[SOLVED] STAT823 Homework 5-Probability Distributions

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Directions

Use the attached RMarkdown Probability Distributions.Rmd to create a pdf report summarizing the common distributions discussed in this lesson. In addition, I have done the first few to show you my expectations for your report, and have given you some leading questions to complete. Partial R-code for the following distributions have been given for you.

  1. Bernoulli
  2. Binomial
  3. Hypergeometric: Qn 1
  4. Poisson: Qn 2
  5. Geometric: Qn 3
  6. Negative Binomial: Qn 4
  7. Normal: Qn 5
  8. Exponential: Qn 6
  9. Chi-square: Qn 7
  10. Student’s t: Qn 8
  11. F: Qn 9 Optional Beta Optional
  12. Logistic Optional

All R-code and output must be clearly shown. Late submission will attract a penalty of 10 points per day after the due date.

If you have any questions, please post them on the lesson discussion board.

1          Discrete Distributions

1.1       Bernoulli

The Bernoulli distribution, named for Jacob Bernoulli, assigns probability to the outcomes of a single Bernoulli experiment—one where the only possible outcomes can be thought of as a “success” or a “failure”

(e.g., a coin toss). Here, the random variable x can take on the values 1 (success) with probability p, or 0

(failure) with probability q = 1≠ p. The plot below contains the pmf of two Bernoulli distributions. The firstp = 0.2 and the second (in black) has a probability of success p = 0.5.

(in gray) has a probability of success

x <- 0:1

plot(x, dbinom(x, 1, 0.2), type = “h”, ylab = “f(x)”, ylim = c(0, 1), lwd = 8, col = “darkgray”, main = “Bernoulli(0.2)”)

lines(x, dbinom(x, 1, 0.5), type = “h”, lwd = 2, col = “black”)

legend(0.7, 1, c(“Bernoulli(0.2)”, “Bernoulli(0.5)”), col = c(“darkgray”, “black”), lwd = c(8, 2))

Bernoulli(0.2)

x

The Bernoulli experiment forms the foundation for many of the next discrete distributions.

1.2       Binomial

The binomial distribution applies when we perform n Bernoulli experiments and are interested in the

total number of “successes” observed. The outcome here, y = x , where P(x = 1) = p and p = 0.5; in blue,

Pgray,(xi = 0) = 1≠ p. The plot below displays three binomial distributions, all forp = 0.1; and in green, p = 0.9.          q i                      in = 10 Bernoulli trials: in

x <- seq(0, 10, 1)

plot(x, dbinom(x, 10, 0.5), type = “h”, ylab = “f(x)”, lwd = 8, col = “dark gray”, ylim = c(0,

0.5), main = “Binomial(10, 0.5) pmf”) lines(x, dbinom(x, 10, 0.1), type = “h”, lwd = 2, col = “blue”)

lines(x, dbinom(x, 10, 0.9), type = “h”, lwd = 2, col = “green”)

legend(3, 0.5, c(“Binomial(10,0.1)”, “Binomial(10,0.5)”,

“Binomial(10,0.9)”), col = c(“blue”,

“dark gray”, “green”), lwd = c(2, 8,

2))

Binomial(10, 0.5) pmf

x

We can see the shifting of probability from low values for p = 0.1 to high values for p = 0.9. This makes sense, as it becomes more likely with p = 0.9 to observe a success for an individual trial. Thus, in 10 trials, more successes (e.g., 8, 9, or 10) are likely. For p = 0.5, the number of successes are likely to be around 5 (e.g., half of the 10 trials).

1.3       Hypergeometric

In the example I have below, I have set the number of balls in the urn to 10, 5 of which are white and 5 of which are black. I have also fixed the number of balls drawn from the urn to 5. Play around with the parameters and describe what you see.

x <- seq(0, 10, 1)

plot(x, dhyper(x, 5, 5, 5), type = “h”, ylab = “f(x)”,

lwd = 2, main = “Hypergeometric(5,10,5) pmf”5 , 5 ,5          )

 

Hypergeometric(5,10,5) pmf5, 5, 5

x

1.4      Poisson

What happens if you increase ⁄? To 2? To 3?

x <- seq(0, 5, 1)

plot(x, dpois(x, 1), type = “h”, ylab = “f(x)”, main = “Poisson(1) pmf”, lwd = 2)

Poisson(1) pmf

x

1.5       Geometric

What happens to the geometric distrbution if you vary p? Show me a few plots and explain.

x <- seq(0, 20, 1)

plot(x, dgeom(x, 0.2), type = “h”, ylab = “f(x)”, lwd = 2, main = “Geometric(0.2) pmf”)

Geometric(0.2) pmf

x

1.6        Negative Binomial

The negative binomial I have below has set r = 1, so it’s identical to the geometric above. Play around with r and see how it changes.

x <- seq(0, 20, 1)

plot(x, dnbinom(x, 1, 0.2), type = “h”, ylab = “f(x)”, lwd = 2, main = “Negative Binomial(0.2) pmf”)

 

Negative Binomial(0.2) pmf

x

2          Continuous Distributions

2.1       Exponential

Vary ⁄ and describe.

x <- seq(0, 10, 0.01)

plot(x, dexp(x, 1), type = “l”, ylab = “f(x)”, lwd = 2, main = “Exponential(1) pdf”)

Exponential(1) pdf

x

2.2      Normal

Vary ‡ and see how the distribution changes. If you make it too big, you may need to adjust the x-axis by

making the sequence span a wider range thanthe proper limits for x for a given ≠5 to 5. You can use a trial-and-error approach to determing ‡.

x <- seq(–5, 5, 0.01)

plot(x, dnorm(x, 0, 1), type = “l”, ylab = “f(x)”, main = “Normal(0, 1) pdf”)

Normal(0, 1) pdf

x

2.3       Chisquare

How do the degrees of freedom change the shape? Plot a few and explain.

x <- seq(0, 20, 0.01)

plot(x, dchisq(x, 6), type = “l”, ylab = “f(x)”, main = “Chi-square(6) pdf”)

 

Chi−square(6) pdf

x

2.4       Students t

How do the degrees of freedom change the shape? Plot a few and explain.

x <- seq(–5, 5, 0.01)

plot(x, dt(x, 6), type = “l”, ylab = “f(x)”, main = “Student s t(6) pdf”)

Student’s t(6) pdf

x

2.5      F

How do the degrees of freedom (numerator and/or denominator) change the shape? Plot a few and explain.

x <- seq(0, 6, 0.01)

plot(x, df(x, 12, 15), type = “l”, ylab = “f(x)”,

main = “F(2, 5) pdf”F(12, 15)  pdf )

F(12, 15) pdfF(2, 5) pdf

x

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