LET Secondary Mathematics — Statistics and ProbabilityStudy Notes

Data presentation is the foundation of statistical literacy. Raw data—a list of numbers—is difficult to interpret. Organized into tables and graphs, the same data reveals patterns instantly. The LET tests your ability to select the correct display for a purpose and to read existing graphs accurately, catching misleading presentations. Five main types of displays serve different purposes: **Frequency Table**: Lists each value (or class interval) with its frequency (count). It is the starting point for all other displays and for computing measures like the mean and median of grouped data. Example: Recording 30 students' heights in 5 cm intervals (140–144 cm, 145–149 cm, etc.) with the number of students in each group. **Bar Graph**: Compares discrete categories (not numerical order). Each category gets a separate bar; bars do not touch. Use it to compare sales of three textbook publishers, enrollment by grade level, or quiz scores by topic. The height of each bar represents the frequency or amount. Reading a bar graph requires identifying the axes (what is being measured and in what units), the scale (the value of each grid line), and reading the bar heights accurately. **Histogram**: Displays the distribution of continuous data divided into intervals (class intervals). Unlike a bar graph, the bars touch, showing a continuous range. The width of each bar represents the class interval; the height represents frequency or relative frequency (percentage). Use histograms to display exam scores grouped in ranges (50–59, 60–69, etc.) or student heights. The shape of a histogram reveals whether data is symmetrical, skewed left, or skewed right. **Line Graph**: Shows change over time. Time is placed on the horizontal axis (x-axis); the measured quantity on the vertical axis (y-axis). Connect the points to show trends. Use line graphs to track monthly rainfall, student attendance over a school year, or improvement in reading fluency across quarterly assessments. A rising line shows increase; a falling line shows decrease; a flat line shows no change. Line graphs are essential for interpreting how a class's average performance changes across quarters. **Pie (Circle) Graph**: Shows parts of a single whole as percentages or angles. The entire pie = 360° or 100%. Each slice's angle = (category value ÷ total) × 360°; its percentage = (category value ÷ total) × 100%. Use pie graphs to show how a school's budget is allocated, how a class's time is spent (instruction, assessment, transition), or the composition of a student body by learning modality. A pie graph cannot compare two separate wholes; for that, use two pie graphs side by side or a bar graph. **Pictograph**: Uses symbols or icons, each representing a fixed count (e.g., one sun symbol = 10 days of sunshine). It is visually engaging, especially for primary grades, but is less precise than bars or numbers for reading exact values. **Reading Graphs Accurately**: Before answering a question, identify (1) the title (what is being shown), (2) the axis labels and units (what is measured), (3) the scale (the value increments), and (4) the legend (if multiple data sets are shown). Common reading errors include misreading the scale (thinking each grid line is 1 when it is 10), confusing categories, or missing a legend. **Spotting Misleading Graphs**: A graph is misleading if (1) the vertical axis does not start at zero (exaggerating differences), (2) intervals on an axis are unequal (distorting the visual impression), (3) the visual size of bars or symbols is not proportional to the data (using 3D effects, varying bar widths, or symbol sizes), or (4) the data is incomplete or out of context. The LET often includes a question asking you to identify why a graph is misleading or to correct it.

