Continuous Probability Distributions Study Guide

Learning Objectives for Continuous Probability Distributions

• LO1: Define, Graph, and Interpret Continuous Uniform Probability Density Functions:

  • Understand the mathematical definition of a continuous uniform probability density function for a random variable assuming values in a specified interval.
  • Construct visual graphs of uniform probability distributions and interpret the height and interval bounds.

• LO2: Calculate Continuous Uniform Probabilities:

  • Compute and interpret the probability that a uniformly distributed continuous random variable assumes a value within any specified sub-interval.

• LO3: Compute Summary Statistics for Uniform Distributions:

  • Calculate and interpret the expected value $E(x)$, variance $\text{Var}(x)$, and standard deviation $\text{sigma}$ for a continuous uniform random variable.

• LO4: Define and Graph Normal Probability Density Functions:

  • Define the probability density function for a normally distributed random variable and interpret its bell-shaped curve and key mathematical characteristics.

• LO5: Calculate Normal Probabilities over Intervals:

  • Compute and interpret the probability that a normally distributed random variable assumes a value within a specified interval using standard normal tables and computer software.

• LO6: Perform Inverse Normal Distribution Calculations:

  • Determine and interpret the value of a continuous normal random variable corresponding to a specified cumulative or tail probability.

• LO7: Calculate Exponential Probabilities over Intervals:

  • Compute and interpret the probability that an exponentially distributed random variable assumes a value within a given interval.

• LO8: Define, Graph, and Interpret Exponential Density Functions:

  • Define the exponential probability density function, visualize its right-skewed curve, and interpret its operational applications in service and arrival processes.

Fundamental Principles of Continuous vs. Discrete Probability Distributions

• Fundamental Difference in Computing Probabilities:

  • For a discrete random variable, the probability function $f(x)$ provides the exact probability that the random variable assumes a specific value $x$.
  • For a continuous random variable, the counterpart to the probability function is the probability density function (PDF), also denoted by $f(x)$.
  • The probability density function $f(x)$ does not directly provide probability values for specific points.

• Area Under the Curve as Probability:

  • Probabilities for continuous random variables are defined by the area under the graph of the probability density function $f(x)$ corresponding to a given interval.
  • Computing probabilities for continuous variables always involves finding the likelihood that the variable falls within a range or interval $[x_1, x_2]$.

• Probability at a Single Point:

  • Because the area under the graph of a continuous probability density function above any single point is zero, the probability of a continuous random variable assuming any exact, specific value is strictly zero:     $P(X = x) = 0$
  • A single point represents an interval of zero width, making its area zero.

• Endpoint Inclusion Independence:

  • Due to the zero-probability point property, the probability that a continuous random variable falls within an interval is identical whether endpoints are included or excluded:     $P(x_1 \text{ ≤ } X \text{ ≤ } x_2) = P(x_1 < X < x_2) = P(x_1 \text{ ≤ } X < x_2) = P(x_1 < X \text{ ≤ } x_2)$

• Mathematical Requirements for Continuous Probability Density Functions:

  • Non-negativity condition: The density function must be non-negative everywhere:     $f(x) \text{ ≥ } 0 \quad \text{for all } x$     This requirement is the continuous analog to $f(x) \text{ ≥ } 0$ for discrete probability functions.
  • Total area condition: The total area under the graph of $f(x)$ over its entire domain must equal $1.0$:     $\int_{-\infty}^{\infty} f(x) \, dx = 1.0$     This requirement is the continuous analog to the discrete condition that the sum of all probabilities equals $1.0$

Continuous Uniform Probability Distribution

• Definition and Characteristics:

  • A continuous random variable $X$ has a uniform probability distribution over an interval $[a, b]$ if every sub-interval of equal length within $[a, b]$ is equally likely.

• Uniform Probability Density Function Formula:

  • The general formula for a continuous uniform PDF is defined as:     $f(x) = \begin{cases} \frac{1}{b - a} & \text{for } a \text{ ≤ } x \text{ ≤ } b \\ 0 & \text{elsewhere} \end{cases}$
  • Parameter $a$ represents the smallest value (lower bound) that the random variable can assume.
  • Parameter $b$ represents the largest value (upper bound) that the random variable can assume.

