3.3 Averages, Weighted Averages & Statistical Word Problems

3.3 Averages, Weighted Averages & Statistical Word Problems

Quick Answer: The key to solving 90% of AFQT average problems is converting means into total sums using $\text{Sum} = \text{Mean} \times N$. When combining groups of different sizes, never take a simple average of the averages; instead, calculate the weighted average by dividing total combined points by total combined count.

Statistical word problems on the AFQT Arithmetic Reasoning subtest evaluate a candidate's ability to manipulate central tendency metrics—mean, median, mode, and range—under realistic operational scenarios.

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The Arithmetic Mean and the "Sum Triangle"

The arithmetic mean (commonly called the average) is defined as the sum of a set of numerical values divided by the total number of values ($N$).

$\text{Mean} = \frac{\text{Sum of All Values}}{N}$

The Master Transformation: $\text{Sum} = \text{Mean} \times N$

On the AFQT, you will rarely be asked to simply add four numbers and divide. Instead, exam questions give you an average and ask you to find a missing value, an added score, or a target requirement.

$\begin{array}{c} \text{Sum} \\ \hline \text{Mean} \quad \vert \quad N \end{array}$

By focusing on the Total Sum, you can solve complex multi-step average problems without getting bogged down in algebraic confusion.

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Missing Value / Target Average Problems

A classic AFQT question provides a series of initial test scores or measurements and asks what score is required on the final evaluation to achieve a target overall average.

Step-by-Step Worked Example: Target Qualification Score

Problem: A soldier completes four qualification rifle shoots with scores of 78, 84, 88, and 92. What score must the soldier achieve on the fifth and final shoot to attain an overall average score of 87?

Step 1: Calculate the Target Total Sum Required

• Target Mean = 87
• Total number of tests $N = 5$ $\text{Required Total Sum} = 87 \times 5 = 435 \text{ points}$

Step 2: Calculate the Current Sum of Completed Tests

$\text{Current Sum} = 78 + 84 + 88 + 92 = 342 \text{ points}$

Step 3: Subtract Current Sum from Required Sum

$\text{Required Score on 5th Test} = 435 - 342 = 93$

The soldier must score 93 on the fifth shoot to achieve an overall 87 average.

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Weighted Averages (Combining Unequal Groups)

A common pitfall on the AFQT is taking the simple average of two group averages when the groups contain different numbers of items. This yields an incorrect result.

The Weighted Average Rule: When combining groups of unequal sizes $N_1$ and $N_2$, you must weight each average by its respective group size.

$\text{Weighted Mean} = \frac{(N_1 \times \text{Mean}_1) + (N_2 \times \text{Mean}_2) + \dots + (N_k \times \text{Mean}_k)}{N_1 + N_2 + \dots + N_k}$

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Step-by-Step Worked Example: Combined Company Test Scores

Problem: Company Alpha has 30 soldiers who scored an average of 80% on a tactical exam. Company Bravo has 20 soldiers who scored an average of 90% on the same exam. What is the combined average score for all 50 soldiers across both companies?

Step 1: Identify the Common Pitfall

Taking the simple average $\frac{80 + 90}{2} = 85%$ is INCORRECT because Company Alpha has more soldiers than Company Bravo. The combined average must be closer to 80% than 90%.

Step 2: Compute Total Points for Each Group

$\text{Company Alpha Total Points} = 30 \text{ soldiers} \times 80\% = 2,400 \text{ points}$ $\text{Company Bravo Total Points} = 20 \text{ soldiers} \times 90\% = 1,800 \text{ points}$

Step 3: Compute Combined Total Points and Total Soldiers

$\text{Combined Points} = 2,400 + 1,800 = 4,200 \text{ points}$ $\text{Combined Count} = 30 + 20 = 50 \text{ soldiers}$

Step 4: Divide Combined Points by Combined Count

$\text{Weighted Mean} = \frac{4,200}{50} = 84\%$

The true combined average score is 84%.

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Dynamic Averages: Adding or Removing Data Points

AFQT questions often describe a scenario where a group's average changes because a new member joins or an existing member leaves.

Step-by-Step Worked Example: New Team Member Weight

Problem: A 6-member search-and-rescue team has an average weight of 180 pounds. When a 7th member joins the team, the team's average weight increases to 184 pounds. What is the weight of the new 7th member?

Step 1: Find the Initial Total Weight ($N = 6$)

$\text{Initial Total Weight} = 6 \times 180 = 1,080 \text{ lbs}$

Step 2: Find the New Total Weight ($N = 7$)

$\text{New Total Weight} = 7 \times 184 = 1,288 \text{ lbs}$

Step 3: Find the Difference

$\text{Weight of 7th Member} = 1,288 - 1,080 = 208 \text{ lbs}$

The new member weighs 208 pounds.

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Descriptive Statistics: Median, Mode, and Range

In addition to the arithmetic mean, AFQT questions test your mastery of other key statistical measures.

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Step-by-Step Worked Example: Full Statistical Profile

Problem: Consider the following recorded completion times (in minutes) for 8 recruits on an obstacle course: $14, 11, 18, 11, 15, 22, 11, 18$. Find the Mean, Median, Mode, and Range.

Step 1: Always Sort the Data First!

Unsorted: $14, 11, 18, 11, 15, 22, 11, 18$ Sorted: $11, 11, 11, 14, 15, 18, 18, 22$ ($N = 8$)

Step 2: Calculate Mean

$\text{Sum} = 11 + 11 + 11 + 14 + 15 + 18 + 18 + 22 = 120$ $\text{Mean} = \frac{120}{8} = 15 \text{ minutes}$

Step 3: Calculate Median

Since $N = 8$ (even), the median is the average of the 4th and 5th terms in the sorted list:

• 4th term = $14$
• 5th term = $15$ $\text{Median} = \frac{14 + 15}{2} = 14.5 \text{ minutes}$

Step 4: Calculate Mode

The value $11$ appears 3 times (more than any other value), so the Mode is 11 minutes.

Step 5: Calculate Range

$\text{Range} = \text{Max} - \text{Min} = 22 - 11 = 11 \text{ minutes}$

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AFQT Exam Traps and Shortcuts

1. Forgetting to Sort for Median: Never attempt to pick the middle number of a list before sorting it in ascending order.
2. Outlier Sensitivity Trap: The mean and range are heavily affected by extreme outliers, whereas the median and mode remain resilient.
3. Deviation Balancing Method: The sum of deviations from the mean always equals zero ($\sum (x_i - \text{Mean}) = 0$). You can use this to quickly verify missing score calculations.