Central tendency describes the "typical" or "middle" value of a data set. Three measures serve different purposes, and choosing the right one is critical. **Mean (Average)**: The sum of all values divided by the count. Formula: Mean = (Σ values) / n, where Σ means "sum of" and n is the number of values. Strengths: Uses all data points; mathematically convenient. Weakness: Sensitive to outliers (extremely high or low values that pull the mean away from the bulk of the data). Example: Class quiz scores 10, 10, 10, 10, 100 (one student scored perfectly). Mean = (40 + 100) ÷ 5 = 28, which misrepresents the typical score (four students scored 10, not 28). **Median**: The middle value when data is arranged in order. For an odd count, it is the single middle value. For an even count, it is the average of the two middle values. Strengths: Resists outliers; best for skewed data (where a few extreme values exist). Weakness: Does not use all data points (ignores how far values are from the middle). Example: Quiz scores 10, 10, 10, 10, 100. In order: 10, 10, 10, 10, 100. The middle (3rd) value is 10, so the median is 10. This better represents the typical score than the mean (28). For an even count (e.g., 6 values), arrange them in order and average the 3rd and 4th values. If scores are 8, 9, 9, 11, 12, 15, the median = (9 + 11) ÷ 2 = 10. **Mode**: The value that occurs most often (has the highest frequency). Strengths: Easy to identify; can be used for categorical (non-numerical) data (e.g., "which learning modality do students prefer?"). Weakness: A set can have no mode (all values appear once), one mode (unimodal), or multiple modes (bimodal or multimodal). The mode alone does not describe spread or typical magnitude. Example: Quiz scores 10, 10, 10, 10, 100. The mode is 10 (appears four times). **When to Use Each Measure**: - **Use the mean** for normally distributed numerical data (symmetric, no outliers) where you want a single summary value, as in computing class averages. - **Use the median** for skewed data, data with outliers, or when you want the "middle" value regardless of extremes. In the Philippines, household income data is skewed (a few very wealthy families pull the mean up); the median income better represents the typical family. - **Use the mode** for categorical data (learning preferences, most common mistake) or when the most frequent outcome is the focus. In a classroom, the mode of error types tells you what to reteach. **Weighted Mean**: A variation used when values have different "weights" or frequencies. Formula: Weighted Mean = (Σ value × weight) / (Σ weight). Example: A final grade is computed as Quizzes (weight 30%) with a score of 80, and Final Exam (weight 70%) with a score of 90. Grade = (80 × 0.30) + (90 × 0.70) = 24 + 63 = 87. This is how teachers compute grades: each component (quizzes, assignments, exams) is multiplied by its weight, summed, and divided by the total weight (which always sums to 1 or 100%). **Mean of Grouped Data (Frequency Table)**: When data is summarized in a frequency table with class intervals, the mean is computed using the midpoint of each interval as the representative value for that class, then using the weighted mean formula. Example: A short quiz gave: Score 5 to 2 students, Score 6 to 3 students, Score 7 to 5 students. Mean = [(5 × 2) + (6 × 3) + (7 × 5)] / (2 + 3 + 5) = (10 + 18 + 35) / 10 = 63 / 10 = 6.3. This is critical for interpreting class distributions: a mean of 6.3 out of 10 suggests the class is below expected performance (typically 75% or 7.5 out of 10 is the passing standard in the K–12 BEC).

While central tendency describes the "center," dispersion describes how **spread out** the data is. Two classes can have the same mean but very different distributions: one consistent, the other highly variable. Dispersion measures reveal this critical difference. **Range**: The simplest measure; the difference between the highest and lowest values. Formula: Range = Maximum - Minimum. Strengths: Easy to compute. Weakness: Sensitive to outliers (a single extreme value inflates the range) and ignores the distribution in the middle. Example: Class A quiz scores: 8, 9, 9, 10, 10. Range = 10 - 8 = 2. Class B scores: 3, 6, 9, 10, 10. Range = 10 - 3 = 7. Class B has a wider range, suggesting greater variability in mastery, despite both having the same mean (9.2). **Variance**: The average of the squared deviations from the mean. It measures how far, on average, each value is from the mean (squared to make all differences positive). Formula: Variance = Σ(value - mean)² / n. Strengths: Uses all data and accounts for the magnitude of deviations. Weakness: Measured in squared units (awkward to interpret); hard to compute by hand. Example: Scores 1, 3, 5, 7, 9. Mean = 5. Deviations: 1-5=-4, 3-5=-2, 5-5=0, 7-5=2, 9-5=4. Squared: 16, 4, 0, 4, 16. Sum = 40. Variance = 40 ÷ 5 = 8. **Standard Deviation (SD)**: The square root of variance. Formula: SD = √Variance. Strengths: Measured in the same units as the data (easy to interpret); standard in statistics. A **larger SD means the data is more spread out** (less consistent). Weakness: More complex to compute. Example (continuing above): SD = √8 ≈ 2.83. This means values deviate from the mean by about 2.83 units, on average. Interpretation: If a class's test scores have a mean of 80 and SD of 5, most scores cluster tightly around 80 (consistent). If another class has a mean of 80 and SD of 15, scores range widely (inconsistent instruction or diverse mastery levels). **Why Dispersion Matters in Teaching**: The K–12 BEC emphasizes **differentiated instruction**. If your class has high SD in reading fluency, you must group students by level and adjust instruction. If SD is low, students are more similar, and whole-class instruction may work. Dispersion directly informs your classroom management and pedagogy. **Practical Classroom Application**: When you analyze quarterly test scores, compute both the mean (typical performance) and SD (consistency). A mean of 75% is passing; but an SD of 20% suggests half your class is far below 75% and half is far above, signaling the need for differentiation (per RA 7836, your duty as a professional teacher is to support all learners equitably). **Computing Standard Deviation Step-by-Step**: 1. Find the mean. 2. Subtract the mean from each value (compute deviations). 3. Square each deviation. 4. Sum the squared deviations. 5. Divide by the number of values (compute variance). 6. Take the square root (compute SD).