• Airplane Flight Time Example:

  • Consider a continuous random variable $X$ representing the flight time of an airplane traveling from Chicago to New York.
  • Flight time can assume any real value in the interval from $120\,\text{minutes}$ to $140\,\text{minutes}$.
  • Actual flight data indicate that the probability of a flight time within any $1\,\text{minute}$ interval is identical to any other $1\,\text{minute}$ interval within the large interval.
  • For this variable, parameter bounds are $a = 120\,\text{minutes}$ and $b = 140\,\text{minutes}$.
  • The continuous uniform density function for flight time is:     $f(x) = \begin{cases} \frac{1}{140 - 120} = \frac{1}{20} = 0.05 & \text{for } 120 \text{ ≤ } x \text{ ≤ } 140 \\ 0 & \text{elsewhere} \end{cases}$

• Area as a Measure of Probability Calculations:

  • Interval Probability Example 1 ($120\,\text{minutes}$ to $130\,\text{minutes}$):
  • Question: What is $P(120 \text{ ≤ } X \text{ ≤ } 130)$?
  • Area calculation: $\text{Area} = \text{Width} \times \text{Height}$
  • $\text{Width} = 130 - 120 = 10$
  • $\text{Height} = f(x) = \frac{1}{20} = 0.05$
  • $\text{Area} = 10 \times \left(\frac{1}{20}\right) = \frac{10}{20} = 0.50$
  • Thus, $P(120 \text{ ≤ } X \text{ ≤ } 130) = 0.50$
  • Interval Probability Example 2 ($128\,\text{minutes}$ to $136\,\text{minutes}$):
  • Question: What is $P(128 \text{ ≤ } X \text{ ≤ } 136)$?
  • $\text{Width} = 136 - 128 = 8$
  • $\text{Height} = f(x) = \frac{1}{20} = 0.05$
  • $\text{Area} = 8 \times \left(\frac{1}{20}\right) = 0.40$
  • Thus, $P(128 \text{ ≤ } X \text{ ≤ } 136) = 0.40$
  • Total Area Verification:
  • $P(120 \text{ ≤ } X \text{ ≤ } 140) = (140 - 120) \times \left(\frac{1}{20}\right) = 20 \times \left(\frac{1}{20}\right) = 1.00$

• Expected Value and Variance Formulas for Uniform Distribution:

  • Expected Value $E(x)$:$E(x) = \frac{a + b}{2}$
  • Variance $\text{Var}(x)$:$\text{Var}(x) = \frac{(b - a)^2}{12}$
  • Flight Time Summary Statistics:
  • $E(x) = \frac{120 + 140}{2} = \frac{260}{2} = 130\,\text{minutes}$
  • $\text{Var}(x) = \frac{(140 - 120)^2}{12} = \frac{20^2}{12} = \frac{400}{12} = 33.33\,\text{minutes}^2$
  • Standard Deviation $\sigma = \sqrt{33.33} = 5.77\,\text{minutes}$

Normal Probability Distribution

• Importance and Applications:

  • The normal probability distribution is the most widely used continuous distribution in statistics.
  • Applicable to natural and industrial measurements such as heights, weights, test scores, scientific measurements, and rainfall amounts.
  • Essential in statistical inference, describing the sampling distributions of estimators.

• Normal Probability Density Function Formula:

  • The mathematical expression for the bell-shaped normal density curve is:     $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left(\frac{x - \mu}{\sigma}\right)^2}$
  • $\mu$ is the distribution mean.
  • $\sigma$ is the distribution standard deviation.
  • $\pi \approx 3.14159$
  • $e \approx 2.71828$

• Seven Key Characteristics of the Normal Distribution:

