Unit 2: Probability, Random Variables, and Probability Distributions Study Guide

Discrete Random Variables & Probability Distributions

• Discrete Random Variable Definition: A discrete random variable $X$ takes on a countable set of distinct numerical values, where each value is associated with a specific probability $P(X = x)$.

• Fundamental Properties of Discrete Probability Distributions:

  • Non-Negativity Constraint: Every individual probability must lie between 0 and 1 inclusive:     $0 \le P(X = x) \le 1$

  • Summation Constraint: The sum of probabilities across all possible outcomes in the sample space must equal exactly 1:     $\sum P(X = x) = 1$

• Weighted Die Problem:

  • Distribution Table:     

    

             | Value | 1 | 2 | 3 | 4 | 5 | 6 |     | :--- | :--- | :--- | :--- | :--- | :--- | :--- |     | Probability | $0.10$ | $0.15$ | $0.20$ | $0.25$ | $0.20$ | $0.10$ |

  • Calculating Probability of Die Sum equal to 10:     When two independent weighted dice are thrown, the event that the sum of the two face values equals 10 can occur through three mutually exclusive outcome pairs: $(4, 6)$, $(6, 4)$, and $(5, 5)$.

  • $P(4 \text{ then } 6) = P(4) \times P(6) = (0.25)(0.10)$

  • $P(6 \text{ then } 4) = P(6) \times P(4) = (0.10)(0.25)$

  • $P(5 \text{ then } 5) = P(5) \times P(5) = (0.20)(0.20) = (0.20)^2$

  • Total probability formula:       $P(\text{sum} = 10) = 2(0.25)(0.10) + (0.20)^2 = 0.050 + 0.040 = 0.090$

• Cookie Decorating Experiment:

  • Distribution Table:     

    

             | $X$ | 1 | 2 | 3 | 4 |     | :--- | :--- | :--- | :--- | :--- |     | $P(X)$ | $0.2$ | $0.4$ | $0.3$ | $0.1$ |

  • Probability of Decorating At Least 3 Cookies:     $P(X \ge 3) = P(X = 3) + P(X = 4) = 0.3 + 0.1 = 0.4$

  • Mean / Expected Value Calculation ($\mu_X$):     $\mu_X = E(X) = \sum x \cdot P(X = x)$     $\mu_X = (1)(0.2) + (2)(0.4) + (3)(0.3) + (4)(0.1) = 0.2 + 0.8 + 0.9 + 0.4 = 2.3\,\text{cookies}$

  • Interpretation of Standard Deviation ($\sigma_X$):     In repeated trials, the number of cookies decorated by participants will typically differ from the mean number of cookies ($2.3\,\text{cookies}$) by approximately one standard deviation.

• Two-Dice Sum Probability Distribution Function:

  • Function Definition: $P(X) = \frac{6 - |X - 7|}{36}$ for $X \in \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

  • Distribution Table:     

    

             | $X$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |     | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |     | $P(X)$ | $\frac{1}{36}$ | $\frac{2}{36}$ | $\frac{3}{36}$ | $\frac{4}{36}$ | $\frac{5}{36}$ | $\frac{6}{36}$ | $\frac{5}{36}$ | $\frac{4}{36}$ | $\frac{3}{36}$ | $\frac{2}{36}$ | $\frac{1}{36}$ |

  • Verification as a True Probability Function:

  • Every probability value $P(X)$ is non-negative and bounded by 1 ($0 \le P(X) \le 1$).

  • Sum of all probabilities across the sample space:       $\sum_{X=2}^{12} P(X) = \frac{1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1}{36} = \frac{36}{36} = 1.00$

  • Physical Modeling Context: This probability function models the sum of the outcomes when rolling two fair, standard six-sided dice.

  • Calculating P(X \ge 9)$format:**\n    P(X \ge 9) = P(9) + P(10) + P(11) + P(12) = \frac{4}{36} + \frac{3}{36} + \frac{2}{36} + \frac{1}{36} = \frac{10}{36} = \frac{5}{18} \approx 0.2778\n * **Determining Expected Value E(X) via Symmetry:**\n * E(X) = 7\n * **Symmetry Justification:** The probability distribution is perfectly symmetric around X = 7$. For any discrete random variable whose distribution is symmetrical, the mean / expected value$E(X) is located at the center of symmetry without requiring explicit calculation.\n\n# Binomial Distributions\n\n* **Binomial Criteria Framework (BINS):**\n * **B - Binary Outcomes:** Each individual trial results in one of exactly two outcomes: a "Success" or a "Failure".\n * **I - Independent Trials:** The outcome of any given trial does not affect or alter the outcome of any other trial.\n * **N - Number Fixed:** There is a predetermined, fixed number of trials n.\n * **S - Same Probability:** The probability of success p remains constant across all trials.\n\n* **Standard Mathematical Formulas for Binomial Random Variable X \sim \text{Bin}(n, p)$format:

