Statistics questions test conceptual understanding more than calculation. You need to know what each measure means and how it responds when the data changes — not just how to compute it.

A) The Three Averages — What They Each Measure

Given a set of numbers, three different "averages" describe different aspects of the data:

• Mean — the arithmetic average. Add all values, divide by count. Sensitive to extreme values (outliers).
  Example: {3, 5, 7, 9, 11} → Mean = (3+5+7+9+11)/5 = 35/5 = 7
• Median — the middle value when sorted. Not affected by extremes.
  Example: {3, 5, 7, 9, 11} → Median = 7 (middle of 5 values).
  Even count: {3, 5, 7, 9} → Median = (5+7)/2 = 6 (average of 2nd and 3rd).
• Mode — the most frequently occurring value. A set can have multiple modes or no mode.
  Example: {2, 3, 3, 5, 7, 7, 7} → Mode = 7.

Key insight: Mean = Sum ÷ Count, so Sum = Mean × Count. This lets you find the total from the average — extremely useful in GMAT word problems.

Example: A class of 30 students has an average score of 72. Total score = 30 × 72 = 2,160. If one student scored 90, the remaining 29 students scored 2160 − 90 = 2070 total, for an average of 2070/29 ≈ 71.4.

B) Frequency Distribution Tables

When data is given as "value | frequency" pairs rather than a raw list, use the frequency to compute weighted statistics.

• Mean: Σ(value × frequency) ÷ total frequency. Do NOT average the values alone.
• Median: find cumulative frequency; median is the value where cumulative count first reaches or exceeds (n+1)/2 for odd n, or the average of positions n/2 and n/2+1 for even n.
• Mode: value with the highest frequency.

Example: Value 2 appears 1×, value 3 appears 2×, value 4 appears 3×, value 5 appears 1×. Total n=7.
Mean = (2×1 + 3×2 + 4×3 + 5×1)/7 = (2+6+12+5)/7 = 25/7 ≈ 3.57.
Median = 4th value (cumulative: 2→1, 3→3, 4→6) = 4. Mode = 4.

C) Weighted Average

When combining groups with different sizes, you cannot simply average the group means — you must weight by group size.

Formula: Weighted mean = Σ(group mean × group size) / Σ(group sizes)

Example: Group A has 40 students with mean score 70. Group B has 60 students with mean score 80. Combined mean = (40×70 + 60×80) / (40+60) = (2800+4800)/100 = 76. Note: the answer is closer to 80 than 70 because Group B is larger — always sanity-check that your weighted mean is pulled toward the larger group.

D) Range and Standard Deviation — Measuring Spread

Range = max − min. Simple but sensitive to a single extreme value.

Standard deviation (SD) measures how far values are from the mean on average. Large SD = values scattered widely. Small SD = values clustered tightly around the mean.

• Set A = {5, 5, 5, 5} → all identical, mean = 5, SD = 0.
• Set B = {1, 3, 7, 9} → mean = 5, values spread far → SD is high.
• Set C = {4, 5, 5, 6} → mean = 5, values close together → SD is low.

The GMAT almost never asks you to compute SD from scratch. It tests directional effects:

• Adding the same constant to every value: mean shifts, SD stays the same.
• Multiplying every value by k: mean multiplies by k, SD multiplies by k.
• Adding a new element y: if |y − mean| > current SD, SD increases. If |y − mean| < current SD, SD decreases.

Example: Set {10, 20, 30} has mean 20. Add 50 to the set. The new value (50) is far from the mean (20), so SD increases. Add 19 instead — that's very close to the mean, so SD decreases.

E) Overlapping Sets

Overlapping sets problems describe a population divided by two or three characteristics. The core challenge is avoiding double-counting people who belong to more than one category.

Sets: Key Vocabulary

• An element is any member of a set. If x is in set S, write x ∈ S.
• |S| denotes the number of elements in S (its cardinality). Example: if S = {2, 4, 6}, then |S| = 3.
• Subset: S ⊆ T means every element of S is also in T. Example: {2, 4} ⊆ {1, 2, 3, 4}.
• Union A ∪ B: all elements in A or B (or both).
• Intersection A ∩ B: only the elements in both A and B simultaneously.
• Disjoint sets: A ∩ B = ∅ — no elements in common; the circles do not overlap.

General addition rule: |S ∪ T| = |S| + |T| − |S ∩ T|. Subtracting the intersection corrects for elements counted twice.

Example: 25 students — 14 study History, 17 study Math, and some study both. Since |H ∪ M| ≤ 25, we get |H ∩ M| = 14 + 17 − |H ∪ M| ≥ 14 + 17 − 25 = 6. At least 6 study both.

Venn diagram

A Venn diagram shows how two or more sets are related. Suppose sets S and T aren't disjoint, and neither is a subset of the other. The diagram below shows their intersection S ∩ T as a shaded area.

Two-Set Formula

|A ∪ B| = |A| + |B| − |A ∩ B|

Equivalently: Total = (A only) + (B only) + (Both) + (Neither)

Example: 100 students. 60 study Math, 40 study Science, 20 study both. At least one subject = 60+40−20 = 80. Neither = 100−80 = 20 students.

The 2×2 Matrix Method

For any problem with two binary characteristics (yes/no on each), draw a 2×2 grid with a Total row and column. Fill known values from the problem, then solve for unknowns algebraically. Every row and every column must sum to its Total — use this as a check.

Example: 200 people surveyed. 120 own a car, 90 own a bike, 50 own both. Find each group.

