Notes on Areas Under Normal Curves, Probability Definitions, and Basic Probability

3.1 Areas Under Normal Curves

• Standard normal distribution: mean 3 bmu = 0, b = 1; written as X \u2261 N(bc, b) with bc = 0, b = 1.
• The normal curve areas correspond to population proportions: the area under the curve over an interval equals the proportion of observations in that interval.
• Rationale: the normal curve is the limit of a sequence of histograms as sample size grows and bin width goes to zero; thus areas converge to probabilities.
• Example: if a student’s arrival time in minutes for class is modeled by a standard normal, then:
  • half the time the student arrives before class starts.
  • 68.27% of the time the student is within $\pm 1$ minute of the start of class.
• Z-transformation (to connect any normal to the standard normal):
  • Define $Z = \frac{X - \mu}{\sigma}$. Then $Z \sim N(0,1)$.
  • Inverse: $X = \mu + Z \sigma$ to go from standard normal to an arbitrary normal.
  • This lets us use the standard normal table to answer questions about any normal distribution.
• Example: IQ distribution for Reggie Jackson
  • IQ given: mean $\mu = 100$, standard deviation $\sigma = 16$, and observed value $X = 140$.
  • Compute z-value: $z = \frac{X - \mu}{\sigma} = \frac{140 - 100}{16} = 2.5$.
  • Probability to the right of 140: P(X > 140) = P(Z > 2.5) = 0.006 \approx 0.6\%.
• Example: top 2% to join Mensa
  • Look up the z-value corresponding to area 0.02 in the upper tail; $z \approx 2.05$.
  • Convert to the IQ score: $X = \mu + z\sigma = 100 + (2.05)(16) = 132.8 \approx 133$.

3.2 Probability Definitions

• Two broad definitions of probability:
  • Frequentist: probability of an event is the long-run relative frequency.
  • Formal: P[A] = \lim_{n \to \infty} \frac{# \text{ times } A \text{ happens}}{n}.
  • Quotation: John Maynard Keynes: “In the long run, we are all dead.”
  • Bayesian: probability reflects personal belief; choose a value for $P[A]$ based on prior information and coherence with other beliefs; update with data via Bayes’ Rule as new information arrives.
  • Requirements for a Bayesian view: must conform to all other personal opinions and must update according to Bayes’ Rule.
• Kolmogorov’s Axioms (probability must satisfy):
  • 0 ≤ $P[A]$ ≤ 1.
  • The probability of some outcome from the sample space equals 1: $P[ ext{some possible event happens}] = 1$.
  • If A and B are incompatible (disjoint), then $P[A \cup B] = P[A] + P[B]$.
• Notes:
  • Axioms provide the foundation from which all other probability rules are derived.
  • Kolmogorov (1903–1987) is cited as a foundational figure for this axiomatization.

3.3 Drawing From A Box

• Box model for randomness requires three pieces of information:
  • What are the tickets (numbers) in the box?
  • How many draws are taken?
  • Are the draws with replacement?
• Applications posed:
  • What is the box model for 50 tosses of a (possibly unfair) coin?
  • What is the box model for drawing a random sample of 50 students from STA 111 to learn about GPA?
  • What is the box model for drawing a random sample of 50 U.S. voters to learn opinion on gun control laws?

3.4 Basic Probability

• Core definitions:

  • A and B are independent iff $P[A \cap B] = P[A] \cdot P[B]$.
  • A and B are disjoint (mutually exclusive) iff $P[A \cap B] = 0$.
  • A finite partition: events $A<em>1, \dots, A</em>n$ are incompatible and $\sum<em>{i=1}^n P[A</em>i] = 1$.

• Key rules:

  • Complement rule: $P[\text{not } A] = 1 - P[A]$.
  • Inclusive OR (union): $P[A \cup B] = P[A] + P[B] - P[A \cap B]$.
  • Conditional probability: $P[A|B] = \frac{P[A \cap B]}{P[B]}$ (provided P[B] > 0).

• Deck of cards example (A = heart, B = red):

  • Independence check: $P[A \cap B] = \tfrac{1}{4} \neq P[A]P[B] = \tfrac{1}{4} \cdot \tfrac{1}{2} = \tfrac{1}{8}$; hence A and B are not independent.
  • Disjoint? No, red cards include hearts, so A and B are not incompatible.
  • Partition of a deck: $A<em>1 = \text{clubs}, A</em>2 = \text{diamonds}, A<em>3 = \text{hearts}, A</em>4 = \text{spades}$.
  • Complement: $P[\text{not heart}] = 1 - P[\text{heart}] = 1 - \tfrac{1}{4} = \tfrac{3}{4}$.
  • Inclusive OR: $P[\text{red} \cup \text{heart}] = P[\text{red}] + P[\text{heart}] - P[\text{red} \cap \text{heart}] = \tfrac{1}{2} + \tfrac{1}{4} - \tfrac{1}{4} = \tfrac{1}{2}.$
  • Conditional: $P[\text{heart} | \text{red}] = \frac{P[\text{heart} \cap \text{red}]}{P[\text{red}]} = \frac{\tfrac{1}{4}}{\tfrac{1}{2}} = \tfrac{1}{2}.$

• DoD draft lottery example (first two draws from December):

  • Approach A (enumeration): There are $365 \times 364 = 132{,}860$ ordered pairs of dates; December pairs number $31 \times 30 = 930$; Probability = $\frac{930}{132{,}860} \approx 0.007$.
  • Approach B (conditional product): Let $A$ = second draw is December, $B$ = first draw is December. Then
    $P[A \cap B] = P[A|B]P[B] = \frac{30}{364} \cdot \frac{31}{365} \approx 0.007.$

• Card-draw problems (two-card sequences and independence):
  1) Two draws from a standard 52-card deck. Probability that the first card is a king (B) and the second card is a queen (A):
  $P[A \cap B] = P[B] \cdot P[A|B] = \frac{4}{52} \cdot \frac{4}{51} = \frac{16}{2652} \approx 0.00603.$
  2) Roll a die; probability of a six (A) given that the result is greater than or equal to 3 (B):
  $P[A|B] = \frac{P[A \cap B]}{P[B]} = \frac{\tfrac{1}{6}}{\tfrac{4}{6}} = \tfrac{1}{4} = 0.25.$

• Die roll pairs (independence reminder):

  • Roll a die twice; probability of a six on the second throw (A) given the first throw is a six (B):
    $P[A|B] = \frac{P[A \cap B]}{P[B]} = \frac{\tfrac{1}{36}}{\tfrac{1}{6}} = \tfrac{1}{6}.$
  • If $P[A|B] = P[A]$, then A and B are independent. Independence means the occurrence of one event does not affect the chance of the other.

• Additional card- or dice-based unions (examples):

  • Probability of a king or a red card in a single draw: $P[A \cup B] = P[A] + P[B] - P[A \cap B] = \tfrac{4}{52} + \tfrac{26}{52} - \tfrac{2}{52} = \tfrac{28}{52} = \tfrac{7}{13}.$
  • Probability of a king or a red card when drawing two cards without replacement can be approached via disjoint cases or via the complement, e.g. for two draws: $P[A \cup B] = 1 - P[\text{not } A \text{ and not } B] = 1 - P[\text{not } B | \text{not } A] \cdot P[\text{not } A].$
  • Practical note: hard calculations often have simplifications or tricks.

• Summary emphasis:

  • Understand the meaning of independence, disjointness, and partition in concrete card/die problems.
  • Use the fundamental rules (complement, union, conditional) to decompose complex events.
  • Recognize when to use the complement or to condition on a known event to simplify calculations.