Lesson 9: Named Distributions

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Discrete Distribution Reminders

Probability Mass Function

\[ P(X=x) = \begin{cases} \dfrac{1}{6}, & x=1, \\[6pt] \dfrac{1}{6}, & x=2, \\[6pt] \dfrac{1}{6}, & x=3, \\[6pt] \dfrac{1}{6}, & x=4, \\[6pt] \dfrac{1}{6}, & x=5, \\[6pt] \dfrac{1}{6}, & x=6, \\[6pt] 0, & \text{otherwise}. \end{cases} \]

Cumulative Distribution Function

\[ F_X(x)=P(X\le x)= \begin{cases} 0, & x<1, \\[6pt] \dfrac{1}{6}, & 1\le x<2, \\[6pt] \dfrac{2}{6}, & 2\le x<3, \\[6pt] \dfrac{3}{6}, & 3\le x<4, \\[6pt] \dfrac{4}{6}, & 4\le x<5, \\[6pt] \dfrac{5}{6}, & 5\le x<6, \\[6pt] 1, & x\ge 6~. \end{cases} \]

\(P(X=2)\)

\[ P(X=2) = \dfrac{1}{6} \]

\(P(X \le 4)\)

\[ P(X \le 4) = P(X=1)+P(X=2)+P(X=3)+P(X=4) = \dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6} = \dfrac{4}{6} = \dfrac{2}{3} \] This is an unnamed distribution.

A named distribution is a probability distribution with a standard formula and notation that’s been given a name because it occurs so often in applications.

The Binomial Distribution

Suppose a basketball player makes a free throw with probability \(p=0.7\) (and misses with probability \(0.3\)).

If they shoot 4 times, what is the probability of exactly 2 made shots?

  • One sequence: Make, Make, Miss, Miss.
    Probability \(= 0.7 \times 0.7 \times 0.3 \times 0.3\).

  • But there are many possible sequences with 2 makes and 2 misses.
    The number of such sequences is
    \[ \binom{4}{2} = 6. \]

So
\[ P(\text{2 makes in 4 shots}) = \binom{4}{2}(0.7^2)(0.3^2). \]

General Pattern → The Binomial Formula

If \(X\) is the number of successes in \(n\) independent trials, each with success probability \(p\), then

\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}, \quad x=0,1,\dots,n. \]

We write this as
\[ X \sim \text{Binomial}(n,p). \]

Visualizing the Example

Figure 1: Binomial(4, 0.7): distribution of made shots in 4 attempts

Expected Value, Variance, and Standard Deviation

For \(X \sim \mathrm{Binomial}(n,p)\):
- \(\mathbb{E}[X] = np\)
- \(\mathrm{Var}(X) = np(1-p)\)
- \(\mathrm{SD}(X) = \sqrt{np(1-p)}\)

For \(n=4\), \(p=0.7\):
- \(\mathbb{E}[X] = 4(0.7) = 2.8\)
- \(\mathrm{Var}(X) = 4(0.7)(0.3) = 0.84\)
- \(\mathrm{SD}(X) = \sqrt{0.84} \approx 0.917\)

\[ P(X = x) = \binom{n}{x} p^x (1-p)^{\,n-x}, \quad x=0,1,\dots,n. \]

Examples with \(n=4\), \(p=0.7\)

  1. \(P(X=2)\)
    \[ \binom{4}{2}(0.7)^2(0.3)^2 \]
dbinom(2, size = 4, prob = .7)
[1] 0.2646
  1. \(P(X \le 2)\)
    \[ \sum_{x=0}^{2} \binom{4}{x}(0.7)^x(0.3)^{4-x} \]
pbinom(2, size = 4, prob = .7)
[1] 0.3483
  1. \(P(X > 2)\) \[ 1 - P(X \le 2) \]
1 - pbinom(2, size = 4, prob = .7)
[1] 0.6517

The Geometric Distribution

Suppose a basketball player makes a free throw with probability \(p=0.7\) (and misses with probability \(0.3\)).

Let \(X\) be the number of failures before the first success. Then:
- \(X=0\): success on the first shot.
- \(X=2\): two misses, then a make on the 3rd shot.

