Statistics on the GMAT Focus comes down to a handful of ideas used cleverly: the average (arithmetic mean), the median, the range and, occasionally, standard deviation. You will not need advanced formulas. You will need to think about sums and about the order of a list.
The most useful habit is to turn every average into a sum. If six numbers average 15, their sum is 90. Removing, adding or replacing a value then becomes simple arithmetic on sums, which avoids the classic trap of subtracting one average from another.
For medians, sort the list and picture it as a row of slots. With an odd number of values the median is the middle slot; with an even number it is the average of the two middle slots. Many hard questions are minimum or maximum problems in disguise: to make the largest value as big as possible, make every other value as small as the rules allow. Read those rules carefully. “Distinct” forbids repeats, “positive” rules out zero, and “integers” rules out fractions, and each one changes the answer.
In Data Sufficiency, remember that the mean reacts to every value while the median only cares about position, so one extreme value can push the mean above or below the median. Test an extreme case before you conclude.
4 practice questions
Problem SolvingStatistics and Sets ProblemsEasy
Question 1
The average (arithmetic mean) of $6$ numbers is $15$. When one of the numbers is removed, the average of the remaining $5$ numbers is $17$. What number was removed?
A$2$
B$5$
C$13$
D$15$
E$17$
Show answer & explanation
Correct answer: B
Key Concept or Approach
Work with sums: sum = mean x count.
Step-by-Step Solution
Step 1: Original sum: $6 \times 15 = 90$.
Step 2: Remaining sum: $5 \times 17 = 85$.
Step 3: Removed number: $90 - 85 = 5$.
Final Answer
Answer: B
Common trap: 2 - subtracting the two averages instead of the two sums.
Data SufficiencyStatistics and Sets ProblemsHard
Question 2
List $S$ consists of the numbers $4$, $7$, $10$, and $x$. Is the average (arithmetic mean) of $S$ greater than the median of $S$?
(1) $x > 5$
(2) $x > 14$
AStatement (1) ALONE is sufficient, but statement (2) ALONE is NOT sufficient.
BStatement (2) ALONE is sufficient, but statement (1) ALONE is NOT sufficient.
CBOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
DEACH statement ALONE is sufficient.
EStatements (1) and (2) TOGETHER are NOT sufficient.
Show answer & explanation
Correct answer: B
Key Concept or Approach
Where $x$ falls in the ordering changes the median. The mean $\frac{21 + x}{4}$ grows with $x$, but the median stops growing once $x \ge 10$.
Step-by-Step Solution
Step 1: Statement (1): $x = 6$ gives ordered set $4, 6, 7, 10$: median $6.5$, mean $6.75$, so yes. $x = 8$ gives $4, 7, 8, 10$: median $7.5$, mean $7.25$, so no. Not sufficient.
Step 2: Statement (2): with $x > 14$ the order is $4, 7, 10, x$, so the median is $8.5$. The mean is $\frac{21 + x}{4} > \frac{35}{4} = 8.75 > 8.5$. Always yes. Sufficient.
Step 3: Only statement (2) is sufficient.
Final Answer
Answer: B
Common trap: D: students test only one value with $x > 5$ and assume a large $x$ always pulls the mean above the median.
Problem SolvingMin/Max ProblemsStatistics and Sets ProblemsHard
Question 3
A set of $7$ distinct positive integers has an average (arithmetic mean) of $20$ and a median of $22$. What is the greatest possible value of the largest integer in the set?
A$65$
B$66$
C$68$
D$69$
E$71$
Show answer & explanation
Correct answer: A
Key Concept or Approach
Fix the sum ($7 \times 20 = 140$) and make every other element as small as the constraints allow.
Step-by-Step Solution
Step 1: The three smallest must be distinct positive integers: $1, 2, 3$. The median is $22$.
Step 2: The 5th and 6th elements must exceed $22$ and be distinct: $23, 24$.
