Discrete Probability Distributions, Expected Value, and the Binomial Distribution

Discrete Probability Distribution Functions (PDF)

• A discrete probability distribution function (PDF) describes the probability associated with each value of a discrete random variable.

• A function qualifies as a discrete probability distribution function if and only if it satisfies two mandatory requirements:

  • Requirement 1: Every individual probability value $P(x)$ must lie between zero and one, inclusive:     $0 \le P(x) \le 1$

  • Requirement 2: The sum of all probabilities across the entire sample space must equal exactly one:     $\sum P(x) = 1$

Worked Examples and Verification of Discrete PDFs

• Example 4.1 (Newborn Crying Study):

  • Context: A child psychologist investigates the frequency with which a newborn baby's crying wakes its mother after midnight each week. A random sample of $n = 50$ mothers was selected.

  • Variable Definition: Let $X = \text{the number of times per week a newborn baby's crying wakes its mother after midnight}$.

  • Possible Values: $x = 0, 1, 2, 3, 4, 5$

  • Sample Data and Probabilities:

  • For $x = 0$: $P(x = 0) = \frac{2}{50}$

  • For $x = 1$: $P(x = 1) = \frac{11}{50}$

  • For $x = 2$: $P(x = 2) = \frac{23}{50}$

  • For $x = 3$: $P(x = 3) = \frac{9}{50}$

  • For $x = 4$: $P(x = 4) = \frac{4}{50}$

  • For $x = 5$: $P(x = 5) = \frac{1}{50}$

  • PDF Verification:

  • Each probability value $\frac{2}{50}, \frac{11}{50}, \frac{23}{50}, \frac{9}{50}, \frac{4}{50}, \frac{1}{50}$ is between $0$ and $1$, inclusive.

  • The sum of all probabilities equals $1$:       $\frac{2}{50} + \frac{11}{50} + \frac{23}{50} + \frac{9}{50} + \frac{4}{50} + \frac{1}{50} = \frac{50}{50} = 1$

• Try It 4.1 (Post-Op Patient Call Button Study):

  • Context: A hospital researcher studies the number of times an average post-op patient rings the nurse during a 12-hour shift across a random sample of $n = 50$ patients.

  • Variable Definition: Let $X = \text{the number of times a patient rings the nurse during a 12-hour shift}$.

  • Possible Values: $x = 0, 1, 2, 3, 4, 5$

  • Probability Distribution Table:

  • For $x = 0$: $P(x = 0) = \frac{4}{50}$

  • For $x = 1$: $P(x = 1) = \frac{8}{50}$

  • For $x = 2$: $P(x = 2) = \frac{16}{50}$

  • For $x = 3$: $P(x = 3) = \frac{14}{50}$

  • For $x = 4$: $P(x = 4) = \frac{6}{50}$

  • For $x = 5$: $P(x = 5) = \frac{2}{50}$

  • Justification as a Discrete PDF:

  • Reason 1: Each probability $P(x)$ is a non-negative value between $0$ and $1$, inclusive.

  • Reason 2: The total sum of probabilities equals $1$:       $\frac{4}{50} + \frac{8}{50} + \frac{16}{50} + \frac{14}{50} + \frac{6}{50} + \frac{2}{50} = \frac{50}{50} = 1$

• Example 4.2 (Student Class Attendance):

  • Context: Nancy has classes scheduled 3 days a week. Over many weeks, she attends class 3 days a week $80\%$ of the time, 2 days $15\%$ of the time, 1 day $4\%$ of the time, and 0 days $1\%$ of the time. One week is randomly selected.

  • Problem Solutions:

  • Part a: Let $X = \text{the number of days Nancy attends class in a randomly chosen week}$.

  • Part b: The random variable $X$ takes on values $x = 0, 1, 2, 3$.