For example, the probability of \(X=2\) is \[ P(X=2) = (1-p)^2\,p = (0.3)^2(0.7). \]

General Pattern → The Geometric Formula

If \(X\) is the number of failures before the first success in independent trials, then \[ P(X = x) = (1-p)^x\,p, \quad x=0,1,2,\dots \]

We write this as \[ X \sim \mathrm{Geometric}(p). \]

Visualizing the Example

Figure 2: Geometric(\(p=0.7\)): probability of \(x\) failures before first success

Expected Value, Variance, and Standard Deviation

For \(X \sim \mathrm{Geometric}(p)\):
- \(\mathbb{E}[X] = \dfrac{1-p}{p}\)
- \(\mathrm{Var}(X) = \dfrac{1-p}{p^2}\)
- \(\mathrm{SD}(X) = \sqrt{\dfrac{1-p}{p^2}}\)

For \(p=0.7\):
- \(\mathbb{E}[X] = \dfrac{0.3}{0.7} \approx 0.429\)
- \(\mathrm{Var}(X) = \dfrac{0.3}{0.49} \approx 0.612\)
- \(\mathrm{SD}(X) = \sqrt{0.612} \approx 0.782\)

Practice Probabilities with \(p=0.7\)

  1. \(P(X=2)\) \[ (1-p)^2\,p \]
dgeom(2, prob = 0.7)
[1] 0.063
  1. \(P(X \le 2)\) \[ \sum_{x=0}^{2} (1-p)^x\,p \]
pgeom(2, prob = 0.7)
[1] 0.973
  1. \(P(X > 2)\) \[ 1 - P(X \le 2) = (1-p)^3 \]
1 - pgeom(2, prob = 0.7)
[1] 0.027

Continuous Distribution Reminder

Probability Density Function (PDF)

\[ f(x) = \begin{cases} 2(1-x), & 0 \le x \le 1, \\ 0, & \text{otherwise}. \end{cases} \]

Cumulative Distribution Function (CDF)

\[ F_X(x) = \begin{cases} 0, & x < 0, \\[6pt] 2x - x^2, & 0 \le x \le 1, \\[6pt] 1, & x > 1~. \end{cases} \]

Example Probabilities

\(P(X \le 0.5)\)

\[ F_X(0.5) = 2(0.5) - (0.5)^2 = 1 - 0.25 = 0.75 \]

\(P(0.25 \le X \le 0.75)\)

\[ F_X(0.75) - F_X(0.25) = (2(0.75) - 0.75^2) - (2(0.25) - 0.25^2) = 0.9375 - 0.4375 = 0.5 \]

The Normal Distribution

Suppose the heights of adult men in a population are approximately Normal with mean \(\mu = 68\) inches and standard deviation \(\sigma = 3\) inches.

What is the probability that a randomly chosen man is taller than 72 inches?

We compute this directly from the Normal distribution:
\[ P(X > 72) = \int_{72}^{\infty} \frac{1}{3\sqrt{2\pi}} \exp\!\left(-\frac{(x-68)^2}{18}\right)\,dx. \]

General Pattern → The Normal Formula

If \(X\) is a continuous random variable with mean \(\mu\) and variance \(\sigma^2\), then

\[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right), \quad -\infty < x < \infty. \]

We write this as
\[ X \sim \text{Normal}(\mu,\sigma^2). \]

Visualizing the Example

Figure 3: Normal(68, 3): density with shading for P(X > 72)

Expected Value, Variance, and Standard Deviation

For \(X \sim \mathrm{Normal}(\mu,\sigma^2)\):
- \(\mathbb{E}[X] = \mu\)
- \(\mathrm{Var}(X) = \sigma^2\)
- \(\mathrm{SD}(X) = \sigma\)

For \(\mu=68\), \(\sigma=3\):
- \(\mathbb{E}[X] = 68\)
- \(\mathrm{Var}(X) = 9\)
- \(\mathrm{SD}(X) = 3\)

Practice Probabilities with \(\mu=68\), \(\sigma=3\)

  1. \(P(X=70)\)
    \[ P(X=70) = 0 \quad \text{(point probability in a continuous distribution is zero)} \]
  1. \(P(X \le 70)\)
    \[ P(X \le 70) = \int_{-\infty}^{70} \frac{1}{3\sqrt{2\pi}} \exp\!\left(-\frac{(x-68)^2}{18}\right)\,dx \]
pnorm(70, mean = 68, sd = 3)
[1] 0.7475075
  1. \(P(X > 72)\)
    \[ P(X > 72) = \int_{72}^{\infty} \frac{1}{3\sqrt{2\pi}} \exp\!\left(-\frac{(x-68)^2}{18}\right)\,dx \]
1 - pnorm(72, mean = 68, sd = 3)
[1] 0.09121122
  1. \(P(65 \le X \le 70)\)
    \[ P(65 \le X \le 70) = \int_{65}^{70} \frac{1}{3\sqrt{2\pi}} \exp\!\left(-\frac{(x-68)^2}{18}\right)\,dx \]
pnorm(70, mean = 68, sd = 3) - pnorm(65, mean = 68, sd = 3)
[1] 0.5888522

Practice Problems

Practice 1

A soldier hits a target with probability \(p=0.6\) on each shot, independently of other shots. Suppose they fire \(n=5\) times. Let \(X\) be the number of hits.