  1. Parameter Differentiation: The entire family of normal distributions is differentiated by two parameters: the mean $\mu$ and the standard deviation $\sigma$
  2. Central Location: The highest point on the normal curve occurs at the mean $\mu$, which is simultaneously the median and mode of the distribution.
  3. Mean Values: The mean $\mu$ can be any real numerical value: negative, zero, or positive (e.g., three distributions with standard deviation $\sigma$ can have distinct means $\mu = -10$, $\mu = 0$, and $\mu = 20$).
  4. Symmetry and Skewness: The normal distribution is perfectly symmetric. The shape to the left of the mean is a exact mirror image of the shape to the right. Skewness is strictly $0$. Tails extend infinitely in both directions, asymptotic to the horizontal axis without touching it.
  5. Standard Deviation Effect: The standard deviation $\sigma$ determines the width and height of the curve. Larger values of $\sigma$ produce flatter, wider curves displaying higher data dispersion.
  6. Total Area and Half-Width Areas: Total area under the normal curve equals $1.0$. By symmetry, area to the left of $\mu$ is $0.50$ and area to the right of $\mu$ is $0.50$
  7. Empirical Rule Percentages:
  • $68.3\%$ of values fall within $\mu \pm 1\sigma$
  • $95.4\%$ of values fall within $\mu \pm 2\sigma$
  • $99.7\%$ of values fall within $\mu \pm 3\sigma$

Standard Normal Probability Distribution

• Definition and Designation:

  • A normal random variable having a mean of $\mu = 0$ and a standard deviation of $\sigma = 1$ is defined as having a standard normal probability distribution.
  • The letter $Z$ is universally reserved to denote this specific standard normal variable.

• Standard Normal Probability Density Function:

  • Substituting $\mu = 0$ and $\sigma = 1$ into the general normal PDF yields:     $f(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}}$

• Conversion Transformation ($Z$‐Score Formula):

  • Any normal random variable $X$ with mean $\mu$ and standard deviation $\sigma$ is converted to standard normal variable $Z$ using:     $Z = \frac{x - \mu}{\sigma}$
  • Interpretation: $Z$ measures the precise number of standard deviations that a given raw value $x$ lies away from its mean $\mu$

• Three Basic Types of Normal Probability Calculations:

  1. Probability that $Z$ is less than or equal to a given value: $P(Z \text{ ≤ } z_1$
  2. Probability that $Z$ is between two given values: $P(z_1 \text{ ≤ } Z \text{ ≤ } z_2)$
  3. Probability that $Z$ is greater than or equal to a given value: $P(Z \text{ ≥ } z_1)$

• Detailed Step-by-Step Standard Normal Table Examples:

  • Lower Tail Calculation $P(Z \text{ ≤ } 1.00)$:
  • Locate $1.0$ row and $0.00$ column in cumulative normal probability table (Table 1, Appendix B).
  • Intersection entry is $0.8413$
  • Result: $P(Z \text{ ≤ } 1.00) = 0.8413$
  • Interval Calculation $P(-0.50 \text{ ≤ } Z \text{ ≤ } 1.25)$:
  • Step 1: Area left of $Z = 1.25$ (Row $1.2$, Column $0.05$) equals $0.8944$
  • Step 2: Area left of $Z = -0.50$ (Row $-0.5$, Column $0.00$) equals $0.3085$
  • Step 3: Subtract lower area from upper area:       $P(-0.50 \text{ ≤ } Z \text{ ≤ } 1.25) = 0.8944 - 0.3085 = 0.5859$
  • Interval within 1 Standard Deviation $P(-1.00 \text{ ≤ } Z \text{ ≤ } 1.00)$:
  • Area left of $Z = 1.00$ equals $0.8413$
  • Area left of $Z = -1.00$ equals $0.1587$
  • $P(-1.00 \text{ ≤ } Z \text{ ≤ } 1.00) = 0.8413 - 0.1587 = 0.6826$
  • Upper Tail Calculation $P(Z \text{ ≥ } 1.58)$:
  • Table value for $Z = 1.58$ (Row $1.5$, Column $0.08$) equals $0.9429$
  • Subtract cumulative left area from $1.0000$:       $P(Z \text{ ≥ } 1.58) = 1.0000 - 0.9429 = 0.0571$

• Inverse Normal Probability Calculations (Finding $Z$ for Given Area):

  • Example: Finding $Z$ for Upper Tail Probability of $0.10$:
  • Cumulative left area required: $1.0000 - 0.1000 = 0.9000$
  • Scan body of standard normal table for probability closest to $0.9000$: found $0.8997$
  • Read leftmost row ($1.2$) and top column ($0.08$).
  • Corresponding value is $Z = 1.28$
  • Key Benchmark $Z$‐Values:
  • Upper tail probability $0.025$ corresponds to cumulative area $0.9750$ and $Z = 1.96$
  • Lower tail probability $0.025$ corresponds to cumulative area $0.0250$ and $Z = -1.96$

Applications of Normal Distribution: Greer Tire Company

• Problem Background and Parameters:

  • Product: New steel-belted radial tire.
  • Decision context: Determining mileage guarantee policy.
  • Continuous random variable $X$ = number of miles tire lasts.
  • Distribution: Normal distribution with mean $\mu = 36,500\,\text{miles}$ and standard deviation $\sigma = 5,000\,\text{miles}$.

