Overview of Probability Distributions

Introduction

• Today we will review probability distributions.
  • Binomial
  • Poisson
  • Uniform
  • Normal
  • Gamma
  • Beta

Basic Definitions

• Discrete random variable: a variable that can assume only a finite or countably infinite number of distinct values.

• Probability distribution of a random variable: collection of probabilities for each value of the random variable.

• Notation:

  • Uppercase letter (e.g., Y) denotes a random variable.
  • Lowercase letter (e.g., y) denotes a particular value that the random variable may assume.
    • The specific observed value, y, is not random.

Probability Distributions for Discrete RV

• Probability function for \boldsymbol Y: sum of the the probabilities of all sample points in S that are assigned the value y

  • P[Y = y] = p(y): the probability that Y takes on the value y.

• Probability distribution for \boldsymbol Y: a formula, table, or graph that provides p(y) \ \forall \ y.

• Theorem: For any discrete probability distribution, the following must be true:

  • 0 \le p(y) \le 1 \ \forall \ y
  • \sum_y p(y) = 1 \ \forall \ p(y) > 0.

Expected Values for Discrete RV

• Expected value: Let Y be a discrete random variable with probability function p(y). Then the expected value of Y, E[Y], is defined to be

E(Y) = \sum_{y} y p(y)

• When p(y) is an accurate characterization of the population frequency distribution, then the expected value is the population mean.

E[Y] = \mu

Expected Values for Discrete RV

• Variance: if Y is a random variable with mean E[Y] = \mu, the variance of a random variable Y is defined to be the expected value of (Y-\mu)^2.

V[Y] = E\left[ (Y-\mu)^2 \right]

• If p(y) is an accurate characterization of the population frequency distribution, then V(Y) is the population variance,

V[Y] = \sigma^2

• Standard deviation: the positive square root of V[Y].

Expected Values for Discrete RV

• There is an alternative (and easier) way to calculate the variance manually,

• Theorem: Let Y be a discrete random variable with probability function p(y) and mean E[Y] = \mu. Then,

V[Y] = \sigma^2 = E\left[(Y-\mu)^2\right] = E\left[Y^2\right] - \mu^2

Binomial Probability Distribution

• Binomial experiment:

  1. The experiment consists of a fixed number, n, of identical trials.

  2. Each trial results in one of two outcomes: success (S) or failure (F).

  3. The probability of success on a single trial is equal to some value p and remains the same from trial to trial.

     • The probability of failure is equal to q = (1-p).

  4. The trials are independent.

  5. The random variable of interest is Y, the number of successes observed during the n trials.

Binomial Probability Distribution

• A random variable Y is said to have a binomial distribution based on n trials with success probability p iff

p(y) = {n \choose y}p^y q^{n-y}, \text{ where } y = 0, 1, 2, ..., n, \text{ and } 0 \le p \le1

• Let Y be a binomial random variable based on n trials and success probability p. Then

E[Y] = \mu = np \ \ \ \text{and} \ \ \ V[Y] = \sigma^2 = npq

• See Wackerly pg. 107 for derivation.

Binomial Probability Distribution

• We can use R to find information related to the binomial distribution.
  • P[X = x]: dbinom(x, size, prob)
  • P[X \le x]: pbinom(q, size, prob)
  • P[X > x]: pbinom(q, size, prob, lower.tail = FALSE)
• In the functions,
  • x or q is the value of X we are interested in
  • size is the sample size (n)
  • prob is the probability of success, \pi
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
    • FALSE returns P[X > x]

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.

• Use R to find:

  1. P[X = 2]
  2. P[X > 2]
  3. P[X < 4]

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.

• Use R to find:

  1. P[X = 2]

dbinom(x = 2, size = 4, prob = 0.5)

[1] 0.375

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.

• Use R to find:

  2. P[X > 2]

pbinom(q = 2, size = 4, prob = 0.5, lower.tail = FALSE)

[1] 0.3125

Binomial Probability Distribution

• The manufacturer of a dairy drink wishes to compare a new formula (B) with that of the standard formula (A). Each of four judges perform a blinded taste test and report which glass he or she most enjoyed. Suppose that the two formulas are equally attractive.

