Chapter 6: Continuous Probability Distributions Notes

Chapter 6: Continuous Probability Distributions

Overview

• This chapter covers three important continuous probability distributions:
  • Uniform Distribution
  • Normal Distribution
  • Exponential Distribution

Continuous Probability Distributions

• A continuous random variable can take any value within a given interval on the real number line or within a collection of intervals.
• The probability of a continuous random variable assuming a specific value is essentially zero.
• Instead, we consider the probability of the variable falling within a certain interval.
• The probability that a random variable falls within an interval $[a, b]$ is the area under the probability density function (PDF) between $a$ and $b$.

Uniform Probability Distribution

• A random variable is uniformly distributed if the probability is proportional to the length of the interval.

• The uniform probability density function is defined as:

  f(x) = \begin{cases}
  \frac{1}{b - a} & \text{for } a \leq x \leq b \
  0 & \text{otherwise}
  \end{cases}

  where:

  • $a$ = smallest value the variable can assume
  • $b$ = largest value the variable can assume

• Expected Value of Uniform Distribution:

  $E(x) = \frac{a + b}{2}$

• Variance of Uniform Distribution:

  $Var(x) = \frac{(b - a)^2}{12}$

Example: Slater's Buffet

• Customers are charged for the amount of salad they take.
• The amount of salad taken is uniformly distributed between 5 ounces and 15 ounces.
• Here, $a = 5$ and $b = 15$.

Example Problem

• What is the probability that a customer will take between 12 and 15 ounces of salad?
• The area under the graph of $f(x)$ and probability are identical for all continuous random variables.
• To find the probability that $x$ falls between $x<em>1$ and $x</em>2$, compute the area under the graph of $f(x)$ over the interval from $x<em>1$ to $x</em>2$.

Normal Probability Distribution

• The normal probability distribution is the most important distribution for describing a continuous random variable.

• Widely used in statistical inference.

• Applications include heights of people, rainfall amounts, test scores, and scientific measurements.

• Abraham de Moivre, a French mathematician, derived the normal distribution in 1733.

• Normal Probability Density Function

  $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x - \mu}{\sigma})^2}$

  where:

  • $\mu$ = mean
  • $\sigma$ = standard deviation
  • $\pi$ ≈ 3.14159
  • $e$ ≈ 2.71828

Characteristics of Normal Distribution

• The distribution is symmetric.

• The highest point on the normal curve is at the mean, which is also the median and mode.

• The mean can be any numerical value: negative, zero, or positive.

• The standard deviation determines the width of the curve.

• The total area under the curve is 1.

• The area under the curve to the left of the mean is 0.5, and the area to the right is also 0.5.

• Empirical Rule:

  • Approximately 68.26% of the values fall within plus or minus one standard deviation of the mean ($\mu \pm 1\sigma$).
  • Approximately 95.44% of the values fall within plus or minus two standard deviations of the mean ($\mu \pm 2\sigma$).
  • Approximately 99.72% of the values fall within plus or minus three standard deviations of the mean ($\mu \pm 3\sigma$).

Standard Normal Probability Distribution

• A special case of the normal distribution where the mean is 0 and the standard deviation is 1.

  • $\mu = 0$
  • $\sigma = 1$

• The standard normal distribution is often denoted by $z$.

• Converting to the Standard Normal Distribution

  • We can convert any normal random variable $x$ to a standard normal random variable $z$ using the formula:

    $z = \frac{x - \mu}{\sigma}$

  • $z$ measures the number of standard deviations that $x$ is away from the mean $\mu$.

Example: Pep Zone

• Pep Zone sells auto parts, including motor oil.
• When the stock drops to 20 gallons, a replenishment order is placed.
• Demand during replenishment lead-time is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons.
• The manager wants to know the probability of a stockout during replenishment lead-time, i.e., what is the probability that demand during lead-time will exceed 20 gallons? P(x > 20) = ?

Solving for the Stockout Probability

• Convert $x = 20$ to $z$.

  $z = \frac{20 - 15}{6} = \frac{5}{6} ≈ 0.83$

• Find the cumulative probability for z < 0.83 from the standard normal table, which is 0.7967.

• The probability of a stockout is P(x > 20) = P(z > 0.83) = 1 - P(z < 0.83) = 1 - 0.7967 = 0.2033.

• If the manager wants the probability of a stockout to be no more than 0.05, we can use the standard normal table in reverse.

Solving for the Reorder Point

• Look up the complement of the tail area (1 - 0.05 = 0.95) in the standard normal table to find the corresponding $z$ value.

• The $z$ value corresponding to 0.95 is approximately 1.645.

• Use the formula to find the reorder point $x$:

  $x = \mu + z\sigma = 15 + 1.645(6) = 15 + 9.87 = 24.87$

• A reorder point of 25 gallons will place the probability of a stockout during lead-time at approximately 0.05.

• By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about 0.20 to 0.05.

Normal Approximation of Binomial Probabilities

• The normal distribution can be used to approximate binomial probabilities when $n$ is large enough.

• Example: Suppose that a company has a history of making errors in 10% of its invoices. A sample of 100 invoices has been taken, and we want to compute the probability that 12 invoices contain errors.

• In this case, we want to find the binomial probability of 12 successes in 100 trials. So, we set:

  • $p = 0.1$

  • $n = 100$

  • $\mu = np = 100(0.1) = 10$

  • $\sigma = \sqrt{np(1-p)} = \sqrt{100(0.1)(0.9)} = 3$

  • P(x = 12) ≈ P(11.5 < x < 12.5)

• $z_1 = \frac{11.5 - 10}{3} ≈ 0.5$

• $z_2 = \frac{12.5 - 10}{3} ≈ 0.83$

• P(z < 0.83) = 0.7967

• P(z < 0.5) = 0.6915

• Therefore, $P(x = 12) = 0.7967 - 0.6915 = 0.1052$

Exponential Probability Distribution

• Useful for describing the time it takes to complete a task.

  • Time between vehicle arrivals at a toll booth.
  • Time required to complete a questionnaire.
  • Distance between major defects in a highway.

• In waiting line applications, the exponential distribution is often used for service times.

• A key property is that the mean and standard deviation are equal.

• The distribution is skewed to the right.

• Density Function

  $f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}}$

  where:

  • $\mu$ = expected or mean
  • $e$ ≈ 2.71828
  • for x > 0

• Cumulative Probabilities

  $P(x \leq x<em>0) = 1 - e^{-\frac{x</em>0}{\mu}}$

  where: $x_0$ = some specific value of $x$

Example: Al’s Full-Service Pump

• The time between arrivals of cars at Al’s full-service gas pump follows an exponential probability distribution with a mean time between arrivals of 3 minutes.

• Al would like to know the probability that the time between two successive arrivals will be 2 minutes or less.

• The probability that the time between two successive arrivals will be 2 minutes or less is:

  $P(x \leq 2) = 1 - e^{-\frac{2}{3}} ≈ 1 - 0.5134 = 0.4866$

Relationship between the Poisson and Exponential Distributions

• The exponential distribution is related to the Poisson distribution.
• If the number of events follows a Poisson distribution, then the time between events follows an exponential distribution.