  1. Write the distribution of \(X\).
  2. Find \(\mathbb{E}[X]\), \(\mathrm{Var}(X)\), and \(\mathrm{SD}(X)\).
  3. Compute:
    • \(P(X=3)\)
    • \(P(X \le 2)\)
    • \(P(X < 2)\)
    • \(P(X > 4)\)
    • \(P(X \ge 4)\)

  1. \(X \sim \mathrm{Binomial}(n=5, p=0.6)\)

  2. \(\mathbb{E}[X] = np = 5(0.6) = 3\)
    \(\mathrm{Var}(X) = np(1-p) = 5(0.6)(0.4) = 1.2\)
    \(\mathrm{SD}(X) = \sqrt{1.2} \approx 1.095\)

  3. Probabilities:

dbinom(3, size=5, prob=0.6)        # P(X=3)
[1] 0.3456
pbinom(2, size=5, prob=0.6)        # P(X ≤ 2)
[1] 0.31744
pbinom(1, size=5, prob=0.6)        # P(X ≤ 2)
[1] 0.08704
1 - pbinom(4, size=5, prob=0.6)    # P(X > 4)
[1] 0.07776
1 - pbinom(3, size=5, prob=0.6)    # P(X > 4)
[1] 0.33696

Practice 2

A cadet passes a marksmanship test with probability \(p=0.4\) on each attempt, independently of other attempts. Let \(Y\) be the number of failures before the first pass.

  1. Write the distribution of \(Y\).
  2. Find \(\mathbb{E}[Y]\), \(\mathrm{Var}(Y)\), and \(\mathrm{SD}(Y)\).
  3. Compute:
    • \(P(Y=2)\)
    • \(P(Y \le 3)\)
    • \(P(Y > 4)\)

  1. \(Y \sim \mathrm{Geometric}(p=0.4)\), where \(Y =\) failures before first success

  2. \(\mathbb{E}[Y] = \dfrac{1-p}{p} = \dfrac{0.6}{0.4} = 1.5\)
    \(\mathrm{Var}(Y) = \dfrac{1-p}{p^2} = \dfrac{0.6}{0.16} = 3.75\)
    \(\mathrm{SD}(Y) = \sqrt{3.75} \approx 1.936\)

  3. Probabilities:

dgeom(2, prob=0.4)        # P(Y=2)
[1] 0.144
pgeom(3, prob=0.4)        # P(Y ≤ 3)
[1] 0.8704
1 - pgeom(4, prob=0.4)    # P(Y > 4)
[1] 0.07776

Practice 3

Suppose the times (in minutes) to complete a training run are approximately Normal with mean \(\mu=50\) and standard deviation \(\sigma=8.\) Let \(Z\) be the time for a randomly chosen runner.

  1. Write the distribution of \(Z\).
  2. Find \(\mathbb{E}[Z]\), \(\mathrm{Var}(Z)\), and \(\mathrm{SD}(Z)\).
  3. Compute:
    • \(P(Z \le 45)\)
    • \(P(Z > 60)\)
    • \(P(40 \le Z \le 55)\)

  1. \(Z \sim \mathrm{Normal}(\mu=50, \sigma^2=64)\)

  2. \(\mathbb{E}[Z] = \mu = 50\)
    \(\mathrm{Var}(Z) = \sigma^2 = 64\)
    \(\mathrm{SD}(Z) = \sigma = 8\)

  3. Probabilities:

pnorm(45, mean=50, sd=8)                         # P(Z ≤ 45)
[1] 0.2659855
1 - pnorm(60, mean=50, sd=8)                     # P(Z > 60)
[1] 0.1056498
pnorm(55, mean=50, sd=8) - pnorm(40, mean=50, sd=8)   # P(40 ≤ Z ≤ 55)
[1] 0.6283647

Problem 4

Let the random variable \(X\) denote the time (in minutes) a patient waits at a doctor’s office before being called.

\[ f_X(x)= \begin{cases} \dfrac{1}{2}, & 2 \le x < 3, \\[6pt] \dfrac{x}{16}, & 3 \le x \le 5, \\[6pt] 0, & \text{otherwise}. \end{cases} \]

  1. Is \(X\) a discrete or continuous random variable? Sketch the PDF and label the axes.
  2. Find the cumulative distribution function (CDF), \(F_X(x)\), in curly-brace notation.
  3. Compute \(P(X \le 2.6)\).
  4. Compute \(P(2.5 \le X \le 4)\).
  5. Verify that this is a valid probability distribution.
  1. Verify Valid PDF \[ \int_{2}^{3}\frac{1}{2}\,dx + \int_{3}^{5}\frac{x}{16}\,dx = \frac{1}{2} + \frac{1}{16}\cdot \frac{25-9}{2} = \frac{1}{2} + \frac{16}{32} = \frac{1}{2} + \frac{1}{2} = 1~. \]
  1. \(X\) is continuous. The PDF is flat at height \(1/2\) on \([2,3)\), then jumps down to \(f(3)=3/16\) and increases linearly to \(f(5)=5/16\).

