6.3 Descriptive Statistics: Mean, Median, Mode, Range, and Weighted Mean

6.3 Descriptive Statistics: Mean, Median, Mode, Range, and Weighted Mean

Quick Summary: GMAT descriptive statistics questions focus heavily on the relationship $\text{Sum} = \text{Mean} \times n$, the resistance of the median to outliers, and the lever principle for weighted averages. Knowing that adding a constant shifts central tendency but leaves range unchanged—while multiplying scales everything—allows rapid Problem Solving without cumbersome arithmetic.

Descriptive statistics questions assess your ability to extract mathematical relationships from collections of data. Rather than testing complex statistical theory, the GMAT evaluates whether you understand how summary statistics behave when data sets are modified, merged, or constrained.

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The Arithmetic Mean and the Sum Formula

The arithmetic mean (commonly called the average, denoted $\bar{x}$ or $\mu$) is the sum of all observations divided by the number of observations:

$\text{Mean} = \frac{\sum x_i}{n} = \frac{x_1 + x_2 + \dots + x_n}{n}$

The Fundamental "Sum Equation"

In virtually every GMAT mean problem, the key algebraic step is converting the mean into a total sum:

$\text{Sum} = \text{Mean} \times n$

Whenever a problem states that "the average of 6 numbers is 18," immediately write down: $\text{Sum} = 6 \times 18 = 108$. If a new number $x$ is added and the average becomes 20 for 7 numbers, the new sum is $7 \times 20 = 140$, meaning $x = 140 - 108 = 32$.

Deviation Property of the Mean

The arithmetic mean acts as the physical center of gravity of a data set. The sum of the deviations of all data points from the mean is identically zero:

$\sum_{i=1}^n (x_i - \bar{x}) = 0$

Points above the mean create positive deviations that perfectly cancel the negative deviations created by points below the mean. This property allows quick mental verification of candidate means without full summation.

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The Median: The Resistant Center

The median is the physical middle value when a data set is ordered from least to greatest.

Determining the Median in Ordered Sets

Always sort the list in ascending order: $x_1 \le x_2 \le x_3 \le \dots \le x_n$.

1. Odd Number of Elements ($n$ is odd): There is a single unique middle number located at position $\frac{n + 1}{2}$: $\text{Median} = x_{(n+1)/2}$ Example: In ${3, 7, 12, 19, 25}$ ($n = 5$), the median is the $\frac{5+1}{2} = 3$rd term, which is 12.

2. Even Number of Elements ($n$ is even): There are two middle numbers, located at positions $\frac{n}{2}$ and $\frac{n}{2} + 1$. The median is the arithmetic mean of these two central numbers: $\text{Median} = \frac{x_{n/2} + x_{(n/2)+1}}{2}$ Example: In ${4, 8, 14, 20, 22, 30}$ ($n = 6$), the two middle values are the 3rd term (14) and 4th term (20). The median is $\frac{14 + 20}{2} = 17$.

Finding the Median from a Frequency Table

When data is presented in a frequency distribution table:

1. Sum the frequencies to determine the total count: $N = \sum f_i$.
2. Compute cumulative frequencies down the table.
3. Identify the value where the median position (the $\frac{N+1}{2}$-th item if odd, or the bridge between $\frac{N}{2}$ and $\frac{N}{2}+1$ if even) falls.

Mean vs. Median: Outlier Sensitivity and Skewness

• The Mean is Sensitive: Extreme outliers pull the mean heavily toward the tail.
• The Median is Resistant: Outliers have zero impact on the median as long as they remain on the same side of the center.

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Mode and Range

Mode

The mode is the data value that appears with the highest frequency in a set.

• A set can have one mode (unimodal): in ${2, 3, 3, 5, 8}$, the mode is 3.
• A set can have multiple modes (bimodal, multimodal): in ${1, 1, 4, 6, 6, 9}$, the modes are 1 and 6.
• A set can have no mode: in ${2, 4, 7, 11, 15}$, every number appears with equal frequency (1 time), so there is no mode.

Range

The range is the simplest measure of data spread:

$\text{Range} = \text{Maximum Value} - \text{Minimum Value}$

Notice that range depends strictly on the two outermost values; it gives no information about how the internal points are distributed.

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Weighted Averages and the Lever / Seesaw Principle

When combining sets of different sizes, a simple average of their individual means yields an incorrect result. You must use a weighted mean:

$\bar{x}_w = \frac{w_1 x_1 + w_2 x_2 + \dots + w_k x_k}{w_1 + w_2 + \dots + w_k}$

where $w_i$ represents the weight or count of each subgroup.

The Lever / Seesaw Shortcut (Balance Rule)

On the calculator-free GMAT, solving weighted averages via the standard formula involves cumbersome multi-digit multiplication. The lever rule solves these problems conceptually in under 30 seconds.

Imagine a seesaw with subgroup 1 on the left (mean $\bar{x}_1$, weight $w_1$) and subgroup 2 on the right (mean $\bar{x}_2$, weight $w_2$). The fulcrum is the combined weighted mean $\bar{x}_w$.

$\text{Weight}_1 \times \text{Distance}_1 = \text{Weight}_2 \times \text{Distance}_2$ $w_1 (\bar{x}_w - \bar{x}_1) = w_2 (\bar{x}_2 - \bar{x}_w)$

This establishes the Inverse Distance Ratio:

$\frac{w_1}{w_2} = \frac{\bar{x}_2 - \bar{x}_w}{\bar{x}_w - \bar{x}_1}$

Core Intuition: The combined average is always pulled closer to the larger group. The ratio of the distances to the combined average is the inverse of the ratio of the group sizes!