Measures of position locate a specific value within an ordered data set. They answer questions like "What score puts a student in the top 25% of the class?" or "At what point does a student's performance exceed 75% of peers?" These are essential for interpreting standardized test results (a student's percentile rank) and for grouping students by performance level. **Quartiles**: Divide an ordered data set into four equal parts, each containing 25% of the data. - **Q1 (Lower Quartile)**: 25% of the data falls below this value. - **Q2 (Median)**: 50% of the data falls below this value (the median we studied earlier). - **Q3 (Upper Quartile)**: 75% of the data falls below this value. - **The Interquartile Range (IQR)** = Q3 - Q1; it captures the middle 50% of the data, resistant to outliers. **Percentiles**: Divide an ordered data set into 100 equal parts. The **pth percentile** is the value below which p% of the data falls. A student at the 90th percentile scores better than 90% of peers; a student at the 25th percentile scores better than 25% of peers (and worse than 75%). Percentiles are widely used to interpret standardized test results (e.g., "Your child scored at the 85th percentile in reading") and to identify gifted or at-risk students. **The Five-Number Summary**: Provides a quick snapshot of a data set's center and spread. It consists of: 1. **Minimum** (lowest value) 2. **Q1** (lower quartile) 3. **Q2** (median) 4. **Q3** (upper quartile) 5. **Maximum** (highest value) Example: Test scores (ordered): 42, 58, 65, 72, 75, 78, 82, 88, 92, 95. n = 10. - Minimum = 42 - Q1 = median of lower half (42, 58, 65, 72, 75) = 65 - Q2 (Median) = (75 + 78) ÷ 2 = 76.5 - Q3 = median of upper half (78, 82, 88, 92, 95) = 88 - Maximum = 95 - IQR = Q3 - Q1 = 88 - 65 = 23 **Box-and-Whisker Plot**: A visual display of the five-number summary. A box spans Q1 to Q3 (IQR), with a line inside marking the median (Q2). Whiskers (lines) extend from Q1 to the minimum and from Q3 to the maximum. This plot immediately shows the center, spread, and any skewness. **Identifying Outliers**: A value is often considered an outlier if it falls below Q1 - 1.5(IQR) or above Q3 + 1.5(IQR). For the example above: Q1 - 1.5(23) = 65 - 34.5 = 30.5, and Q3 + 1.5(23) = 88 + 34.5 = 122.5. Since all values fall between 30.5 and 122.5, there are no outliers. **Classroom Application**: When you administer a unit test to your 30 students, computing quartiles identifies three groups: - **Students at Q3 and above** (top 25%): Ready for enrichment. - **Students between Q1 and Q3** (middle 50%): Meeting grade-level standards; continue current instruction. - **Students below Q1** (bottom 25%): Needing remediation; provide additional support (per DepEd's Learner Progress Monitoring System and your ethical duty under RA 7836). This is data-driven differentiation and is a key competency tested on the LET.