  • Part c: PDF Table construction:

    • $x = 0 \implies P(x) = 0.01$

    • $x = 1 \implies P(x) = 0.04$

    • $x = 2 \implies P(x) = 0.15$

    • $x = 3 \implies P(x) = 0.80$

    • Sum of the $P(x)$ column:         $0.01 + 0.04 + 0.15 + 0.80 = 1.00$

• Try It 4.2 (Basketball Practice Attendance):

  • Context: Jeremiah has basketball practice scheduled two days a week. $90\%$ of the time he attends both practices, $8\%$ of the time he attends one practice, and $2\%$ of the time he attends neither practice.

  • Variable Definition: Let $X = \text{the number of basketball practices Jeremiah attends in a week}$.

  • Possible Values: $X$ takes on values $x = 0, 1, 2$

Mean, Expected Value, and Standard Deviation of Discrete Random Variables

Theoretical Foundation and the Law of Large Numbers

• Expected Value Concept:

  • The expected value represents the long-term average outcome or mean of a probability experiment repeated across a vast number of trials.

  • Short-term results are inherently variable and do not necessarily match theoretical probabilities. For example, flipping a fair coin ten times might yield nine heads, but this deviation reflects short-term random variability.

• Historical Illustration:

  • Karl Pearson tossed a fair coin $24000$ times and recorded $12012$ heads. His relative frequency of heads was $\frac{12012}{24000} = 0.5005$, approaching the theoretical probability of $0.5$

• Law of Large Numbers:

  • As the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero.

• Population Mean and Standard Deviation Symbols:

  • Mean / Expected Value: $\mu$

  • Standard Deviation: $\sigma$

  • Variance: $\sigma^2$

Mathematical Formulas for Discrete Probability Distributions

• Mean or Expected Value Formula:   $\mu = \sum \left(x \cdot P(x)\right)$

• Variance Formula:   $\sigma^2 = \sum \left[(x - \mu)^2 \cdot P(x)\right]$

• Standard Deviation Formula:   $\sigma = \sqrt{\sum \left[(x - \mu)^2 \cdot P(x)\right]}$

• Calculation Procedure:

  • To compute the expected value $\mu$, multiply each numerical value $x$ of the random variable by its corresponding probability $P(x)$, and sum all resulting products.

  • To compute the variance $\sigma^2$, subtract $\mu$ from each value $x$ to obtain the deviation $(x - \mu)$, square each deviation to get $(x - \mu)^2$, multiply each squared deviation by its probability $P(x)$, and add all resulting products.

  • To compute the standard deviation $\sigma$, take the square root of the variance.

Applied Worked Examples of Expected Value and Standard Deviation

• Example 4.3 (Soccer Team Days Played):

  • Context: A soccer team plays $0$, $1$, or $2$ days a week with probabilities $0.2$, $0.5$, and $0.3$, respectively.

  • Variable Definition: Let $X = \text{the number of days per week the soccer team plays soccer}$. Possible values: $x = 0, 1, 2$

  • Expected Value Table:

  • $x = 0 \implies P(x) = 0.2 \implies x \cdot P(x) = (0)(0.2) = 0$

  • $x = 1 \implies P(x) = 0.5 \implies x \cdot P(x) = (1)(0.5) = 0.5$

  • $x = 2 \implies P(x) = 0.3 \implies x \cdot P(x) = (2)(0.3) = 0.6$

  • Expected Value Calculation:     $\mu = 0 + 0.5 + 0.6 = 1.1\,\text{days}$

  • Interpretation: Over the long term, if the soccer team plays week after week, they expect to play an average of $1.1\,\text{days}$ per week.

• Try It 4.3 (Water Bottle Orders):

  • Context: A customer orders $1$, $2$, or $3$ bottles of water with probabilities $0.6$, $0.3$, and $0.1$, respectively.

  • Variable Definition: Let $X = \text{the number of bottles of water a customer orders}$.

