at what values (X) can actually have.

    • The problem says 0 < x < 10, so distribution only exists between 0 and 10.

  • Step 2: Interpret $X\geq0$

    • This is asking what is the probability of X being bigger or equal to 0. Because the distribution is bigger than 0, you’re finding the probability of the entire distribution of 0 < x < 10.

  • Step 3: Use the PDF Formula:

    • $P\left(0<X<10\right)=\int_0^{10}0.02x\,dx$

      Step 1: Take the function 0.02x and integrate it:

      • $\int x\,dx=\frac{x^2}{2}=\int0.02xdx$

        • Which becomes $0.02\frac{x^2}{x},$ then simplifies to $0.01x^2$

    • Step 2: Use your boundaries:

      • $\int_0^{10}0.02x\,dx$ is evaluated to:

        • $\left\lbrack0.01x^2\right\rbrack_{}^{10}0$

          • Thus is simplified to = $\left(0.01(10)^2\right)-\left(0.01\left(0\right)^2\right)$

            • Now solve to get = 1

              
              

• How to solve this problem? $P(X\geq10)=$

  • If a problem is asking for the probability of a number bigger than the maximum of the range, in this case it is 10, it is always 0 because distribution can only happen between 0 and 10.