Counting is the foundation of probability: before you can compute a probability, you must know how many possible outcomes exist. The **Fundamental Counting Principle** systematizes this. **The Fundamental Counting Principle**: If one event can happen in **m** ways and a second independent event can happen in **n** ways, then both events together can happen in **m × n** ways. This extends to any number of sequential choices: multiply the number of choices at each stage. Example: A student must choose an outfit for school: 4 shirts and 3 pairs of pants. Total outfits = 4 × 3 = 12. (The choice of shirt does not affect which pants are available; they are independent.) Example: Rolling a die, then flipping a coin. Die outcomes = 6; coin outcomes = 2. Total outcomes = 6 × 2 = 12. (The die result does not change the coin's probability.) **When Order Matters: Permutations** A **permutation** is an arrangement of items where **order matters**. If you rank items (1st place, 2nd place, 3rd place) or arrange them in a sequence, order is significant. Formula for permutations: **P(n, r)** = the number of ways to arrange r items chosen from n distinct items. P(n, r) = n × (n-1) × (n-2) × ... × (n-r+1) [multiply r factors starting from n]. Special case: **n!** (n factorial) = the number of ways to arrange all n items. n! = n × (n-1) × (n-2) × ... × 1. For example, 4! = 4 × 3 × 2 × 1 = 24. Example: A teacher must assign three students (Alex, Bella, Carlos) as first-presenter, second-presenter, and third-presenter from a class of 10. This is P(10, 3) = 10 × 9 × 8 = 720 ways. (The choice of 1st limits the 2nd, which limits the 3rd; order matters because the roles differ.) Example: Arranging 5 books on a shelf. P(5, 5) = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways. **When Order Does Not Matter: Combinations** A **combination** is a selection of items where **order does not matter**. If you choose a committee or a group (where all members have equal roles), the order of selection is irrelevant. Formula for combinations: **C(n, r)** = the number of ways to choose r items from n distinct items. C(n, r) = P(n, r) ÷ r! = [n × (n-1) × ... × (n-r+1)] ÷ [r × (r-1) × ... × 1]. Example: A teacher must form a 3-person committee from a class of 10 students. This is C(10, 3) = (10 × 9 × 8) ÷ (3 × 2 × 1) = 720 ÷ 6 = 120 ways. (Choosing Alex, Bella, Carlos is the same committee as Carlos, Bella, Alex because all have equal roles; order does not matter.) Note: C(10, 3) is much smaller than P(10, 3) because order is ignored. **The Key Distinction**: If swapping two selected items produces a **different outcome**, use **permutations**. If swapping produces the **same outcome**, use **combinations**. Example (Permutation): Assigning three students to different roles—president, treasurer, secretary. Swapping Alex (president) with Bella (treasurer) creates a different outcome; order matters. Use P(n, r). Example (Combination): Choosing three students for a group project. Swapping which of Alex, Bella, or Carlos is listed first does not change the group; order does not matter. Use C(n, r). **Practical Classroom Scenario**: - **Permutation question**: "In how many ways can you arrange 6 students in a line for a class photo?" Answer: 6! = 720. - **Combination question**: "In how many ways can you choose 3 students from 6 to form a learning group?" Answer: C(6, 3) = 20. **LET Exam Tips**: 1. Always ask: Does the order of selection or arrangement matter? 2. If roles or positions are distinct (1st, 2nd, 3rd; president, vice-president), it is a permutation. 3. If the selection is a group with no ranks, it is a combination. 4. Write out P(n, r) or C(n, r) carefully; computational errors are common.