  • Expected Value Calculation:     $\mu = (1)(0.6) + (2)(0.3) + (3)(0.1) = 0.6 + 0.6 + 0.3 = 1.5\,\text{bottles}$

• Example 4.4 (Full Calculation of Newborn Crying Mean and Standard Deviation):

  • Context: Using the data from Example 4.1 ($n = 50$ mothers sample), calculate the expected value $\mu$ and standard deviation $\sigma$

  • Step 1: Calculate Mean $\mu$:     $\mu = \frac{(0)(2) + (1)(11) + (2)(23) + (3)(9) + (4)(4) + (5)(1)}{50} = \frac{105}{50} = 2.1$

  • Step 2: Compute Deviation Table for Variance:

  • For $x = 0$: $P(x) = 0.04 \implies (0 - 2.1)^2 \cdot 0.04 = (-2.1)^2 \cdot 0.04 = 4.41 \cdot 0.04 = 0.1764$

  • For $x = 1$: $P(x) = 0.22 \implies (1 - 2.1)^2 \cdot 0.22 = (-1.1)^2 \cdot 0.22 = 1.21 \cdot 0.22 = 0.2662$

  • For $x = 2$: $P(x) = 0.46 \implies (2 - 2.1)^2 \cdot 0.46 = (-0.1)^2 \cdot 0.46 = 0.01 \cdot 0.46 = 0.0046$

  • For $x = 3$: $P(x) = 0.18 \implies (3 - 2.1)^2 \cdot 0.18 = (0.9)^2 \cdot 0.18 = 0.81 \cdot 0.18 = 0.1458$

  • For $x = 4$: $P(x) = 0.08 \implies (4 - 2.1)^2 \cdot 0.08 = (1.9)^2 \cdot 0.08 = 3.61 \cdot 0.08 = 0.2888$

  • For $x = 5$: $P(x) = 0.02 \implies (5 - 2.1)^2 \cdot 0.02 = (2.9)^2 \cdot 0.02 = 8.41 \cdot 0.02 = 0.1682$

  • Step 3: Sum Squared Deviation Products to Find Variance $\sigma^2$:     $\sigma^2 = 0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05$

  • Step 4: Take Square Root to Find Standard Deviation $\sigma$:     $\sigma = \sqrt{1.05} \approx 1.0247$

• Try It 4.4 (Expected Value of Hospital Call Button Rings):

  • Context: Using the data from Try It 4.1 ($n = 50$ post-op patients sample), calculate the expected number of nurse calls per shift.

  • Expected Value Calculation:     $\mu = \frac{(0)(4) + (1)(8) + (2)(16) + (3)(14) + (4)(6) + (5)(2)}{50} = \frac{0 + 8 + 32 + 42 + 24 + 10}{50} = \frac{116}{50} = 2.32$

• Soccer Team Standard Deviation Example (Table 4.12 Reference):

  • Context: Computing standard deviation for the soccer team in Example 4.3 with $\mu = 1.1$

  • Computation Table:

  • $x = 0 \implies P(x) = 0.2 \implies (0 - 1.1)^2 \cdot 0.2 = (-1.1)^2 \cdot 0.2 = 1.21 \cdot 0.2 = 0.242$

  • $x = 1 \implies P(x) = 0.5 \implies (1 - 1.1)^2 \cdot 0.5 = (-0.1)^2 \cdot 0.5 = 0.01 \cdot 0.5 = 0.005$

  • $x = 2 \implies P(x) = 0.3 \implies (2 - 1.1)^2 \cdot 0.3 = (0.9)^2 \cdot 0.3 = 0.81 \cdot 0.3 = 0.243$

  • Variance Sum:     $\sigma^2 = 0.242 + 0.005 + 0.243 = 0.490$

  • Standard Deviation:     $\sigma = \sqrt{0.49} = 0.7$

Expected Profit and Financial Applications in Games of Chance

• Example 4.5 (5-Digit Number Selection Game):

  • Rules: A computer selects 5 numbers from $0$ to $9$ with replacement. Entry fee is $\$2$. Winning jackpot pays $\$100000$ profit (player receives $\$2$ refund plus $\$100000$ profit).