  • Probability Mass Function (PMF):     $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

  • Mean / Expected Value:     $\mu = n \times p$

  • Variance:     $\sigma^2 = n \times p \times (1 - p)$

  • Standard Deviation:     $\sigma = \sqrt{n \times p \times (1 - p)}$

• SAT Multiple Choice Random Guessing Application:

  • Parameters: $n = 25$ questions, probability of correct guessing on a 5-option question $p = \frac{1}{5} = 0.20$.

  • Probability of answering 6 or more questions correctly (P(X \ge 6)$format):\n    P(X \ge 6) = 1 - P(X \le 5) = 1 - \text{binomcdf}(25, 0.20, 5)\n    P(X \le 5) = \text{binomcdf}(25, 0.20, 5) \approx 0.6167\n    P(X \ge 6) = 1 - 0.6167 = 0.3833\n\n* **Clemson University Freshmen Graduation Case Study:**\n * Parameters: Sample size n = 16$freshmen living on a dorm floor, baseline 4-year graduation rate$p = 0.63.\n * Figure showing handwritten solutions:\n    \n\n * **Verification of Binomial Settings (BINS):**\n * **Binary:** Each freshman either graduates in 4 years (Success) or does not graduate in 4 years (Failure).\n * **Independent:** The graduation outcome of one freshman is independent of the graduation outcome of another freshman on the floor.\n * **Number Fixed:** There is a fixed sample size of n = 16 freshmen.\n * **Same Probability:** Each freshman has an identical probability p = 0.63 of graduating in 4 years.\n * **Probability Exactly Half Graduate (X = 8):**\n    P(X = 8) = \text{binompdf}(16, 0.63, 8) \approx 0.1118\,\text{(or } 11.18\% \text{)}\n * **Probability More Than 10 Freshmen Graduate (P(X > 10)$format):**     $P(X > 10) = P(X \ge 11) = 1 - \text{binomcdf}(16, 0.63, 10) = 1 - 0.5770 = 0.4230\,\text{(or } 42.30\% \text{)}$

  • Probability Less Than 5 Freshmen Graduate (P(X < 5)$format):**\n    P(X < 5) = P(X \le 4) = \text{binomcdf}(16, 0.63, 4) = 0.0023\,\text{(or } 0.23\% \text{)}\n * **Probability At Least 6 Freshmen Graduate (P(X \ge 6)$format):     $P(X \ge 6) = 1 - P(X \le 5) = 1 - \text{binomcdf}(16, 0.63, 5) = 1 - 0.0100 = 0.9900\,\text{(or } 99.00\% \text{)}$

  • Probability At Most 7 Freshmen Graduate (P(X \le 7)$format):**\n    P(X \le 7) = \text{binomcdf}(16, 0.63, 7) = 0.0926\,\text{(or } 9.26\% \text{)}\n * **Expected Number of Freshmen to Graduate (\mu):**\n    \mu = n \times p = 16 \times 0.63 = 10.08\n * **Interpretation:** Over time, an average of 10.08 freshmen on a dorm floor of 16 are expected to graduate in 4 years.\n * **Standard Deviation Calculation (\sigma) and Interpretation:**\n    \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{16 \times 0.63 \times 0.37} = \sqrt{3.7296} \approx 1.93\n * **Interpretation:** In repeated random selections of 16 freshmen, the number of freshmen who graduate in 4 years will typically vary from the expected mean (10.08$) by approximately$1.93 students.\n\n# Expected Value & Decision Analysis\n\n* **Carnival Coin Selection Game Analysis:**\n * **Game Rules:** A blindfolded player selects one coin at random from a container containing 60 total coins. Game entry costs \$0.25, and the player keeps the drawn coin.\n * **Coin Composition & Individual Probabilities:**\n * Pennies (\$0.01$): 9 coins,$P = \frac{9}{60}\n * Nickels (\$0.05$): 8 coins,$P = \frac{8}{60}\n * Dimes (\$0.10$): 12 coins,$P = \frac{12}{60}\n * Quarters (\$0.25$): 16 coins,$P = \frac{16}{60}\n * Half-Dollars (\$0.50$): 11 coins,$P = \frac{11}{60}\n * Dollar pieces (\$1.00$): 4 coins,$P = \frac{4}{60}\n * Total Coins: 9 + 8 + 12 + 16 + 11 + 4 = 60\,\text{coins}\n * **Expected Value Computation (E(V)):**\n    E(V) = (0.01)\left(\frac{9}{60}\right) + (0.05)\left(\frac{8}{60}\right) + (0.10)\left(\frac{12}{60}\right) + (0.25)\left(\frac{16}{60}\right) + (0.50)\left(\frac{11}{60}\right) + (1.00)\left(\frac{4}{60}\right)\n    E(V) = \frac{0.09 + 0.40 + 1.20 + 4.00 + 5.50 + 4.00}{60} = \frac{15.19}{60} \approx \$0.2532\n * **Financial Decision Conclusion:**\n * Expected Net Gain per Play: E(V) - \text{Cost} = \$0.2532 - \$0.2500 = +\$0.0032\n * Decision: A player should expect to gain money in the long run because the expected value of the coin selected (\$0.2532$) is greater than the$\$0.25 cost to play.\n\n# Continuous Random Variables & Standard Normal Distributions\n\n* **Standard Normal Distribution Variable Z \sim N(0, 1)$format:

