Understanding Central Tendency: Mean, Median, and Mode

Statistics & Data 7th-8th Grade
Problem

The following table shows how many siblings a sample of people reported they had.

Number of Siblings012345
Frequency653102

A. Find the mean. [1]
B. Find the median. [1]
C. Find the mode. [1]
D. Which of the above best represents the number of siblings? Explain.

What You Will Learn

How to calculate mean from frequency data using weighted averages
Finding median by locating the middle position in grouped data
Identifying mode as the value with highest frequency
Comparing measures of central tendency to choose the most representative
Understanding when data distribution affects which measure is best

Visualizing the Data

Let's look at what this frequency table tells us. We can see the distribution clearly in a bar chart:

The following table shows how many siblings a sample of people reported they had. Number of Siblings 0 1 2 3 4 5...

Notice how 0 siblings (highlighted in gold) is the most common response, while the data tapers off for larger families. This shape will help us understand which measure of central tendency tells the most meaningful story.

Solution: Method 1 — The Frequency Table Approach

Step 1 — Find the total number of people surveyed

Add up all the frequencies: 6 + 5 + 3 + 1 + 0 + 2 = 17 people

Step 2 — Calculate the mean using weighted averages

For frequency data, the mean is the sum of (value × frequency) divided by total frequency:

Mean = (0×6 + 1×5 + 2×3 + 3×1 + 4×0 + 5×2) ÷ 17
= (0 + 5 + 6 + 3 + 0 + 10) ÷ 17
= 24 ÷ 17
= 1.41 siblings

Step 3 — Find the median position

With 17 people, the median is the 9th value when data is arranged in order (since (17+1)÷2 = 9). Now we count through the frequencies:

  • Positions 1-6: 0 siblings
  • Positions 7-11: 1 sibling
  • Positions 12-14: 2 siblings

The 9th person has 1 sibling, so the median is 1.

Step 4 — Identify the mode

The mode is the value that appears most frequently. Looking at our table, 0 siblings has the highest frequency (6 people), so the mode is 0 siblings.

Step 5 — Determine which measure best represents the data

The median (1 sibling) best represents this data because:

  • It's not affected by the few families with many siblings (like the 2 families with 5 siblings)
  • It represents where the "middle" family falls
  • The mean (1.41) is pulled upward by the larger families and doesn't represent a typical response
  • The mode (0) represents the single most common response but doesn't account for the overall distribution

Solution: Method 2 — The Cumulative Distribution Method

Step 1 — Build a cumulative frequency table

SiblingsFrequencyCumulative FrequencyValue × Frequency
0660
15115
23146
31153
40150
521710

Step 2 — Calculate mean from the products column

Mean = Sum of (Value × Frequency) ÷ Total Frequency
= (0 + 5 + 6 + 3 + 0 + 10) ÷ 17 = 1.41 siblings

Step 3 — Use cumulative frequency to find median

The median position is 9. Looking at cumulative frequencies, position 9 falls where the cumulative frequency first reaches or exceeds 9. Since cumulative frequency reaches 11 at "1 sibling," the median is 1 sibling.

Step 4 — Compare all measures

  • Mean: 1.41 siblings (pulled up by larger families)
  • Median: 1 sibling (middle value, unaffected by outliers)
  • Mode: 0 siblings (most common single response)

The median provides the best representation because this data is right-skewed (more families have fewer siblings, with a tail extending toward larger families). The median is resistant to this skewness.

A. Mean = 1.41 siblings
B. Median = 1 sibling
C. Mode = 0 siblings
D. The median best represents the data because it's not affected by the skewed distribution toward larger families.

Verification

Check the mean calculation:

Total siblings = 0×6 + 1×5 + 2×3 + 3×1 + 4×0 + 5×2 = 24
Total people = 6 + 5 + 3 + 1 + 0 + 2 = 17
Mean = 24 ÷ 17 = 1.41... ✓

Check the median: Write out all 17 data points in order:

0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 3, 5, 5

The 9th value is 1 ✓

Check the mode: 0 appears 6 times, which is more than any other value ✓

Check that frequencies sum correctly:6 + 5 + 3 + 1 + 0 + 2 = 17

What Trips Students Up

✗ Mistake #1: Treating the table like raw data

Some students calculate mean as (0 + 1 + 2 + 3 + 4 + 5) ÷ 6 = 2.5, forgetting that each value appears multiple times. You must multiply each value by its frequency before adding.