  • Variable Definition: Let $X = \text{net profit in dollars}$. Possible values: $x = 100000$ or $x = -2$

  • Probability Calculations:

  • Probability of selecting 1 correct digit: $\frac{1}{10}$

  • Probability of winning:       $P(\text{win}) = \left(\frac{1}{10}\right)^5 = 10^{-5} = 0.00001$

  • Probability of losing:       $P(\text{loss}) = 1 - 0.00001 = 0.99999$

  • Expected Value Table:

  • Loss: $x = -2 \implies P(x) = 0.99999 \implies x \cdot P(x) = (-2)(0.99999) = -1.99998$

  • Profit: $x = 100000 \implies P(x) = 0.00001 \implies x \cdot P(x) = (100000)(0.00001) = 1.00000$

  • Expected Value Calculation:     $\mu = -1.99998 + 1.00000 = -0.99998 \approx -1.00$

  • Interpretation: On average, a player expects to lose approximately $\$1$ per game over long-term play.

• Try It 4.5 (Card Suit Guessing Game):

  • Rules: Draw 4 cards from a standard 52-card deck with replacement, guessing the suit each time. Cost is $\$1$. Correctly guessing all 4 suits returns the $\$1$ plus $\$256$ profit.

  • Variable Definition: Let $X = \text{net profit in dollars}$.

  • Probability Calculations:

  • Probability of guessing 1 card's suit correctly: $\frac{13}{52} = \frac{1}{4}$

  • Probability of winning:       $P(\text{win}) = \left(\frac{1}{4}\right)^4 = \frac{1}{256}$

  • Probability of losing:       $P(\text{loss}) = 1 - \frac{1}{256} = \frac{255}{256}$

  • Expected Value Calculation:     $\mu = (256)\left(\frac{1}{256}\right) + (-1)\left(\frac{255}{256}\right) = 1 - \frac{255}{256} = \frac{1}{256} \approx 0.0039$

• Example 4.6 (Biased Coin Game):

  • Rules: Toss a biased coin with $P(\text{heads}) = \frac{2}{3}$ and $P(\text{tails}) = \frac{1}{3}$. Tossing a head requires paying $\$6$ ($x = -6$). Tossing a tail wins $\$10$ ($x = 10$).

  • Variable Definition: Let $X = \text{net profit/loss from one coin toss}$.

  • Expected Value Table:

  • WIN: $x = 10 \implies P(x) = \frac{1}{3} \implies x \cdot P(x) = \frac{10}{3}$

  • LOSE: $x = -6 \implies P(x) = \frac{2}{3} \implies x \cdot P(x) = \frac{-12}{3}$

  • Expected Value Calculation:     $\mu = \frac{10}{3} + \frac{-12}{3} = -\frac{2}{3} \approx -0.67$

  • Conclusion: You do not come out ahead; over many plays, you expect to lose an average of $\$0.67$ per game.

• Try It 4.6 (Spinner Game):

  • Rules: Spin a spinner once with $P(\text{red}) = \frac{2}{5}$, $P(\text{blue}) = \frac{2}{5}$, and $P(\text{green}) = \frac{1}{5}$. Red costs $\$10$ ($x = -10$), Blue pays/costs $\$0$ ($x = 0$), Green wins $\$10$ ($x = 10$).

  • Expected Value Table:

  • Red: $x = -10 \implies P(x) = \frac{2}{5} \implies x \cdot P(x) = -\frac{20}{5} = -4$

  • Blue: $x = 0 \implies P(x) = \frac{2}{5} \implies x \cdot P(x) = 0$

  • Green: $x = 10 \implies P(x) = \frac{1}{5} \implies x \cdot P(x) = \frac{10}{5} = 2$

  • Expected Value Calculation:     $\mu = -4 + 0 + 2 = -2.00$

Symmetric and Equal Probability Examples

• Example 4.7 (Die Roll Even Numbers):

  • Context: Toss a fair six-sided die twice. Let $X = \text{the number of faces showing an even number}$.