• Question 1: Percentage of Tires Lasting Exceeding $40,000\,\text{Miles}$:

  • Target probability: $P(X > 40,000)$
  • Convert $x = 40,000$ to standard normal variable $Z$:     $Z = \frac{40,000 - 36,500}{5,000} = \frac{3,500}{5,000} = 0.70$
  • Standard normal table cumulative probability for $Z = 0.70$ is $0.7580$
  • Upper tail probability calculation:     $P(X > 40,000) = P(Z > 0.70) = 1.0000 - 0.7580 = 0.2420$
  • Conclusion: Approximately $24.2\%$ of tires will last more than $40,000\,\text{miles}$.

• Question 2: Mileage Guarantee for Maximum $10\%$ Discount Eligibility:

  • Requirement: Find $x$ such that $P(X < x) = 0.10$
  • Find $Z$ value cutting off lower tail area of $0.10$: table gives $Z = -1.28$
  • Transform $Z$ back to original continuous variable $x$:     $Z = \frac{x - \mu}{\sigma} \implies x = \mu + Z \sigma$$x = 36,500 + (-1.28)(5,000) = 36,500 - 6,400 = 30,100\,\text{miles}$
  • Recommendation: Setting a mileage guarantee of $30,100\,\text{miles}$ (or rounded to $30,000\,\text{miles}$) ensures approximately $10\%$ of tires are eligible for replacement discount.

Computing Normal Probabilities Using Excel and R

• Excel Standard Normal Distribution Functions:

  • NORM.S.DIST(z, cumulative):
  • Inputs: z value, cumulative (TRUE for cumulative area, FALSE for PDF curve height).
  • Example $P(Z \text{ ≤ } 1.00)$: =NORM.S.DIST(1.00, TRUE) yields $0.8413$
  • Example $P(-0.50 \text{ ≤ } Z \text{ ≤ } 1.25)$: =NORM.S.DIST(1.25, TRUE) - NORM.S.DIST(-0.50, TRUE) yields $0.5859$
  • Example $P(Z \text{ ≥ } 1.58)$: =1 - NORM.S.DIST(1.58, TRUE) yields $0.0571$
  • NORM.S.INV(probability):
  • Inputs: Cumulative lower tail probability.
  • Example for upper tail $0.10$ (lower tail $0.90$): =NORM.S.INV(0.90) yields $1.28$
  • Example for upper tail $0.025$: =NORM.S.INV(0.975) yields $1.96$
  • Example for lower tail $0.025$: =NORM.S.INV(0.025) yields $-1.96$

• Excel General Normal Distribution Functions:

  • NORM.DIST(x, mean, standard_dev, cumulative):
  • Inputs: x, mean, standard_dev, cumulative (TRUE for cumulative probability).
  • Greer Tire Example $P(X \text{ ≤ } 20,000)$: =NORM.DIST(20000, 36500, 5000, TRUE) yields $0.0005$ (almost all tires last over $20,000\,\text{miles}$).
  • Greer Tire Interval $P(20,000 \text{ ≤ } X \text{ ≤ } 40,000)$: =NORM.DIST(40000, 36500, 5000, TRUE) - NORM.DIST(20000, 36500, 5000, TRUE) yields $0.7576$
  • Greer Tire Upper Tail $P(X \text{ ≥ } 40,000)$: =1 - NORM.DIST(40000, 36500, 5000, TRUE) yields $0.2420$
  • NORM.INV(probability, mean, standard_dev):
  • Inputs: Lower tail probability, mean, standard_dev.
  • Greer Tire $10\%$ Guarantee: =NORM.INV(0.10, 36500, 5000) yields $30,092.24\,\text{miles}$.
  • Minimum mileage for top $2.5\%$ of tires (lower area $0.975$): =NORM.INV(0.975, 36500, 5000) yields $46,299.82\,\text{miles}$.