✗ Mistake #2: Finding median of the frequencies instead of the data

The median is not the middle frequency value (which would be between 1 and 3). The median is the middle value in the actual dataset of 17 responses.

✗ Mistake #3: Confusing mode with median position

The mode is the most frequently occurring data value (0), not the value that appears in the middle frequency position. Don't mix up "most common value" with "middle value."

The Pattern Behind This

This problem demonstrates the general principle for analyzing frequency distributions:

Mean from frequency data = Σ(value × frequency) ÷ Σ(frequency)

Median position = (n + 1) ÷ 2, where n = total frequency

Mode = value with highest frequency

The key insight is understanding when each measure is most useful:

  • Mean: Best for symmetric data, but sensitive to outliers
  • Median: Best for skewed data or data with outliers
  • Mode: Best for categorical data or when you need the most typical response

In this problem, the data is right-skewed (long tail toward higher values), making the median the most representative measure of central tendency.

When Numbers Get Messy

In this problem, we got a clean median (1) and mode (0), but a decimal mean (1.41). This is completely normal for real data — don't round the mean to a whole number just because we're counting siblings.

If our frequency table had been slightly different — say, 7 people with 0 siblings instead of 6 — our calculations would be:

Mean = (0×7 + 1×5 + 2×3 + 3×1 + 4×0 + 5×2) ÷ 18 = 23 ÷ 18 = 1.28
Median = 9.5th position = average of 9th and 10th values = (1 + 1) ÷ 2 = 1
Mode = 0 (still the most frequent)

The method stays exactly the same — the only difference is that we might get messier decimals or need to average two middle values for the median.

What If?

1
Different Distribution
The frequency table now shows: 0 siblings (2 people), 1 sibling (4 people), 2 siblings (6 people), 3 siblings (3 people), 4 siblings (2 people), 5 siblings (1 person). Find the mean, median, and mode. Which best represents this data?
Step 1 — Calculate total people

Total = 2 + 4 + 6 + 3 + 2 + 1 = 18 people

Step 2 — Find the mean

Mean = (0×2 + 1×4 + 2×6 + 3×3 + 4×2 + 5×1) ÷ 18
= (0 + 4 + 12 + 9 + 8 + 5) ÷ 18 = 38 ÷ 18 = 2.11 siblings

Step 3 — Find the median

Median position = 9.5 (average of 9th and 10th values)
Positions 1-2: 0 siblings, 3-6: 1 sibling, 7-12: 2 siblings
Both 9th and 10th people have 2 siblings, so median = 2

Step 4 — Find the mode

2 siblings appears 6 times (highest frequency), so mode = 2 siblings

Step 5 — Best representation

Mean = 2.11, Median = 2, Mode = 2. Since the distribution is more symmetric now, both the median and mode (which happen to be equal) best represent the data at 2 siblings.

2
Adding an Outlier
Using the original data, suppose we add one person who has 12 siblings. Now we have 18 people total. How do the mean, median, and mode change? Which measure is most affected?
Step 1 — New frequency table

0 siblings (6), 1 sibling (5), 2 siblings (3), 3 siblings (1), 4 siblings (0), 5 siblings (2), 12 siblings (1)
Total people = 18

Step 2 — New mean

Mean = (0×6 + 1×5 + 2×3 + 3×1 + 4×0 + 5×2 + 12×1) ÷ 18
= (0 + 5 + 6 + 3 + 0 + 10 + 12) ÷ 18 = 36 ÷ 18 = 2.00 siblings

Step 3 — New median

Median position = 9.5 (average of 9th and 10th values)
The ordered list now has the value 12 at the end, but the 9th and 10th values are still both 1 sibling
New median = 1 sibling (unchanged)

Step 4 — New mode

0 siblings still appears most frequently (6 times)
New mode = 0 siblings (unchanged)

Step 5 — Which changed most?