Lever Walkthrough

Problem: Class A has 30 students with an average score of 80. Class B has 20 students with an average score of 90. What is the combined average?

1. Find the ratio of weights: $\frac{w_A}{w_B} = \frac{30}{20} = \frac{3}{2}$.
2. The total distance between the means is $90 - 80 = 10$ points.
3. Divide this distance into $3 + 2 = 5$ equal parts: $\frac{10}{5} = 2$ points per part.
4. Class A has weight 3, so the combined mean is 2 parts away from 80 (toward 90): $80 + (2 \times 2) = 84$. (Check: distance from 90 is 3 parts: $90 - (3 \times 2) = 84$.) Combined average is 84.

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Effects of Operations on Summary Statistics

A favorite GMAT testing area is determining the effect of adding or multiplying a constant on statistical metrics.

1. Adding / Subtracting a Constant $c$ to Every Element ($x_i \to x_i + c$)

Every data point shifts by exactly $c$ units along the number line:

• $\text{New Mean} = \text{Old Mean} + c$
• $\text{New Median} = \text{Old Median} + c$
• $\text{New Mode} = \text{Old Mode} + c$
• $\text{New Range} = (\text{Max} + c) - (\text{Min} + c) = \text{Max} - \text{Min} = \mathbf{\text{UNCHANGED}}$

2. Multiplying / Dividing Every Element by Positive Constant $k$ ($x_i \to k x_i$)

The entire distribution is stretched or compressed by a factor of $k$:

• $\text{New Mean} = k \times \text{Old Mean}$
• $\text{New Median} = k \times \text{Old Median}$
• $\text{New Mode} = k \times \text{Old Mode}$
• $\text{New Range} = (k \times \text{Max}) - (k \times \text{Min}) = \mathbf{k \times \text{Old Range}}$

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Worked Problem Solving Examples

Example 1: Finding an Unknown Term with Mean, Median, and Range

Problem: A set of 5 distinct integers has a mean of 14, a median of 12, and a range of 16. If the smallest integer in the set is 6, what is the maximum possible value of the second-largest integer in the set?

Solution:

1. Let the 5 sorted integers be $x_1 < x_2 < x_3 < x_4 < x_5$.
2. Identify known values from the problem statement:
   • Smallest integer: $x_1 = 6$.
   • Range is 16: $x_5 - x_1 = 16 \implies x_5 = 6 + 16 = 22$.
   • Median is the 3rd term (since $n = 5$): $x_3 = 12$.
   • Mean is 14: $\text{Sum} = 5 \times 14 = 70$.
3. Express the total sum algebraically: $x_1 + x_2 + x_3 + x_4 + x_5 = 70$ $6 + x_2 + 12 + x_4 + 22 = 70$ $x_2 + x_4 + 40 = 70 \implies x_2 + x_4 = 30$
4. Apply the strict inequality constraints for distinct integers: $x_1 < x_2 < x_3 < x_4 < x_5$: $6 < x_2 < 12 < x_4 < 22$
5. We need to find the value of $x_4$ (the second-largest integer). From $x_2 + x_4 = 30$, we have $x_2 = 30 - x_4$.
6. Substitute into the bound $6 < x_2 < 12$: $6 < 30 - x_4 < 12$ Subtract 30 from all sides: $-24 < -x_4 < -18$ Multiply by $-1$ (flipping the inequalities): $18 < x_4 < 24$
7. Combine with the upper bound $x_4 < 22$: $18 < x_4 < 22$
8. The feasible integer values are $x_4 \in {19, 20, 21}$. The greatest is $21$. It is attainable with $x_2 = 9$, producing the valid ordered set ${6, 9, 12, 21, 22}$, whose sum is 70. Therefore, the maximum possible second-largest integer is 21.

Example 2: Weighted Mean in Financial Portfolio Blends

Problem: An investor allocates $40,000 to Fund X earning a 6% annual return, and $60,000 to Fund Y earning an 11% annual return. What is the overall percentage return of the combined $100,000 portfolio?

Solution:

1. Calculate the weight ratio: $\frac{w_X}{w_Y} = \frac{40,000}{60,000} = \frac{2}{3}$.
2. The distance between the percentage returns is $11% - 6% = 5%$.
3. Divide the distance into $2 + 3 = 5$ equal parts: $\frac{5%}{5} = 1%$ per part.
4. Fund Y has weight 3, so Fund X receives a distance of 3 parts from the combined mean. The combined return is 3 parts above 6%: $\bar{x}_w = 6\% + (3 \times 1\%) = 9\%$ (Verification: $\frac{(40,000 \times 0.06) + (60,000 \times 0.11)}{100,000} = \frac{2,400 + 6,600}{100,000} = \frac{9,000}{100,000} = 9%$.)

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High-Frequency GMAT Traps

• Trap 1: Finding the Median Without Sorting First
  Selecting the physical middle number of an unsorted list is the most common careless error on statistics questions. Always reorder the values from smallest to largest.

• Trap 2: Simple Averaging of Subgroup Averages
  Never calculate $\frac{\bar{x}_1 + \bar{x}_2}{2}$ unless the two subgroups have identical sample sizes ($n_1 = n_2$). If group sizes differ, the larger group dominates the weighted average.

• Trap 3: Believing Adding a Constant Changes the Range
  Because adding a constant shifts the maximum and minimum by the exact same amount, the difference $(\text{Max} + c) - (\text{Min} + c)$ remains strictly invariant.