CEUET Mathematics — Statistics & ProbabilityStudy Notes

Statistics is the science of collecting, organizing, analyzing, and interpreting data. Understanding basic terminology is crucial for solving statistical problems. **Key Definitions:** - **Data**: A collection of quantitative or qualitative information used for reasoning or calculation - **Variable**: Any characteristic, number, or quantity that can be measured or counted - **Population**: The complete set of all individuals sharing a common characteristic - **Sample**: A subset selected from a population to represent that population **Types of Variables:** 1. **Qualitative Variables**: Can be categorized but not numerically measured (e.g., gender, eye color) 2. **Quantitative Variables**: Numerical values that can be ordered - **Discrete**: Countable values (e.g., number of students) - **Continuous**: Infinite possible values within a range (e.g., height, weight) **Levels of Measurement:** - **Nominal**: Categories with no order (gender, nationality) - **Ordinal**: Ranked categories with no precise differences (grades A, B, C) - **Interval**: Ranked with precise differences, no true zero (temperature in °C) - **Ratio**: Ranked with precise differences and true zero (height, weight, age)

Measures of central tendency describe the center or typical value of a dataset. The three main measures are mean, median, and mode. **1. MEAN (Arithmetic Average)** The mean is the sum of all values divided by the number of values. Formula: Mean = (Sum of all values) ÷ (Number of values) **Step-by-Step Example:** Find the mean of test scores: 85, 90, 78, 92, 88, 85, 95 Step 1: Add all values 85 + 90 + 78 + 92 + 88 + 85 + 95 = 613 Step 2: Count the number of values Number of scores = 7 Step 3: Divide sum by count Mean = 613 ÷ 7 = 87.57 **2. MEDIAN** The median is the middle value when data is arranged in order. **For Odd Number of Values:** Step 1: Arrange data in ascending order Step 2: Find the middle position: (n+1)÷2 Step 3: The value at this position is the median **For Even Number of Values:** Step 1: Arrange data in ascending order Step 2: Find the two middle values Step 3: Calculate their average **Example with Odd Values:** Data: 12, 15, 18, 20, 25, 30, 35 Median position = (7+1)÷2 = 4th position Median = 20 **Example with Even Values:** Data: 10, 15, 20, 25, 30, 35 Middle positions: 3rd and 4th values Median = (20 + 25)÷2 = 22.5 **3. MODE** The mode is the value that appears most frequently. **Types of Mode:** - **Unimodal**: One mode - **Bimodal**: Two modes - **Multimodal**: More than two modes - **No mode**: All values appear equally **Example:** Data: 3, 5, 7, 5, 9, 5, 12 The value 5 appears 3 times (most frequent) Mode = 5 (unimodal)

Measures of variability describe how spread out the data values are from the center. **1. RANGE** Range is the difference between the maximum and minimum values. Formula: Range = Maximum value - Minimum value **Example:** Test scores: 65, 70, 75, 80, 85, 90, 95 Range = 95 - 65 = 30 **2. VARIANCE AND STANDARD DEVIATION** Variance measures the average squared deviation from the mean. Standard deviation is the square root of variance. **Steps to Calculate Variance and Standard Deviation:** Step 1: Find the mean Step 2: Find each deviation from the mean Step 3: Square each deviation Step 4: Find the average of squared deviations (variance) Step 5: Take the square root of variance (standard deviation) **Detailed Example:** Data: 10, 12, 14, 16, 18 Step 1: Mean = (10+12+14+16+18)÷5 = 70÷5 = 14 Step 2: Deviations from mean: 10-14 = -4 12-14 = -2 14-14 = 0 16-14 = 2 18-14 = 4 Step 3: Square the deviations: (-4)² = 16 (-2)² = 4 (0)² = 0 (2)² = 4 (4)² = 16 Step 4: Variance = (16+4+0+4+16)÷5 = 40÷5 = 8 Step 5: Standard deviation = √8 = 2.83 **Interpretation:** - Small variance/standard deviation = data points close to mean - Large variance/standard deviation = data points spread out from mean