  • Probabilities for single roll: $P(\text{even}) = 0.5$, $P(\text{odd}) = 0.5$

  • Distribution for 2 rolls:

  • $P(x = 0) = 0.25$

  • $P(x = 1) = 0.50$

  • $P(x = 2) = 0.25$

  • Mean Calculation:     $\mu = (0)(0.25) + (1)(0.50) + (2)(0.25) = 1.0$

  • Variance and Standard Deviation Calculation:     $\sigma^2 = (0 - 1)^2(0.25) + (1 - 1)^2(0.50) + (2 - 1)^2(0.25) = 0.25 + 0 + 0.25 = 0.50$     $\sigma = \sqrt{0.50} \approx 0.7071$

• Try It 4.7 (Card Stacks Odd Numbers):

  • Context: Two stacks of cards numbered 1 to 8 are placed on a table. One card is picked from each stack. Let $X = \text{the number of cards showing an odd number}$.

  • Probabilities per stack: Odd cards (1, 3, 5, 7) account for 4 out of 8 cards ($p = 0.5$).

  • Distribution for 2 draws:

  • $P(x = 0) = 0.25$

  • $P(x = 1) = 0.50$

  • $P(x = 2) = 0.25$

  • Mean and Standard Deviation:     $\mu = 1.0$     $\sigma = \sqrt{0.50} \approx 0.7071$

Real-World Event Risk Betting Examples

• Example 4.8 (Earthquake Risk Betting in Iran):

  • Context: On May 11, 2013 at 9:30 PM, the probability of a moderate earthquake occurring in Iran within 48 hours was $21.42\%$ ($0.2142$). A bet is placed: Win $\$50$ profit if an earthquake occurs, Lose $\$20$ if it does not.

  • Variable Definition: Let $X = \text{the amount of profit from a bet}$.

  • Probabilities:

  • $P(\text{win}) = 0.2142$

  • $P(\text{loss}) = 100\% - 21.42\% = 78.58\% = 0.7858$

  • Expected Value Calculation:     $\mu = (50)(0.2142) + (-20)(0.7858) = 10.71 - 15.716 = -5.006 \approx -5.01$

  • Explanation: You do not come out ahead because you expect to lose an average of $\$5.01$ per bet over many trials.

  • Variance and Standard Deviation:     $\sigma^2 = (50 - (-5.006))^2(0.2142) + (-20 - (-5.006))^2(0.7858) = (55.006)^2(0.2142) + (-14.994)^2(0.7858) \approx 648.096 + 176.664 = 824.76$     $\sigma = \sqrt{824.76} \approx 28.718$

• Try It 4.8 (Earthquake Risk Betting in Japan):

  • Context: On May 11, 2013 at 9:30 PM, the probability of a moderate earthquake in Japan within 48 hours was $1.08\%$ ($0.0108$). A bet is placed: Win $\$100$ profit if an earthquake occurs, Lose $\$10$ if it does not.

  • Variable Definition: Let $X = \text{the amount of profit from a bet}$.

  • Probabilities:

  • $P(\text{win}) = 0.0108$

  • $P(\text{loss}) = 1 - 0.0108 = 0.9892$

  • Mean Calculation:     $\mu = (100)(0.0108) + (-10)(0.9892) = 1.08 - 9.892 = -8.812 \approx -8.81$

  • Variance and Standard Deviation:     $\sigma^2 = (100 - (-8.812))^2(0.0108) + (-10 - (-8.812))^2(0.9892) = (108.812)^2(0.0108) + (-1.188)^2(0.9892) \approx 127.87 + 1.396 = 129.266$     $\sigma = \sqrt{129.266} \approx 11.37$

The Binomial Distribution

Characteristics of a Binomial Experiment

• A probability experiment is defined as a binomial experiment if it satisfies three criteria:

  1. Fixed Number of Trials: The experiment consists of a fixed number of repeated trials, denoted by $n$.

  2. Binary Outcomes: Each trial has only two possible outcomes, classified as a "success" or a "failure".

  • $p = \text{probability of a success on any single trial}$

  • $q = \text{probability of a failure on any single trial}$

  • Complement Rule:        $p + q = 1 \implies q = 1 - p$

  1. Independent and Identical Trials: All $n$ trials are independent and conducted under identical conditions. Consequently, $p$ and $q$ remain strictly constant across all trials.

Bernoulli Trials

• A Bernoulli Trial is defined as an experiment that meets the binary outcome and independence conditions, but consists of exactly $n = 1$ trial.

