Discrete Probability Distributions and Normal Distribution Review

Exam Structure and Assessment Guidelines

• Point Distribution and Weighting:

  • Discrete probability distribution questions account for approximately $10$ points on the exam.
  • The total exam value is $80$ points, meaning discrete probability represents $10$ out of the $80$ total points ($12.5\%$ of the examination).
  • The discrete section may be presented as a single comprehensive multi-part problem or split across two separate questions.

• Examination Composition:

  • The examination contains four multiple-choice questions:
  • Two questions focusing on histograms.
  • One question on response and explanatory variables.
  • One question covering an additional fundamental statistical concept.
  • The remainder of the exam consists of structured free-response problems formatted identically to homework assignments stacked together.
  • Covered chapters on the exam: Chapters $1$, $2$, $3$, $4$, $5$, $8$, $9$, and $12$.
  • Chapter $3$ and Chapter $12$ material is assessed through a five-part question structure:
  • Parts (a), (b), and (c): Finding proportions or probabilities under a normal distribution.
  • Parts (d) and (e): "Backwards problems" solving for unknown observation values ($x$) given a percentile or probability.

• Course Logistics and Review Session:

  • Homework assignments covering discrete probability are scheduled for submission on Tuesday.
  • A formal review session is conducted on Tuesday with an open question-and-answer format utilizing authentic problems from previous exams.
  • The exam is administered exactly one week following the lecture.

• Partial Credit Protocol:

  • All intermediate values and specific numbers being summed must be written explicitly in the solution.
  • Stating an isolated numerical answer without the accompanying arithmetic setup precludes the award of partial credit in the event of an addition error.

Discrete Probability Distributions

• Fundamental Principles:

  • Discrete finite probability distributions consist of a fixed, countable, and limited number of possible outcomes.
  • To calculate the probability of any event, identify all individual outcomes that comprise the event and sum their corresponding probabilities.
  • The individual outcomes in these distributions are mutually exclusive (disjoint) and non-overlapping.
  • The sum of all probabilities across the entire sample space must equal exactly $1$.

• High School Foreign Language Enrollment Problem:

  • Scenario: A student in a United States high school is chosen at random and surveyed regarding foreign language study.
  • Data Definition and Structure:
  • Spanish indicates the student studies English and Spanish.
  • French indicates the student studies English and French.
  • German indicates the student studies English and German.
  • All Other indicates the student studies English alongside another language (such as Swahili or any other non-listed language).
  • English indicates the student studies solely English and no secondary language.
  • Given Probabilities:
  • Probability of studying solely English: $P(\text{English}) = 0.57$.
  • Combined probability of foreign language study: $0.43$.
  • Part (a) — Probability of Studying a Language Other Than English:
  • Method 1 (Direct Addition): Sum all non-English language categories:       $P(\text{Other than English}) = P(\text{Spanish}) + P(\text{French}) + P(\text{German}) + P(\text{All Other}) = 0.43$
  • Method 2 (Complement Rule): Subtract the probability of the complementary event (studying English only) from $1$:       $P(\text{Other than English}) = 1 - P(\text{English}) = 1 - 0.57 = 0.43$
  • Part (b) — Probability of Studying French, German, or Spanish:
  • Because the language categories are disjoint events, calculate the union by summing the discrete probabilities directly:       $P(\text{French or German or Spanish}) = P(\text{French}) + P(\text{German}) + P(\text{Spanish}) = 0.40$

• Motor Vehicle Color Distribution Problem:

  • Scenario: A newly sold motor vehicle is chosen at random and its exterior paint color is recorded.
  • Listed Categories: White, black, silver, gray, red, brown, blue.
  • Part (a) — Evaluating the Need for an "Other" Category:
  • Sum the probabilities of all explicitly listed vehicle colors:       $\sum P(\text{Listed}) = 0.98$
  • Determine the remaining probability required to satisfy the total probability axiom (summing to $1$):       $P(\text{Other}) = 1 - \sum P(\text{Listed}) = 1 - 0.98 = 0.02$
  • An "Other" category is required and carries a probability of $0.02$.
  • Part (b) — Probability of Selecting Neither White Nor Silver:
  • Event Definition: The vehicle is "not white nor silver" (neither white nor silver).
  • Method 1 (Complement Rule):       $P(\text{Neither White nor Silver}) = 1 - (P(\text{White}) + P(\text{Silver})) = 1 - 0.53 = 0.47$(Note: where the combined probability of white and silver is $0.35 + 0.18 = 0.53$)
  • Method 2 (Direct Addition): Sum all remaining disjoint color outcomes:       $P(\text{Neither White nor Silver}) = P(\text{Black}) + P(\text{Gray}) + P(\text{Red}) + P(\text{Brown}) + P(\text{Blue}) + P(\text{Other}) = 0.47$

• Undergraduate Essay Error Analysis Problem:

  • Scenario: Spell-checking software is applied to assess errors in a $250$-word essay composed by undergraduate students.
  • Random Variable: Let $X$ denote the count of spelling errors caught by software, representing a finite discrete random variable.
  • Quantitative Literacy of Directional Inequalities:
  • "At least $1$" ($X \ge 1$): Indicates that $1$ is the minimum boundary value; the outcome must be $1$ or larger.
  • "At most $2$" ($X \le 2$): Indicates that $2$ is the maximum boundary value; the outcome must be $2$ or smaller.
  • Part (a) — Probability of At Least One Error ($X \ge 1$):
  • Sum all probabilities for error counts $X = 1, 2, 3, 4$:       $P(X \ge 1) = 0.2 + 0.3 + 0.3 + 0.1 = 0.9$
  • Equivalent Complement Rule:       $P(X \ge 1) = 1 - P(X = 0) = 1 - 0.1 = 0.9$
  • Part (b) — Probability of At Most Two Errors ($X \le 2$):
  • Sum all probabilities where $2$ is the greatest possible value ($X = 0, 1, 2$).