  1. From PDF to CDF

We want the cumulative distribution function (CDF):

\[ F_X(x) = P(X \leq x) = \int_{-\infty}^{x} f_X(t)\,dt \]

where the PDF is

\[ f_X(t) = \begin{cases} \dfrac{1}{2}, & 2 \le t < 3, \\[6pt] \dfrac{t}{16}, & 3 \le t \le 5, \\[6pt] 0, & \text{otherwise}. \end{cases} \]

Case 1: \(x < 2\)

No probability has accumulated yet:

\[ F_X(x) = 0. \]

Case 2: \(2 \le x < 3\)

We integrate from 2 up to \(x\):

\[ F_X(x) = \int_{2}^{x} \frac{1}{2}\,dt = \left[\frac{t}{2}\right]_{t=2}^{t=x} = \frac{x}{2} - \frac{2}{2} = \frac{x}{2} - 1. \]

Case 3: \(3 \le x \le 5\)

We add all mass up to 3 plus the integral from 3 to \(x\):

\[ F_X(x) = \int_{2}^{3} \frac{1}{2}\,dt \;+\; \int_{3}^{x} \frac{t}{16}\,dt. \]

First part:

\[ \int_{2}^{3} \frac{1}{2}\,dt = \left[\frac{t}{2}\right]_{2}^{3} = \frac{3}{2} - \frac{2}{2} = 0.5. \]

Second part (leaving the constant inside):

\[ \int_{3}^{x} \frac{t}{16}\,dt = \left[\frac{t^2}{32}\right]_{t=3}^{t=x} = \frac{x^2}{32} - \frac{9}{32} = \frac{x^2 - 9}{32}. \]

So together:

\[ F_X(x) = 0.5 + \frac{x^2 - 9}{32} = \frac{16}{32} + \frac{x^2 - 9}{32} = \frac{x^2 + 7}{32}. \]

Case 4: \(x > 5\)

All probability mass is included:

\[ F_X(x) = 1. \]

Final CDF

\[ F_X(x) = \begin{cases} 0, & x < 2, \\[6pt] \dfrac{x}{2} - 1, & 2 \le x < 3, \\[6pt] \dfrac{x^2 + 7}{32}, & 3 \le x \le 5, \\[6pt] 1, & x > 5~. \end{cases} \]

  1. Since \(2.6 \in [2,3)\), \[ P(X \le 2.6)=F_X(2.6)=\frac{2.6-2}{2}=0.3. \]
  1. Split at \(3\): \[ \begin{aligned} P(2.5 \le X \le 4) &= \int_{2.5}^{3} \frac{1}{2}\,dx + \int_{3}^{4} \frac{x}{16}\,dx \\[6pt] &= \frac{1}{2}(0.5) + \frac{1}{16}\cdot \frac{4^2-3^2}{2} \\[6pt] &= 0.25 + \frac{1}{16}\cdot \frac{16-9}{2} = 0.25 + \frac{1}{16}\cdot \frac{7}{2} = 0.25 + \frac{7}{32} = \frac{8}{32} + \frac{7}{32} = \frac{15}{32} \approx 0.46875~. \end{aligned} \]

Lesson 10

Resources:

  1. Computer with blank script in RStudio
  2. Course Guide
  3. Tidyverse Tutorial
  4. One Page (Front and Back) of personally handwritten notes
  5. Issued Calculator

Outline

Probability Rules

  • Venn Diagrams
  • Definitions of Union/Intersection
  • Addition Rule
  • Complement Rule
  • Conditional Probability
  • Multiplication Rule
  • Law of Total Probability
  • Bayes’ Theorem
  • Tree Diagrams
  • Rules of Independence and Mutually Exclusive Events
  • Take Scenarios and Answer Questions Related to Probability

Distributions

  • Identify appropriate distribution (named or unnamed)
  • Go from PDF to CDF for unnamed distributions
  • Find probabilities
  • Find expected values
  • Find variances
  • Linear Transformation Rules (Adding/Subtracting Expected Values and Variances)

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