• Use R to find:

  3. P[X < 4] = P[X \le 3]

pbinom(q = 3, size = 4, prob = 0.5)

[1] 0.9375

Poisson Probability Distribution

• We often use the Poisson distribution to model count data.

• A random variable Y is said to have a Poisson probability distribution iff

p(y) = \frac{\lambda^y}{y!}e^{-\lambda}, \text{ where } y=0,1,2,..., \text{ and } \lambda > 0

• If Y is a random variable with a Poisson distribution with parameter \lambda, then

E[Y] = \mu = \lambda \text{ and } V[Y] = \sigma^2 = \lambda

• See Wackerly pg. 134 for derivation.

Poisson Probability Distribution

• We can use R to find information related to the Poisson distribution.
  • P[X = x]: dpois(x, lambda)
  • P[X \le x]: ppois(q, lambda)
  • P[X > x]: ppois(q, lambda, lower.tail = FALSE)
• In the functions:
  • x or q is the value of X we are interested in
  • lambda is the rate of occurrence
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
    • FALSE returns P[X > x]

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. Use R to find the following probabilities.
  1. No more than three customers arrive.
  2. At least two customers arrive.
  3. Exactly five customers arrive.

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. Use R to find the following probabilities.
  1. No more than three customers arrive.

ppois(q = 3, lambda = 7)

[1] 0.08176542

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. Use R to find the following probabilities.
  2. At least two customers arrive.

ppois(q = 1, lambda = 7, lower.tail = FALSE)

[1] 0.9927049

Poisson Probability Distribution

• Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. Use R to find the following probabilities.
  3. Exactly five customers arrive.

dpois(x = 5, lambda = 7)

[1] 0.1277167

Probability Distributions for Continuous RV

• Theorem: If a continuous random variable Y has density function f(y) and a < b, then the probability that Y falls in the interval [a, b] is

P[a \le Y \le b] = \int_a^b f(y) dy.

Expected Values for Continuous RV

• Expected value: The expected value of a continuous variable Y is

E[Y] = \int_{-\infty}^{\infty} y f(y) \ dy

• This is the continuous version of the expected value for a discrete random variable,

E[Y] = \sum_y y p(y)

Expected Values for Continuous RV

• Theorem: Let g(Y) be a function of Y; then the expected value of g(Y) is given by

E\left[ g(Y) \right] = \int_{-\infty}^{\infty} g(y) f(y) \ dy

• Theorem: Let c be a constant and let g(Y), g_1(Y), g_2(Y), …, g_k(Y) be functions of a continuous random variable, Y. Then the following results hold:
  • E[c] = c
  • E\left[cg(Y)] = cE[g(Y)\right]
  • E\left[g_1(Y)+...+g_k(Y)\right] = E\left[ g_1(Y) \right] + ... + E\left[ g_k(Y) \right]

Uniform Probability Distribution

• Uniform Distribution

Uniform Probability Distribution

• A random variable Y is said to have a uniform distribution iff

f(y) = \frac{1}{\theta_2 - \theta_1}, \ \theta_1 \le y \le \theta_2

• If \theta_1 < \theta_2 and Y is a uniformly distributed r.v. on the interval (\theta_1, \theta_2), then

E[Y] = \mu = \frac{\theta_1+\theta_2}{2} \ \ \ \text{and} \ \ \ V[Y] = \sigma^2 = \frac{(\theta_2-\theta_1)^2}{12}

• See Wackerly pg. 176 for derivation.

Uniform Probability Distribution

• We can use R to find information related to the uniform distribution:
  • P[X \le x]: punif(q, min, max)
  • P[X \ge x]: punif(q, min, max, lower.tail = FALSE)
• In the functions:
  • q is the value of X we are interested in
  • min is the lower bound of the distribution
  • max is the upper bound of the distribution
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
    • FALSE returns P[X \ge x]

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine the following probabilities:
  1. A worker takes fewer than 13 minutes.
  2. A worker takes at least 11 minutes.
  3. A worker takes between 14 and 15 minutes.