• A binomial experiment takes place when counting the total number of successes across $n \ge 1$ independent Bernoulli trials.

Notation and Standard Formulas

• Binomial Notation:   $X \sim B(n, p)$

  • Read as: "$X$ is a random variable with a binomial probability distribution."

  • Parameters: $n = \text{number of trials}$, $p = \text{probability of success on each trial}$.

  • Random Variable Definition: $X = \text{number of successes in } n \text{ independent trials}$.

  • Possible Values: $x = 0, 1, 2, 3, \dots, n$

• Binomial Formulas:

  • Mean (Expected Value):     $\mu = np$

  • Variance:     $\sigma^2 = npq$

  • Standard Deviation:     $\sigma = \sqrt{npq}$

  • Probability Density Function (PDF) Formula:     $P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x}$     where the binomial coefficient is given by:     $\binom{n}{x} = \frac{n!}{x!(n - x)!}$

Identifying Binomial Variables and Success Definitions

• Example 4.9 (Course Withdrawal Rates):

  • Context: At ABC College, the withdrawal rate for elementary physics is $30\%$ ($0.30$) per term, meaning $70\%$ ($0.70$) stay the entire term.

  • Variable Definition: Defining "success" as a student who withdrew, let $X = \text{the number of students who withdraw from a randomly selected elementary physics class}$ .

• Try It 4.9 (School Fruit Provision):

  • Context: $48\%$ of state schools offer fruit in lunch daily ($52\%$ do not).

  • Success Definition: A "success" is defined as a school that offers fruit in its lunches every day.

• Example 4.10 (Game Wins Mathematical Translation):

  • Context: A game has two outcomes: win ($p = 0.55$) or lose ($q = 0.45$). Played $n = 20$ independent times.

  • Variable Definition: Let $X = \text{the number of wins}$. Values $x = 0, 1, 2, \dots, 20$

  • Mathematical Translation: The question "probability of winning 15 of 20 times" is written as:     $P(x = 15)$

• Try It 4.10 (Dolphin Fish Catching Mathematical Translation):

  • Context: A rescued dolphin has a $35\%$ success rate ($p = 0.35$) and $65\%$ failure rate ($q = 0.65$) in catching live fish across $n = 20$ attempts.

  • Mathematical Translation: The question "probability of succeeding 12 times out of 20 attempts" is written as:     $P(x = 12)$

Skewness, Graphing, and Addition Rule Applications

• Example 4.11 (Altered Coin Toss and Distribution Skewness):

  • Context: An altered coin with head probability $p = 0.25$ ($q = 0.75$) is flipped $n = 5$ times.

  • Variable Definition: Let $X = \text{the number of heads in 5 flips}$. Values $x = 0, 1, 2, 3, 4, 5$

  • Mathematical Question: Finding the probability of getting more than 3 heads is written as $P(x > 3)$.

  • Application of Addition Rule:     $P(x > 3) = P(x = 4) + P(x = 5) = 0.0146 + 0.0007 = 0.0153$

  • Mean Calculation:     $\mu = np = (5)(0.25) = 1.25$

  • Formula Setup:     $P(x = x_0) = \binom{5}{x_0} (0.25)^{x_0} (0.75)^{5 - x_0}$

  • Effect of $p \ne 0.5$ on Shape: When $p \ne 0.5$, the binomial distribution is asymmetrical. Because $p = 0.25 < 0.5$, this distribution is skewed right with its mean at $\mu = 1.25$

• Try It 4.11 (Die Rolling Mathematical Translation):

  • Context: A fair six-sided die is rolled $n = 10$ independent times. The probability of rolling a one on any single roll is $p = \frac{1}{6}$.

  • Mathematical Translation: The probability of rolling a one more than three times is written as:     $P(x > 3)$

Population Sampling and Verbal Inequality Translations

• Example 4.12 (Statistics Homework Completion):

  • Context: $70\%$ of statistics students complete homework on time ($p = 0.70$). A sample of $n = 50$ students is chosen.

  • Problem Solutions:

  • Part a: Binomial criteria met because there are only two outcomes (success or failure), a fixed trial count ($n = 50$), and constant probability ($p = 0.70$).