**Probability** measures the likelihood of an event occurring. It is a number from **0 (impossible) to 1 (certain)**. A probability of 0.5 means the event is equally likely to happen or not happen (like a fair coin flip). **Definition of Probability (Classical Approach)**: For an experiment with equally likely outcomes, **P(Event E) = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)** Example: Drawing a card from a standard 52-card deck. There are 13 spades out of 52 total cards. P(drawing a spade) = 13/52 = 1/4 = 0.25. **Properties of Probability**: 1. **0 ≤ P(E) ≤ 1**: Probability is always between 0 and 1 (inclusive). 2. **P(E) = 0**: The event is impossible (e.g., rolling a 7 on a standard die). 3. **P(E) = 1**: The event is certain to happen (e.g., rolling a number 1–6 on a die). 4. **Sum of all outcomes = 1**: If an experiment has mutually exclusive outcomes that cover all possibilities, their probabilities sum to 1. Example: P(heads) + P(tails) = 0.5 + 0.5 = 1. **The Complement Rule**: The complement of event E (written E' or "not E") is the event that E does not happen. **P(not E) = 1 - P(E)**. This is often the fastest way to solve "at least one" problems. Example: A bag holds 5 red, 3 blue, and 2 green marbles (10 total). Find P(not red). - P(red) = 5/10 = 1/2. - P(not red) = 1 - 1/2 = 1/2. Alternatively, direct count: P(not red) = (3 blue + 2 green) / 10 = 5/10 = 1/2. Both methods match. **The Complement Rule is Powerful for "At Least One" Problems**: Example: Two fair coins are tossed. Find P(at least one head). - Direct method: Outcomes are HH, HT, TH, TT (4 total). Favorable (at least one H): HH, HT, TH (3). P = 3/4. - Complement method: P(at least one H) = 1 - P(no heads) = 1 - P(TT) = 1 - 1/4 = 3/4. Faster! Example: A student guesses on a 5-question multiple-choice test (each with 4 options). Find P(at least one correct). - P(one question correct by guessing) = 1/4. P(one question wrong) = 3/4. - P(all 5 wrong) = (3/4)^5 ≈ 0.237. - P(at least one correct) = 1 - 0.237 ≈ 0.763 or about 76%. **Equally Likely Outcomes Assumption**: The definition of probability assumes all outcomes are equally likely. A fair die has each face with probability 1/6. A biased die (weighted) violates this; you'd need to know the actual probabilities. The LET specifies "fair" or "unbiased" to signal equal likelihood. **Practical Examples in a Classroom Context**: 1. **Lottery or drawing**: A classroom has 15 girls and 10 boys (25 students total). If a student is chosen at random for a presentation, what is the probability it is a girl? P(girl) = 15/25 = 3/5 = 0.6 = 60%. 2. **Assessment**: A unit test has 20 multiple-choice questions. A student guesses randomly on 5 questions (4 options each). What is P(at least one guess is correct)? P(all 5 wrong) = (3/4)^5 ≈ 0.237. P(at least one right) = 1 - 0.237 ≈ 0.763 or 76%. 3. **Learning preference (categorical)**: A teacher surveys 24 students on preferred learning modality: Modular Distance Learning (10), In-Person (12), Hybrid (2). If you randomly select a student, what is P(they prefer In-Person)? P(In-Person) = 12/24 = 1/2 = 0.5. **LET-Style Question**: A spinner has 8 equally spaced sections: 3 red, 2 blue, 2 green, 1 yellow. When spun, what is P(red or blue)? P(red) = 3/8. P(blue) = 2/8. Since a spin cannot be both red and blue simultaneously (mutually exclusive), P(red or blue) = 3/8 + 2/8 = 5/8. (This introduces compound events, covered next.)