Counting principles help us determine the number of ways events can occur, which is fundamental for probability calculations. **1. FUNDAMENTAL COUNTING PRINCIPLE** If event E₁ can occur in m ways and event E₂ can occur in n ways, then both events together can occur in m × n ways. **Example:** A student has 4 shirts and 3 pants. How many different outfits are possible? Solution: 4 × 3 = 12 different outfits **Extended Example:** A phone number format: 09XX-XXX-XXXX - First two digits: fixed as 09 - Third digit: can be 1,2,3,4,5,6,7,8,9 (9 choices) - Fourth digit: can be 0,1,2,3,4,5,6,7,8,9 (10 choices) - Remaining 7 digits: each can be 0-9 (10 choices each) Total possibilities = 1 × 1 × 9 × 10 × 10⁷ = 9 × 10⁸ **2. FACTORIAL NOTATION** n! = n × (n-1) × (n-2) × ... × 3 × 2 × 1 Examples: 3! = 3 × 2 × 1 = 6 4! = 4 × 3 × 2 × 1 = 24 5! = 5 × 4 × 3 × 2 × 1 = 120 Note: 0! = 1 (by definition) **3. PERMUTATIONS** A permutation is an arrangement where ORDER MATTERS. **Formula for n objects taken r at a time:** ₙPᵣ = n!/(n-r)! **Step-by-Step Example:** In how many ways can 5 students be arranged in a row for a photo? Solution: n = 5 (total students) r = 5 (all students) ₅P₅ = 5!/(5-5)! = 5!/0! = 5!/1 = 120 ways **Another Example:** How many ways can we select and arrange 3 officers (President, VP, Secretary) from 10 candidates? Solution: n = 10 (total candidates) r = 3 (positions to fill) ₁₀P₃ = 10!/(10-3)! = 10!/7! = 10×9×8 = 720 ways **4. PERMUTATIONS WITH REPETITION** When objects are identical, we divide by the factorial of repeated objects. Formula: n!/(n₁! × n₂! × n₃! × ...) **Example:** How many distinct arrangements of the letters in BANANA? Solution: Total letters = 6 A appears 3 times N appears 2 times B appears 1 time Arrangements = 6!/(3! × 2! × 1!) = 720/(6 × 2 × 1) = 720/12 = 60

Combinations count arrangements where order doesn't matter, and probability measures the likelihood of events. **1. COMBINATIONS** A combination is a selection where ORDER DOESN'T MATTER. **Formula:** ₙCᵣ = n!/[(n-r)! × r!] = ₙPᵣ/r! **Step-by-Step Example:** A committee of 3 students is chosen from 8 students. How many combinations are possible? Step 1: Identify n and r n = 8 (total students) r = 3 (students to choose) Step 2: Apply formula ₈C₃ = 8!/[(8-3)! × 3!] = 8!/(5! × 3!) Step 3: Calculate ₈C₃ = (8×7×6×5!)/(5! × 3×2×1) = (8×7×6)/(3×2×1) = 336/6 = 56 **Key Difference: Permutation vs Combination** - Selecting President, VP, Secretary from 8 people: ₈P₃ = 336 (order matters) - Selecting 3-person committee from 8 people: ₈C₃ = 56 (order doesn't matter) **2. BASIC PROBABILITY** Probability measures the likelihood of an event occurring. **Formula:** P(E) = Number of favorable outcomes / Total number of possible outcomes **Basic Concepts:** - **Experiment**: A process that produces outcomes - **Sample Space (S)**: Set of all possible outcomes - **Event (E)**: Subset of sample space - **Outcome**: Individual result of an experiment **Properties of Probability:** 1. 0 ≤ P(E) ≤ 1 2. P(certain event) = 1 3. P(impossible event) = 0 4. Sum of all probabilities in sample space = 1 **Step-by-Step Example:** A fair die is rolled. What is the probability of getting an even number? Step 1: Identify sample space S = {1, 2, 3, 4, 5, 6} Total outcomes = 6 Step 2: Identify favorable outcomes Even numbers: E = {2, 4, 6} Favorable outcomes = 3 Step 3: Calculate probability P(even) = 3/6 = 1/2 = 0.5 or 50% **3. PROBABILITY WITH COMBINATIONS** **Example:** A bag contains 5 red balls and 3 blue balls. What is the probability of selecting 2 red balls when drawing 2 balls without replacement? Step 1: Calculate total ways to select 2 balls from 8 ₈C₂ = 8!/(6!×2!) = (8×7)/(2×1) = 28 Step 2: Calculate ways to select 2 red balls from 5 ₅C₂ = 5!/(3!×2!) = (5×4)/(2×1) = 10 Step 3: Calculate probability P(2 red balls) = 10/28 = 5/14 ≈ 0.357 or 35.7%