  • Part b: Variable definition: Let $X = \text{the number of students who do their homework on time}$.

  • Part c: Values $x$ takes on: $x = 0, 1, 2, \dots, 50$

  • Part d: Failure in words: A student who does not do their homework on time.

  • Part e: Failure probability: $q = 1 - p = 1 - 0.70 = 0.30$

  • Part f: Verbal inequality translation: "At least 40" translates to the inequality:       $P(x \ge 40)$

• Try It 4.12 (Driver's License Examination):

  • Context: $65\%$ of people pass the driver's exam on the first try ($p = 0.65$). A sample of $n = 50$ individuals is selected.

  • Two Reasons why this is a Binomial Problem:

  1. Fixed number of trials ($n = 50$) with exactly two possible outcomes (pass or fail).

  2. Trials are independent with a constant probability of success ($p = 0.65$) for each individual.

Cumulative Probabilities and TI-83/84 Calculator Commands

• TI-83, TI-83+, TI-84, and TI-84+ Calculator Commands:

  • Command Access: Press 2nd DISTR.

  • Exact Probability $P(X = x)$:     $\text{binompdf}(n, p, x)$

  • Note: If $x$ is omitted, the calculator outputs the entire binomial probability table.

  • Cumulative Probability $P(X \le x)$:     $\text{binomcdf}(n, p, x)$

  • Note: Computes cumulative probability from $0$ up to $x$

  • Right-Tail Probability $P(X > x)$:     $1 - \text{binomcdf}(n, p, x)$

• Example 4.13 (Adult Workers High School Diploma Case Study):

  • Context: About $41\%$ of adult workers have a high school diploma as their highest education ($p = 0.41$, $q = 0.59$). A random sample of $n = 20$ adult workers is selected.

  • Distribution: $X \sim B(20, 0.41)$. Values $x = 0, 1, 2, \dots, 20$

  • Probability at most 12 workers:     $P(x \le 12) = \text{binomcdf}(20, 0.41, 12) = 0.9738$

  • Probability more than 12 workers:     $P(x > 12) = 1 - \text{binomcdf}(20, 0.41, 12) = 1 - 0.9738 = 0.0262$

  • Mean Calculation:     $\mu = np = (20)(0.41) = 8.2$

  • Variance Calculation:     $\sigma^2 = npq = (20)(0.41)(0.59) = 4.838$

  • Standard Deviation Calculation:     $\sigma = \sqrt{npq} = \sqrt{(20)(0.41)(0.59)} \approx 2.20$

• Try It 4.13 (Community Volunteer Program):

  • Context: $32\%$ of students participate in a community volunteer program ($p = 0.32$). Sample size $n = 30$

  • Probability at most 14 students:     $P(x \le 14) = \text{binomcdf}(30, 0.32, 14) \approx 0.9691$

Comprehensive Case Studies

• Example 4.14 (Art Supplies Catalog Signature Artists):

  • Context: In a 560-page art supply catalog, 8 pages feature signature artists. A random sample of $n = 100$ pages is drawn.

  • Parameters: Success probability $p = \frac{8}{560} = \frac{1}{70} \approx 0.014286$

  • Distribution: $X \sim B\left(100, \frac{8}{560}\right)$. Possible values $x = 0, 1, 2, \dots, 100$

  • Probability Computations:

  • Probability exactly two pages:       $P(x = 2) = \text{binompdf}\left(100, \frac{8}{560}, 2\right) \approx 0.2458$

  • Probability at most six pages:       $P(x \le 6) = \text{binomcdf}\left(100, \frac{8}{560}, 6\right) \approx 0.9989$

  • Probability more than three pages:       $P(x > 3) = 1 - \text{binomcdf}\left(100, \frac{8}{560}, 3\right) \approx 0.0520$

  • Mean and Standard Deviation:     $\mu = np = 100 \times \frac{8}{560} \approx 1.43$     $\sigma = \sqrt{100 \times \frac{8}{560} \times \frac{552}{560}} \approx 1.188$

• Try It 4.14 (Gallup Poll Saving vs. Spending Preferences):

  • Context: Gallup poll indicates $60\%$ of American adults prefer saving over spending ($p = 0.60$). Sample size $n = 50$

  • Distribution: $X \sim B(50, 0.60)$.