• Car only = 120 − 50 = 70. Bike only = 90 − 50 = 40. Neither = 200 − 120 − 90 + 50 = 40.
• The "neither" cell (bottom-right before the Total column) is part of the population — do not omit it.
• Strategy: fill the four inner cells first, then verify rows and columns match the margin totals.

Three-Set Formula (Inclusion-Exclusion)

|A ∪ B ∪ C| = |A| + |B| + |C| − |A∩B| − |A∩C| − |B∩C| + |A∩B∩C|

Memory cue: add the singles, subtract the pairs, add back the triple. The pattern alternates: add, subtract, add.

Example: 100 students. French=40, Spanish=35, German=25. French∩Spanish=15, French∩German=10, Spanish∩German=8. All three=5. How many study none?
At least one = 40+35+25 − 15−10−8 + 5 = 72. None = 100 − 72 = 28.

Venn Diagram — Fill from the Center Out

For 3-set problems, label the 7 regions of a Venn diagram from inside out:

1. Center (A∩B∩C): fill in the "all three" value first.
2. Three pair-only regions: A∩B only = (given A∩B) − center. Repeat for A∩C and B∩C.
3. Three outer regions: A only = |A| − (A∩B only) − (A∩C only) − center. Repeat for B and C.
4. Neither = Total − sum of all 7 interior regions.

F) Probability Fundamentals

Probability measures the likelihood of an uncertain outcome. An experiment is any process with an uncertain result (rolling a die, drawing a card). Each possible result is an outcome, and the set of all possible outcomes is the sample space. An event is any subset of the sample space — a specific outcome or group of outcomes you care about.

For an event E: P(E) = (number of outcomes in E) / (total number of equally likely outcomes). P(E) is always between 0 and 1 (inclusive).

• Impossible event: P(E) = 0 — E can never occur.
• Certain event: P(E) = 1 — E always occurs.
• Complement "not E": P(not E) = 1 − P(E). Critical for "at least one" problems.
• Union "E or F": P(E or F) = P(E) + P(F) − P(E and F).
• Intersection "E and F": P(E and F) — probability both occur simultaneously.
• Mutually exclusive: E and F cannot both occur → P(E and F) = 0 → P(E or F) = P(E) + P(F).
• Independent events: occurrence of E does not affect F → P(E and F) = P(E) × P(F).
• Dependent events: P(A and B) = P(A|B) × P(B), where P(A|B) = P(A and B) / P(B) is the conditional probability of A given B. The condition B shrinks your sample space to only outcomes where B is true.

Example (independent): Flip a coin and roll a die. P(heads and 6) = 1/2 × 1/6 = 1/12.

Example (dependent): A bag has 3 red and 2 blue balls. Draw without replacement. P(second red | first red) = 2/4 = 1/2. After removing one red, the sample space is now 4 balls with 2 red.

G) The "At Least One" Strategy

P(at least one) = 1 − P(none). Almost always faster than counting all favorable cases directly.

Example: P(at least one head in 3 coin flips) = 1 − P(all tails) = 1 − (1/2)³ = 1 − 1/8 = 7/8.

H) Binomial Probability

For exactly k successes in n independent trials, each with probability p:

P = C(n, k) × p^k × (1 − p)^n−k

Example: P(exactly 2 heads in 4 fair coin flips) = C(4,2) × (1/2)² × (1/2)² = 6 × 1/4 × 1/4 = 6/16 = 3/8.

I) Standard Deviation: Formal Computation

Standard deviation (SD) measures how spread out the values in a data set are around the mean. The GMAT Official Guide provides the formal 5-step computation procedure. Most GMAT questions test interpretation (what happens to SD when data changes), but understanding the computation deepens your intuition for those questions.

The 5-Step SD Calculation Procedure (from the GMAT OG)

1. Find the arithmetic mean of the n numbers.
2. Find the difference (x − mean) for each number x in the set.
3. Square each difference (x − mean)².
4. Find the average of the squared differences (divide their sum by n). This value is called the variance.
5. Take the nonnegative square root of that average. This is the standard deviation.

Formula summary: SD = √[ Σ(xᵢ − x̄)² / n ], where x̄ is the mean.

Worked Example (from GMAT OG)

Find the standard deviation of {0, 7, 8, 10, 10}.

• Step 1: Mean = (0 + 7 + 8 + 10 + 10)/5 = 35/5 = 7.
• Steps 2–3: Differences and squared differences shown in table above.
• Step 4: Variance = 68/5 = 13.6.
• Step 5: SD = √13.6 ≈ 3.7.

The second set {6, 6, 6.5, 7, 9} also has mean 7 but its squared differences sum to only 5, giving SD ≈ 1.1 — far smaller because these numbers cluster closely around the mean.

Frequency Distribution and SD

When data is given in a frequency table, the SD calculation still uses the same 5 steps, but each value's squared difference is weighted by its frequency.

Example: Data set of 20 numbers with frequency distribution:

Mean = 0/20 = 0. For SD, compute f × (x − 0)² for each row: 2(16) + 3(4) + 7(0) + 5(1) + 3(9) = 32 + 12 + 0 + 5 + 27 = 76. Variance = 76/20 = 3.8. SD = √3.8 ≈ 1.9.

SD Computation Traps

• Variance ≠ SD: variance is the average of squared differences; SD is the square root of variance. The GMAT may ask for either — read carefully.
• Adding a constant shifts the mean, not the spread: every student scoring 10 points more on a test shifts the mean by 10 but leaves SD identical.
• SD can only be zero or positive. It is never negative. SD = 0 if and only if all values are the same.
• Comparing two sets: higher range does not automatically mean higher SD. A set can have large range but low SD if all values cluster in the middle except two extremes.