Advanced probability concepts include compound events, conditional probability, and independence. **1. COMPLEMENT OF AN EVENT** The complement of event E (written as Ē or E') consists of all outcomes not in E. **Formula:** P(Ē) = 1 - P(E) **Example:** If P(passing exam) = 0.85, then P(failing exam) = 1 - 0.85 = 0.15 **2. ADDITION RULE** For any two events A and B: P(A or B) = P(A) + P(B) - P(A and B) **For Mutually Exclusive Events** (cannot occur simultaneously): P(A or B) = P(A) + P(B) **Step-by-Step Example:** A card is drawn from a standard deck. What is the probability of drawing a red card or a face card? Step 1: Identify probabilities P(red card) = 26/52 = 1/2 P(face card) = 12/52 = 3/13 P(red face card) = 6/52 = 3/26 Step 2: Apply addition rule P(red or face) = P(red) + P(face) - P(red and face) P(red or face) = 26/52 + 12/52 - 6/52 = 32/52 = 8/13 **3. MULTIPLICATION RULE** **For Independent Events:** P(A and B) = P(A) × P(B) **Example:** Two coins are tossed. What is the probability of getting two heads? Solution: P(first head) = 1/2 P(second head) = 1/2 P(two heads) = 1/2 × 1/2 = 1/4 **For Dependent Events:** P(A and B) = P(A) × P(B|A) where P(B|A) is the probability of B given that A occurred. **4. CONDITIONAL PROBABILITY** P(B|A) = P(A and B) / P(A) **Step-by-Step Example:** A bag contains 3 red and 2 blue marbles. Two marbles are drawn without replacement. What is the probability the second marble is red given the first was red? Step 1: After first red marble is drawn Remaining marbles: 2 red, 2 blue (total 4) Step 2: Calculate conditional probability P(second red | first red) = 2/4 = 1/2 **Alternative method using formula:** P(both red) = (3/5) × (2/4) = 6/20 = 3/10 P(first red) = 3/5 P(second red | first red) = (3/10) ÷ (3/5) = (3/10) × (5/3) = 1/2 **5. TREE DIAGRAMS** Tree diagrams help visualize sequential probability problems. **Example:** A bag contains 4 red and 6 blue balls. Two balls are drawn with replacement. Find P(at least one red). **Method 1: Direct calculation** P(at least one red) = P(RR) + P(RB) + P(BR) = (4/10)(4/10) + (4/10)(6/10) + (6/10)(4/10) = 16/100 + 24/100 + 24/100 = 64/100 = 16/25 **Method 2: Using complement** P(at least one red) = 1 - P(no red) = 1 - P(BB) = 1 - (6/10)(6/10) = 1 - 36/100 = 64/100 = 16/25

Sampling involves selecting a subset of a population to make inferences about the entire population. Proper sampling techniques are crucial for obtaining reliable results. **1. WHY SAMPLE?** - **Cost-effective**: Cheaper than surveying entire population - **Time-efficient**: Faster data collection - **Practical**: Some populations are infinite or inaccessible - **Accuracy**: Well-designed samples can be very representative **2. SAMPLING METHODS** **A. SIMPLE RANDOM SAMPLING (SRS)** Every member has equal chance of selection. **Step-by-Step Process:** Step 1: Create a complete list of population members Step 2: Assign each member a unique number Step 3: Use random number generator to select sample **Example:** Select 50 students from 1,000 high school students. - Number students 0001-1000 - Use random number table to select 50 numbers - Survey students with selected numbers **B. STRATIFIED SAMPLING** Divide population into homogeneous groups (strata), then randomly sample from each group. **Step-by-Step Example:** Survey student satisfaction in a university with 40% freshmen, 30% sophomores, 20% juniors, 10% seniors. For sample size of 200: Step 1: Calculate strata sizes - Freshmen: 40% × 200 = 80 students - Sophomores: 30% × 200 = 60 students - Juniors: 20% × 200 = 40 students - Seniors: 10% × 200 = 20 students Step 2: Randomly sample from each stratum **C. SYSTEMATIC SAMPLING** Select every kth member from ordered population. **Formula:** k = Population size ÷ Sample size **Example:** Select 100 students from 2,000 students. k = 2,000 ÷ 100 = 20 Start randomly (say #7), then select every 20th: 7, 27, 47, 67, ... **D. CLUSTER SAMPLING** Divide population into clusters, randomly select clusters, survey all members in selected clusters. **Example:** Study Manila public schools. - Clusters: individual schools - Randomly select 10 schools - Survey all students in selected schools **E. CONVENIENCE SAMPLING** (Non-probability) Sample easily accessible members. **Warning:** May introduce significant bias! **3. SAMPLE SIZE CONSIDERATIONS** **Factors Affecting Sample Size:** - **Population size**: Larger populations need larger samples - **Desired precision**: Higher precision needs larger samples - **Population variability**: More variable populations need larger samples - **Confidence level**: Higher confidence needs larger samples **Rule of Thumb:** - Sample size of 30+ usually adequate for normal approximation - For proportions, need at least 5 successes and 5 failures **4. SAMPLING BIAS** **Types of Bias:** **A. Selection Bias** Sample doesn't represent population. **Example:** Surveying only smartphone users about internet usage excludes those without smartphones. **B. Non-response Bias** Selected participants don't respond. **Example:** Mail survey gets 20% response rate; non-responders may differ from responders. **C. Response Bias** Participants give inaccurate answers. **Example:** Asking about illegal behavior - people may lie. **D. Volunteer Bias** Volunteers differ from general population. **5. MARGIN OF ERROR** Indicates precision of sample estimate. **Interpretation:** If poll shows 55% ± 3% support, true population percentage is likely between 52% and 58%. **Factors Reducing Margin of Error:** - Larger sample size - Less population variability - Better sampling method