• R Language Commands for Normal Distribution:

  • Application Example: Greer Aircraft Engine Lifetime ($\mu = 36,500\,\text{hours}$, $\sigma = 5,000\,\text{hours}$).
  • Compute cumulative probability $P(X \text{ ≤ } 40,000)$: pnorm(40000, 36500, 5000) returns 0.7580.
  • Compute upper tail probability $P(X > 40,000) = 1 - 0.7580 = 0.2420$ ($24.2\%$).
  • Inverse quantile computation for lower $10\%$ guarantee: qnorm(0.10, 36500, 5000) returns 30092.24\,\text{hours}.

Continuous Exponential Probability Distribution

• Applications and Context:

  • Used for continuous random variables measuring time between arrivals at service facilities, loading/service times, and distances between physical defects.

• Exponential Density Function Formula:

  • Defined mathematically as:     $f(x) = \begin{cases} \frac{1}{\mu} e^{-\frac{x}{\mu}} & \text{for } x \text{ ≥ } 0 \\ 0 & \text{for } x < 0 \end{cases}$
  • Parameter $\mu$ is the mean or average length of the interval.

• Cumulative Exponential Probability Formula:

  • The formula providing the cumulative probability of obtaining a value less than or equal to $x_0$ is:     $P(X \text{ ≤ } x_0) = 1 - e^{-\frac{x_0}{\mu}}$

• Summary Statistics Properties:

  • Mean: $E(x) = \mu$
  • Standard Deviation: $\sigma = \mu$ (A unique property: standard deviation equals the mean).
  • Variance: $\text{Var}(x) = \sigma^2 = \mu^2$

• Ship Loading Dock Application Example:

  • Random variable $X$ = loading time for a grey truck at dock.
  • Mean loading time $\mu = 15\,\text{minutes}$.
  • PDF Formula: $f(x) = \frac{1}{15} e^{-\frac{x}{15}}$
  • Standard deviation $\sigma = 15\,\text{minutes}$, Variance $\sigma^2 = 15^2 = 225\,\text{minutes}^2$
  • Calculation 1 ($6\,\text{minutes}$ or less):$P(X \text{ ≤ } 6) = 1 - e^{-\frac{6}{15}} = 1 - e^{-0.40} = 1 - 0.6703 = 0.3297$
  • Calculation 2 ($18\,\text{minutes}$ or less):$P(X \text{ ≤ } 18) = 1 - e^{-\frac{18}{15}} = 1 - e^{-1.20} = 1 - 0.3012 = 0.6988$
  • Calculation 3 (Between $6\,\text{minutes}$ and $18\,\text{minutes}$):$P(6 \text{ ≤ } X \text{ ≤ } 18) = P(X \text{ ≤ } 18) - P(X \text{ ≤ } 6) = 0.6988 - 0.3297 = 0.3691$

Relationship Between Poisson and Exponential Distributions

• Discrete vs. Continuous Complementarity:

  • The discrete Poisson distribution models the number of occurrence events over a specified unit interval of time or space.
  • Poisson probability function: $f(x) = \frac{\mu^x e^{-\mu}}{x!}$
  • The continuous exponential distribution models the length of the interval between consecutive occurrences.

• Mathematical Equivalence and Parameter Conversion:

  • If arrivals follow a Poisson distribution with arrival rate $\lambda$ occurrences per unit time, the expected time between arrivals is $\mu = \frac{1}{\lambda}$.

• Car Wash Arrival Example:

  • Arrival rate: Poisson distribution with mean $\lambda = 10\,\text{cars per hour}$.
  • Average time between car arrivals: $\mu = \frac{1\,\text{hour}}{10\,\text{cars}} = 0.1\,\text{hours per car}$.
  • Continuous exponential PDF for time between arrivals $X$:     $f(x) = \frac{1}{0.1} e^{-\frac{x}{0.1}} = 10 e^{-10x}$

Computing Exponential Probabilities Using Excel and R

• Excel EXPON.DIST Function:

  • Syntax: EXPON.DIST(x, lambda, cumulative)
  • Input parameters: x value, lambda (defined as $\frac{1}{\mu}$), cumulative (TRUE for cumulative area).
  • Ship Loading Dock Examples ($\mu = 15\,\text{minutes}$, lambda = 1/15):
  • $P(X \text{ ≤ } 18)$: =EXPON.DIST(18, 1/15, TRUE) yields $0.6988$
  • $P(6 \text{ ≤ } X \text{ ≤ } 18)$: =EXPON.DIST(18, 1/15, TRUE) - EXPON.DIST(6, 1/15, TRUE) yields $0.3691$
  • $P(X \text{ ≥ } 8)$: =1 - EXPON.DIST(8, 1/15, TRUE) yields $0.5866$

• R Language Commands for Exponential Distributions:

  • Application Example: Time between business loan defaults ($\mu = 15\,\text{months}$).
  • Parameter rate in R is 1/mu ($1/15$
  • Calculate $P(6 \text{ ≤ } X \text{ ≤ } 18)$: pexp(18, 1/15) - pexp(6, 1/15) yields 0.3691.

Case Study Analysis: Specialty Toys, Inc. (Weather Teddy)

• Business Context & Problem Statement:

  • Product: "Weather Teddy" (talking bear with built-in barometer predicting weather with 5 distinct phrases).
  • Financial parameters: Selling Price = $\$24$, Cost = $\$16$, Clearance price for leftover stock = $\$5$
  • Forecasted Demand: Expected demand $\mu = 20,000\,\text{units}$ with a $0.95$ probability that demand falls between $10,000$ and $30,000\,\text{units}$.
  • Candidate order quantities suggested by management team: $15,000$, $18,000$, $24,000$, and $28,000\,\text{units}$.

• Managerial Report Requirements:

  1. Normal Distribution Model:
  • Mean $\mu = 20,000\,\text{units}$.
  • Standard deviation estimation: Since $95\%$ of area ($Z = 1.96$) lies between $10,000$ and $30,000$, $\mu \pm 1.96\sigma = 20,000 \pm 10,000 \implies 1.96\sigma = 10,000 \implies \sigma = 5,102\,\text{units}$ (or rounded $\sigma \approx 5,000\,\text{units}$ using $2\sigma$
  1. Stockout Probabilities: Compute $P(D > Q)$ for order quantities $Q \in \{15,000; 18,000; 24,000; 28,000\}$.
  2. Profit Projections across Scenarios: Evaluate profit for worst case ($10,000\,\text{units}$), most likely ($20,000\,\text{units}$), and best case ($30,000\,\text{units}$).
  3. Policy Analysis ($70\%$ Demand Satisfaction):
  • Find $Q$ such that $P(D \text{ ≤ } Q) = 0.70$
  • Corresponding $Z = 0.524$
  • Target Quantity: $Q = 20,000 + 0.524(5,102) \approx 22,673\,\text{units}$.
  1. Final Recommendation and Justification: Optimal order quantity selection balancing stockout risk and inventory clearance losses.

Case Study Analysis: Gephardt Electronics Wire Quality

• Business Context & Technical Parameters:

  • Component: Three-foot segment ($36\,\text{inches}$) of $0.20\,\text{mm}$ diameter solid pure oxygen-free electronic (OFE) copper wire.
  • Supplier claim: Spools average $50\,\text{inches}$ between internal flaws.

• Managerial Assessment Questions:

  1. Flaw Distribution Identification: The continuous variable length between consecutive flaws follows an exponential probability distribution with mean $\mu = 50\,\text{inches}$.
  2. Probability of Flawless Wire Segment:
  • Segment length $x = 36\,\text{inches}$.
  • Probability of distance to first flaw exceeding $36\,\text{inches}$:        $P(X > 36) = 1 - P(X \text{ ≤ } 36) = e^{-\frac{36}{50}} = e^{-0.72} \approx 0.4868$
  1. Minimum Mean Length for Quality Criterion: Determine $\mu$ required to meet specified benchmark flawless probabilities.
  2. Target Mean Length for $1\,\text{in}\,100$ Flaw Probability:
  • Requirement: Probability of flaw in $36\text{ inches}$ is $0.01$, so $P(X > 36) = 0.99$
  • Formula: $e^{-\frac{36}{\mu}} = 0.99 \implies -\frac{36}{\mu} = \ln(0.99) \approx -0.01005$
  • Result: $\mu = \frac{36}{0.01005} \approx 3,582.1\,\text{inches}$