Continuous Distributions and Normal Distribution Review

• Conceptual Equivalences across Course Modules:

  • In Chapter $3$, the metric is designated as a proportion.
  • In Chapter $12$, the identical metric is designated as a probability.
  • In Chapter $17$, the identical metric is designated as a $p$-value.
  • Proportions, probabilities, and percentages quantify the identical mathematical area under a probability density curve.

• Standardization via $z$-Score Transformation:

  • Formula for converting an observation $x$ from a normal distribution $N(\mu, \sigma)$ to standard units:     $z = \frac{x - \mu}{\sigma}$
  • Calculator entry best practice: Enclose the numerator in explicit parentheses: (x - mean) / standard_deviation.
  • Rounding requirement: All $z$-scores must strictly be rounded to two decimal places ($0.01$) for compatibility with Table A.

• Table A Cumulative Probability Rules:

  • Lower-tail Area ("Less than", $P(X < x)$):     $\text{Probability} = \text{Table A value for } z$
  • Upper-tail Area ("Greater than", $P(X > x)$):     $\text{Probability} = 1 - (\text{Table A value for } z)$
  • Interval Area ("Between", $P(a < X < b)$):     $\text{Probability} = (\text{Table A value for larger } z) - (\text{Table A value for smaller } z)$

• Normal Distribution Application — Candidate Voting Proportions:

  • Distribution Parameters:
  • Proportion planning to vote for the candidate: $X \sim N(\mu = 0.51, \sigma = 0.012)$
  • Part (a) — Finding $P(X < 0.475)$:
  • Compute the $z$-score:       $z = \frac{0.475 - 0.51}{0.012} = \frac{-0.035}{0.012} = -2.92$
  • Table A lookup procedure:
    • Navigate to row $-2.9$.
    • Navigate to column $0.02$.
    • Table intersection value: $0.0018$.
  • Final calculation:       $P(X < 0.475) = P(z < -2.92) = 0.0018$
  • Part (b) — Finding $P(X > 0.53)$:
  • Compute the $z$-score:       $z = \frac{0.53 - 0.51}{0.012} = \frac{0.02}{0.012} = 1.67$
  • Table A lookup procedure:
    • Navigate to row $1.6$.
    • Navigate to column $0.07$.
    • Table intersection value: $0.9525$.
  • Apply the complement rule for upper-tail probabilities:       $P(X > 0.53) = 1 - P(z < 1.67) = 1 - 0.9525 = 0.0475$
  • Part (c) — Finding $P(0.48 < X < 0.52)$:
  • Compute the lower-bound $z$-score ($z_1$):       $z_1 = \frac{0.48 - 0.51}{0.012} = \frac{-0.03}{0.012} = -2.50$
  • Compute the upper-bound $z$-score ($z_2$):       $z_2 = \frac{0.52 - 0.51}{0.012} = \frac{0.01}{0.012} = 0.83$
  • Table A lookups:
    • Larger $z$ ($z_2 = 0.83$): Row $0.8$, column $0.03$ yields $0.7967$.
    • Smaller $z$ ($z_1 = -2.50$): Row $-2.5$, column $0.00$ yields $0.0062$.
  • Subtract the lower cumulative probability from the upper cumulative probability:       $P(0.48 < X < 0.52) = 0.7967 - 0.0062 = 0.7905$

• Inverse Normal Computations ("Backwards Problems"):

  • Evaluated in parts (d) and (e) of the corresponding examination section.
  • Requires reversing the standard normal procedure: locating a given cumulative probability inside the body of Table A to identify the corresponding $z$-score.
  • Algebraic formula used to convert $z$ back into original observation units ($x$):     $x = \mu + z \times \sigma$

Questions and Discussion

• Question Regarding Subtracting From One in Part (a):

  • Question: Should the Table A probability value be subtracted from $1$ in part (a)?
  • Response: No. Subtraction from $1$ is solely reserved for "greater than" probabilities. Table A natively provides cumulative area to the left (less than the specified value). Because the problem specifies "less than $0.475$", the Table A value is taken directly without subtraction.

• Question Regarding Upper-Tail Areas in Part (b):

  • Question: If a problem asks for a value greater than a threshold, is the Table A lookup always subtracted from $1$?
  • Response: Yes. Because standard normal distribution tables measure cumulative area from the negative infinity tail up to $z$, finding the area above a cutoff requires applying the complement rule ($1 - \text{Table A value}$).

• Question Regarding Terminology Equivalence:

  • Question: How does homework asking for a proportion compare to an exam problem asking for a probability or percentage?
  • Response: They are functionally and mathematically identical. Whether framed as a proportion (Chapter $3$), a probability (Chapter $12$), or later as a $p$-value (Chapter $17$), the quantitative process and resulting decimal answer remain exactly the same.