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine the following probabilities:
  1. A worker takes fewer than 13 minutes.

punif(13, 9, 15)

[1] 0.6666667

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine the following probabilities:
  2. A worker takes at least 11 minutes.

punif(11, 9, 15, lower.tail = FALSE)

[1] 0.6666667

Uniform Probability Distribution

• An industrial psychologist has determined that it takes a worker between 9 and 15 minutes to complete a task on an automobile assembly line. If the time to complete the task is uniformly distributed over the interval 9 \le y \le 15, then determine the following probabilities:
  3. A worker takes between 14 and 15 minutes.

punif(15, 9, 15) - punif(14, 9, 15)

[1] 0.1666667

Normal Probability Distribution

• Normal Distribution

Normal Probability Distribution

• A random variable Y is said to have a normal distribution iff, for \sigma > 0 and -\infty < \mu < \infty,

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

• If Y is a random variable normally distributed with parameters \mu and \sigma, then

E[Y] = \mu \ \ \ \text{and} \ \ \ V[Y] = \sigma^2

Normal Probability Distribution

• We can use R to find information related to the normal distribution.
  • P[X \le x]: pnorm(q, mean, sd)
  • P[X \ge x]: pnorm(q, mean, sd, lower.tail = FALSE)
• In the functions:
  • q is the value of X we are interested in
  • mean is the population mean \mu
  • sd is the standard deviation \sigma
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
    • FALSE returns P[X \ge x]

Normal Probability Distribution

• A random variable Y is said to have a standard normal distribution iff

Y \sim N(\mu=0,\sigma=1)

• The normal distribution is then simplified to

f(y) = \frac{1}{\sqrt{2\pi}} e^{-y^2/2}

• Note that in all cases of the normal distribution, we assume -\infty < y < \infty.

Normal Probability Distribution

• When using pnorm(), the default values for mean and sd are 0 and 1.

• Thus, if we have the standard normal our R functions simplify to:

  • P[Z \le z]: pnorm(z)
  • P[Z \ge z]: pnorm(z, lower.tail = FALSE)

• In the functions:

  • q is the z-score value of interest
  • lower.tail = TRUE returns P[Z \le z]
  • lower.tail = FALSE returns P[Z \ge z]

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm.
  1. What is the probability that a carrot will be between 10 and 13 cm?
  2. What is the probability that a carrot will be less than 9 cm?
  3. What is the probability that a carrot will be 12 cm or larger?

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm.
  1. What is the probability that a carrot will be between 10 and 13 cm?

pnorm(q = 13, mean = 11.5, sd = 1.15) - pnorm(q = 10, mean = 11.5, sd = 1.15)

[1] 0.807885

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm.
  2. What is the probability that a carrot will be less than 9 cm?

pnorm(q = 9, mean = 11.5, sd = 1.15)

[1] 0.01485583

Normal Probability Distribution

• A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5000 of these carrots, it can be said that the carrot length, Y, is normally distributed with \mu = 11.5 cm and \sigma = 1.15 cm.
  3. What is the probability that a carrot will be 12 cm or larger?

pnorm(q = 12, mean = 11.5, sd = 1.15, lower.tail = FALSE)

[1] 0.3318601

Gamma Probability Distribution

• Gamma Distribution

Gamma Probability Distribution

• A random variable Y is said to have a gamma distribution with parameters \alpha > 0 and \beta > 0 iff,

f(y) = \frac{y^{\alpha-1} e^{-y/\beta}}{\beta^{\alpha} \Gamma(\alpha)}, \ 0 \le y < \infty

• Note that \Gamma(\alpha) = \int_{0}^{\infty} y^{\alpha-1} e^{-y} \ dy.

• If Y has a gamma distribution with parameters \alpha and \beta, then

E[Y] = \mu = \alpha\beta \ \ \ \text{and} \ \ \ V[Y] = \sigma^2 = \alpha\beta^2

• See Wackerly pg. 187 for derivation.