  • Probability Comparison & Ordering:

  • Event I: $P(Z < 1.3) \approx 0.9032$

  • Event II: $P(-1.2 < Z < 1.2) = P(Z < 1.2) - P(Z < -1.2) \approx 0.8849 - 0.1151 = 0.7698$

  • Event III: $P(Z > -1) = P(Z < 1) \approx 0.8413$

  • Ordered from smallest to largest probability: II ($0.7698$), III ($0.8413$), I ($0.9032$).

• Tennis Serve Speed Percentile Evaluation:

  • Distribution: Serve speeds $X \sim N(\mu = 115\,\text{mph}, \sigma = 8\,\text{mph})$.

  • Target observation: Taylor Fritz serve speed $X = 127\,\text{mph}$.

  • Standardized $Z$-score calculation:     $Z = \frac{X - \mu}{\sigma} = \frac{127 - 115}{8} = \frac{12}{8} = 1.50$

  • Percentile derivation:     $P(Z < 1.50) \approx 0.9332 \implies 93\text{rd percentile}$

• Adult Male Cholesterol Level Parameter Derivation:

  • Distribution: Cholesterol levels $X \sim N(\mu, \sigma = 17\,\text{mg/dL})$.

  • High cholesterol threshold condition: $X > 200\,\text{mg/dL}$.

  • Given high cholesterol proportion: $P(X > 200) = 0.08 \implies P(X \le 200) = 0.92$

  • Standard score for the 92nd percentile:     $Z = \text{invNorm}(0.92, 0, 1) \approx 1.4051$

  • Solving for mean cholesterol level $\mu$:     $Z = \frac{X - \mu}{\sigma} \implies 1.4051 = \frac{200 - \mu}{17}$     $200 - \mu = 1.4051 \times 17 = 23.8867$     $\mu = 200 - 23.8867 = 176.1133 \approx 176.1\,\text{mg/dL}$

• High School Daily Phone Screen Time Study:

  • Distribution: Screen time $X \sim N(\mu = 325\,\text{minutes}, \sigma = 25\,\text{minutes})$.

  • Proportion Spending Longer Than 370 Minutes (P(X > 370)$format):**\n    Z = \frac{370 - 325}{25} = \frac{45}{25} = 1.80\n    P(X > 370) = P(Z > 1.80) = 1 - P(Z \le 1.80) = 1 - 0.9641 = 0.0359\,\text{(or } 3.59\% \text{)}\n * **Probability Spending Between 300 and 400 Minutes (P(300 < X < 400)$format):     $Z_1 = \frac{300 - 325}{25} = -1.00$     $Z_2 = \frac{400 - 325}{25} = 3.00$     $P(300 < X < 400) = P(-1.00 < Z < 3.00) = P(Z < 3.00) - P(Z < -1.00) = 0.9987 - 0.1587 = 0.8400\,\text{(or } 84.00\% \text{)}$

  • Minimum Screen Time Cutoff for the Top 12% of Students:     $P(X > x) = 0.12 \implies P(X \le x) = 0.88$     $Z = \text{invNorm}(0.88, 0, 1) \approx 1.1750$     $x = \mu + Z \sigma = 325 + (1.1750 \times 25) = 325 + 29.375 = 354.375\,\text{minutes}$

Sampling Distributions & Central Limit Theorem

• Disney World Space Mountain Wait Time Comparison:

  • Individual wait time $X \sim N(\mu = 50\,\text{minutes}, \sigma = 4\,\text{minutes})$.