  • Probability Computations:

  • Probability exactly 25 adults prefer saving:       $P(x = 25) = \text{binompdf}(50, 0.60, 25) \approx 0.0415$

  • Probability at most 20 adults prefer saving:       $P(x \le 20) = \text{binomcdf}(50, 0.60, 20) \approx 0.0028$

  • Probability more than 30 adults prefer saving:       $P(x > 30) = 1 - \text{binomcdf}(50, 0.60, 30) \approx 0.4465$

  • Mean and Standard Deviation:     $\mu = np = (50)(0.60) = 30$     $\sigma = \sqrt{50 \times 0.60 \times 0.40} = \sqrt{12} \approx 3.464$

• Example 4.15 (Lifetime Risk of Cancer Sample Study):

  • Context: Lifetime risk of developing cancer is 1 in 67 ($1.5\% = 0.015$). Sample size $n = 200$ people.

  • Variable Definition: Let $X = \text{the number of people who will develop cancer}$.

  • Distribution: $X \sim B(200, 0.015)$.

  • Mean and Standard Deviation:     $\mu = np = (200)(0.015) = 3$     $\sigma = \sqrt{200 \times 0.015 \times 0.985} = \sqrt{2.955} \approx 1.719$

  • Probability at most 8 people:     $P(x \le 8) = \text{binomcdf}(200, 0.015, 8) \approx 0.9972$

  • Comparing Likelihood of 5 vs 6 People:     $P(x = 5) = \text{binompdf}(200, 0.015, 5) \approx 0.1002$     $P(x = 6) = \text{binompdf}(200, 0.015, 6) \approx 0.0486$

  • Numerical Justification: Since $0.1002 > 0.0486$, it is more likely that five people will develop cancer than six people.

• Try It 4.15 (NBA Field Goal Shooting Performance):

  • Context: An NBA player for the Los Angeles Clippers completed $61.3\%$ of his shots ($p = 0.613$). Sample size $n = 80$ shots.

  • Variable Definition: Let $X = \text{the number of shots that scored points}$.

  • Distribution: $X \sim B(80, 0.613)$.

  • Mean and Standard Deviation:     $\mu = np = (80)(0.613) = 49.04$     $\sigma = \sqrt{80 \times 0.613 \times 0.387} = \sqrt{18.97848} \approx 4.356$

  • Probability scored with 60 shots:     $P(x = 60) = \text{binompdf}(80, 0.613, 60) \approx 0.0042$

  • Probability scored with more than 50 shots:     $P(x > 50) = 1 - \text{binomcdf}(80, 0.613, 50) \approx 0.3670$

Non-Binomial Scenarios and Independence Violations

• Example 4.16 (Selection Without Replacement in Advisory Committee):

  • Context: An advisory committee has 10 staff members and 6 students (total 16 members). Two names are drawn without replacement to pick a chairperson and a recorder.

  • Analysis of Independence:

  • Probability of selecting a student on Trial 1:       $P(\text{Student}_1) = \frac{6}{16}$

  • Probability of selecting a student on Trial 2 depends on Trial 1 outcome:

    • If a student was drawn on Trial 1:         $P(\text{Student}_2 \mid \text{Student}_1) = \frac{5}{15}$

    • If a staff member was drawn on Trial 1:         $P(\text{Student}_2 \mid \text{Staff}_1) = \frac{6}{15}$

  • Conclusion: Because the trials are conducted without replacement, the probability of success changes between trials. This violates the condition of independence, so this experiment is NOT binomial.

• Try It 4.16 (Lacrosse Captain Selection Without Replacement):

  • Context: Selecting 3 captains from senior player names in a hat without replacement.

  • Analysis: This is NOT a binomial problem because sampling without replacement alters the probability of picking players of specific positions across consecutive draws, violating the independence requirement.