Compound events combine two or more simple events using "and" or "or." The calculation depends on whether the events are independent (one does not affect the other) or dependent (one affects the other), and whether they are mutually exclusive (cannot both happen) or overlapping (can both happen). **The Multiplication Rule ("And"): P(A and B)** When finding the probability that both event A and event B occur, we multiply probabilities. However, the formula differs depending on independence. **Independent Events**: Events where one outcome does not affect the other. Examples: rolling a die, then flipping a coin; drawing a card, replacing it, then drawing again; two separate student performances. Formula: **P(A and B) = P(A) × P(B)** (for independent events) Example: A fair die is rolled, then a fair coin is flipped. Find P(rolling 3 and flipping heads). - P(rolling 3) = 1/6. - P(flipping heads) = 1/2. - P(3 and heads) = 1/6 × 1/2 = 1/12 ≈ 0.083. Example: Two students independently guess on a multiple-choice question (4 options, probability of correct = 1/4 each). Find P(both guess correctly). - P(student 1 correct) = 1/4. - P(student 2 correct) = 1/4. - P(both correct) = 1/4 × 1/4 = 1/16 = 0.0625. **Dependent Events**: Events where the outcome of one affects the outcome of the other. The most common scenario is drawing without replacement (once an item is chosen, it is not returned, changing the total and favorable counts for the next draw). Formula: **P(A and B) = P(A) × P(B|A)**, where P(B|A) is the conditional probability of B given that A has occurred. Example: A bag holds 5 red and 3 blue marbles (8 total). Two marbles are drawn without replacement. Find P(both red). - P(first red) = 5/8. - After removing one red, 4 red and 3 blue remain (7 total). - P(second red | first was red) = 4/7. - P(both red) = 5/8 × 4/7 = 20/56 = 5/14 ≈ 0.357. Note the denominator changes from 8 to 7 (one fewer marble remains). Example: A teacher selects 2 students from a class of 10 (without replacement) to present. Find P(both are boys) if 6 of the 10 are boys. - P(first is boy) = 6/10. - P(second is boy | first was boy) = 5/9 (9 students left, 5 boys left). - P(both boys) = 6/10 × 5/9 = 30/90 = 1/3 ≈ 0.333. **The Addition Rule ("Or"): P(A or B)** When finding the probability that at least one of event A or event B occurs, we add probabilities. However, we must subtract the overlap (probability of both) to avoid double-counting. Formula: **P(A or B) = P(A) + P(B) - P(A and B)** **Mutually Exclusive Events**: Events that cannot both happen (no overlap). Examples: rolling a die and getting either a 3 or a 5 (not both on one roll); a card being a king or a queen (not both at once). For mutually exclusive events, P(A and B) = 0, so the formula simplifies: **P(A or B) = P(A) + P(B)** (for mutually exclusive events) Example: A card is drawn from a 52-card deck. Find P(king or queen). - P(king) = 4/52. - P(queen) = 4/52. - A card cannot be both king and queen simultaneously (mutually exclusive). - P(king or queen) = 4/52 + 4/52 = 8/52 = 2/13 ≈ 0.154. **Overlapping (Non-Mutually Exclusive) Events**: Events that can both happen. Example: drawing a red card or a face card from a deck (some red face cards satisfy both conditions). Formula: **P(A or B) = P(A) + P(B) - P(A and B)** Example: A card is drawn. Find P(red or face card). - P(red) = 26/52 (13 hearts + 13 diamonds). - P(face card) = 12/52 (3 face cards × 4 suits). - P(red and face card) = 6/52 (red jacks, queens, kings = 6 cards). - P(red or face card) = 26/52 + 12/52 - 6/52 = 32/52 = 8/13 ≈ 0.615. If we had added without subtracting the overlap, we would get 26/52 + 12/52 = 38/52 = 19/26, which double-counts the 6 red face cards. **Summary Table**: | Scenario | Formula | Example | |----------|---------|----------| | Independent "and" | P(A) × P(B) | Die roll, then coin flip | | Dependent "and" | P(A) × P(B\|A) | Draw two cards without replacement | | Mutually exclusive "or" | P(A) + P(B) | Roll 3 or 5 on one die | | Overlapping "or" | P(A) + P(B) - P(A and B) | Red or face card from deck | **Practical Classroom Scenario**: A teacher draws names from a class roster to form a small reading group. If 12 of 20 students are proficient readers and 8 are struggling: - P(first selected is proficient) = 12/20 = 0.6. - P(second selected is proficient | first was proficient) = 11/19 ≈ 0.579. - P(both proficient) = (12/20) × (11/19) = 132/380 ≈ 0.347 or 35%. This informs how the teacher groups students: to ensure mixed-ability groups for peer support, the teacher should recognize that two independent "proficient" draws have only a 35% probability. **LET Exam Tips**: 1. Identify whether events are independent (separate, unaffected) or dependent (sequential, affecting one another). 2. For "and," multiply; for "or," add (and subtract the overlap if not mutually exclusive). 3. Watch the denominator in dependent (without replacement) scenarios; it changes after each draw. 4. If "at least one," consider using the complement rule: P(at least one) = 1 - P(none).