Gamma Probability Distribution

• We can use R to find information related to the Gamma distribution.
  • P[X \le x]: pgamma(q, shape, rate)
    
  • P[X \ge x]: pgamma(q, shape, rate, lower.tail = FALSE)
• In the functions:
  • q is the value of X we are interested in
    
  • shape is the shape parameter, \alpha
    
  • scale is the scale parameter, \beta
    • Alternatively, can parameterize with rate = 1/\beta, rate = 1 / scale
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
      
    • FALSE returns P[X \ge x]

Gamma Probability Distribution

• Annual incomes for heads of household in an affluent section of a city have approximately a gamma distribution with \alpha=32 and \beta=2500.
  1. What proportion have incomes in excess of $100,000?
  2. What proportion have incomes between $75,000 and $150,000?

Gamma Probability Distribution

• Annual incomes for heads of household in a section of a city have approximately a gamma distribution with \alpha=32 and \beta=2500.
  1. What proportion have incomes in excess of $100,000?

pgamma(q = 100000, shape = 32, scale = 2500, lower.tail = FALSE)

[1] 0.08552057

Gamma Probability Distribution

• Annual incomes for heads of household in an affluent section of a city have approximately a gamma distribution with \alpha=32 and \beta=2500.

  2. What proportion have incomes between $75,000 and $150,000?
     

pgamma(q = 150000, shape = 32, scale = 2500) - pgamma(q = 75000, shape = 32, scale = 2500)

[1] 0.6186147

Beta Probability Distribution

• Beta Distribution

Beta Probability Distribution

• A random variable Y is said to have a beta distribution with parameters \alpha > 0 and \beta > 0 iff,

f(y) = \frac{y^{\alpha-1}(1-y)^{\beta-1}}{B(\alpha,\beta)}, \ 0 \le y \le 1

• Note: B(\alpha,\beta) = \int_0^1 y^{\alpha-1}(1-y)^{\beta-1} \ dy = \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}.

• If Y has a beta distribution with parameters \alpha > 0 and \beta > 0, then

E[Y] = \mu = \frac{\alpha}{\alpha+\beta} \ \ \ \text{and} \ \ \ V[Y] = \sigma^2 = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}

Beta Probability Distribution

• We can use R to find information related to the Beta distribution.
  • P[X \le x]: pbeta(q, shape1, shape2)
    
  • P[X \ge x]: pbeta(q, shape1, shape2, lower.tail = FALSE)
• In the functions:
  • q is the value of X we are interested in – must be in [0, 1]!
  • shape1 is the first shape parameter, \alpha
    
  • shape2 is the second shape parameter, \beta
    
  • lower.tail has two options:
    • TRUE (default) returns P[X \le x]
      
    • FALSE returns P[X \ge x]

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. What is the probability that:
  1. fewer than 60% of respondents like the new flavor?
  2. more than 90% of respondents like the new flavor?
  3. somewhere between 70% and 90% of respondents like the new flavor?

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. What is the probability that:
  1. fewer than 60% of respondents like the new flavor?

pbeta(q = 0.6, shape1 = 8, shape2 = 2)

[1] 0.07054387

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. What is the probability that:
  2. more than 90% of respondents like the new flavor?

pbeta(q = 0.9, shape1 = 8, shape2 = 2, lower.tail = FALSE)

[1] 0.225159

Beta Probability Distribution

• In a survey of cupcake preferences, 8 respondents liked the new cupcake flavor and 2 did not. We will model the proportion of all respondents who would like the cupcake flavor using a Beta distribution with \alpha = 8 and \beta = 2. What is the probability that:
  3. somewhere between 70% and 90% of respondents like the new flavor?

pbeta(q = 0.9, shape1 = 8, shape2 = 2) - pbeta(q = 0.7, shape1 = 8, shape2 = 2)

[1] 0.5788377

Wrap Up

• We reviewed basic probability rules and distributions today.

  • We are setting up for our next lecture(s).

• Wednesday: Thinking Like a Bayesian