  • Sample size $n = 16$ riders, with sample mean wait time $\bar{x}$.

  • Sampling distribution parameters:     $\mu_{\bar{x}} = \mu = 50\,\text{minutes}$     $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{16}} = \frac{4}{4} = 1\,\text{minute}$

  • Probability evaluation for wait times under 46 minutes:

  • Individual rider (P(X < 46)$format):\n      Z = \frac{46 - 50}{4} = -1.00 \implies P(Z < -1.00) \approx 0.1587\n * Sample mean of 16 riders (P(\bar{x} < 46)$format):       $Z = \frac{46 - 50}{1} = -4.00 \implies P(Z < -4.00) \approx 0.00003$

  • Conclusion: $P(X < 46)$ is larger than $P(\bar{x} < 46)$ because the sampling distribution of $\bar{x}$ has a smaller standard deviation ($\sigma_{\bar{x}} = 1\,\text{minute}$) than the individual wait time distribution $X$ ($\sigma = 4\,\text{minutes}$).

• Central Limit Theorem (CLT) Population Normality Requirements:

  • The population distribution does NOT need to be normally distributed for the sampling distribution of $\bar{x}$ to be approximately normal.

  • Central Limit Theorem Core Principle: When the sample size $n$ is sufficiently large ($n \ge 30$), the sampling distribution of the sample mean $\bar{x}$ will be approximately normal regardless of the shape of the underlying population distribution.

• Bluffton High School Teacher Email Analysis:

  • Individual teacher daily email output $X \sim N(\mu = 12\,\text{emails}, \sigma = 2.45\,\text{emails})$.

  • Random sample size $n = 50$ teachers.

  • Sampling Distribution Mean (\mu_{\bar{x}}$format):**\n    \mu_{\bar{x}} = \mu = 12\,\text{emails}\n * **Justification:** The sample mean \bar{x}$is an unbiased estimator of the population mean$\mu.\n * **Sampling Distribution Standard Deviation (\sigma_{\bar{x}}$format) & Condition Checking:     $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{2.45}{\sqrt{50}} \approx 0.3465\,\text{emails}$

  • Condition to Check: The 10% Condition (Independence Condition). The total population size $N$ must be at least 10 times larger than the sample size ($N \ge 10n = 10 \times 50 = 500$). There must be at least 500 teachers in the population (or sampling performed with replacement) to ensure individual teacher observations are independent.

  • Probability One Selected Teacher Sends Between 9 and 11 Emails (P(9 < X < 11)$format):**\n    Z_1 = \frac{9 - 12}{2.45} \approx -1.2245\n    Z_2 = \frac{11 - 12}{2.45} \approx -0.4082\n    P(9 < X < 11) = \text{normalcdf}(9, 11, 12, 2.45) \approx 0.2309\,\text{(or } 23.09\% \text{)}\n * **Probability Sample Mean of 50 Teachers is Between 9 and 11 Emails (P(9 < \bar{x} < 11)$format):     $Z_1 = \frac{9 - 12}{0.3465} \approx -8.66$     $Z_2 = \frac{11 - 12}{0.3465} \approx -2.89$     $P(9 < \bar{x} < 11) = \text{normalcdf}(9, 11, 12, 0.3465) \approx 0.0019\,\text{(or } 0.19\% \text{)}$

  • Comparative Likelihood Analysis:

  • Selecting one teacher who sends between 9 and 11 emails ($23.09\%$) is significantly more likely than finding a sample average of 50 teachers between 9 and 11 emails ($0.19\%$).

  • Reasoning: Individual data values exhibit far greater variability than sample averages. Increasing the sample size shrinks the standard deviation of the sample mean ($\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$), tightly packing the distribution of $\bar{x}$ around the mean $\mu = 12$. Because the interval $(9, 11)$ excludes the mean of 12, the probability of the sample average falling in that interval approaches zero.

  • Impact of Increasing Sample Size to $n = 75$:

  • When sample size increases to $n = 75$, standard deviation decreases to $\sigma_{\bar{x}} = \frac{2.45}{\sqrt{75}} \approx 0.2829\,\text{emails}$.

  • As $\sigma_{\bar{x}}$ decreases, the sampling distribution concentrates even more tightly around $\mu = 12$. Consequently, the probability $P(9 < \bar{x} < 11)$ decreases even further toward 0.