The mean changed most dramatically — from 1.41 to 2.00 siblings (increased by 42%). The median and mode were unaffected by the outlier, showing their resistance to extreme values.

3
Missing Data Point
A researcher tells you that 20 people were surveyed about siblings, and the mean is exactly 1.5 siblings. You know that 5 people have 0 siblings, 4 people have 1 sibling, 6 people have 2 siblings, and 3 people have 3 siblings. How many people must have 4 siblings, and how many have 5 siblings?
Step 1 — Set up what we know

20 people total, mean = 1.5 siblings
Known: 0 siblings (5), 1 sibling (4), 2 siblings (6), 3 siblings (3)
Unknown: 4 siblings (x), 5 siblings (y)
We know: 5 + 4 + 6 + 3 + x + y = 20, so x + y = 2

Step 2 — Use the mean formula

Total of all sibling values = mean × number of people = 1.5 × 20 = 30
(0×5 + 1×4 + 2×6 + 3×3 + 4×x + 5×y) = 30
(0 + 4 + 12 + 9 + 4x + 5y) = 30
25 + 4x + 5y = 30
4x + 5y = 5

Step 3 — Solve the system

We have: x + y = 2 and 4x + 5y = 5
From first equation: y = 2 - x
Substituting: 4x + 5(2 - x) = 5
4x + 10 - 5x = 5
-x = -5
x = 5, but this is impossible since x + y = 2

Step 4 — Try integer solutions

Since x + y = 2 and both must be non-negative integers:
Try x = 0, y = 2: 4(0) + 5(2) = 10 ≠ 5
Try x = 1, y = 1: 4(1) + 5(1) = 9 ≠ 5
Try x = 2, y = 0: 4(2) + 5(0) = 8 ≠ 5

Step 5 — Conclusion

This problem has no solution with the given constraints. The mean of 1.5 is inconsistent with having exactly 20 people and the given frequency distribution. This illustrates why checking for consistency is important in statistics problems.

4
Weighted Survey
A more complex survey asked 100 families about siblings, but the responses were grouped: 30 families had 0-1 siblings per child, 45 families had 2-3 siblings per child, and 25 families had 4-5 siblings per child. Using the midpoint of each range, estimate the mean number of siblings and explain why this is just an approximation.
Step 1 — Identify midpoints

0-1 siblings: midpoint = (0 + 1) ÷ 2 = 0.5
2-3 siblings: midpoint = (2 + 3) ÷ 2 = 2.5
4-5 siblings: midpoint = (4 + 5) ÷ 2 = 4.5

Step 2 — Calculate estimated mean

Using midpoint method:
Mean ≈ (0.5×30 + 2.5×45 + 4.5×25) ÷ 100
= (15 + 112.5 + 112.5) ÷ 100
= 240 ÷ 100 = 2.4 siblings

Step 3 — Why this is an approximation

This is only an estimate because we don't know the exact distribution within each range. For example:
• If most families in "0-1" actually have 0 siblings, the true mean would be lower
• If most families in "4-5" actually have 5 siblings, the true mean would be higher

Step 4 — Range of possible means

Minimum mean (everyone at lower bounds): (0×30 + 2×45 + 4×25) ÷ 100 = 1.9
Maximum mean (everyone at upper bounds): (1×30 + 3×45 + 5×25) ÷ 100 = 2.9
So the true mean is between 1.9 and 2.9, with 2.4 as our best estimate.

Verification

Check frequencies sum to 100: 30 + 45 + 25 = 100
Check calculation: 15 + 112.5 + 112.5 = 240, 240 ÷ 100 = 2.4

Frequently Asked Questions

How do you calculate the mean from a frequency table?+

Multiply each value by its frequency, add all those products together, then divide by the total number of data points. In this problem: (0×6 + 1×5 + 2×3 + 3×1 + 4×0 + 5×2) ÷ 17 = 1.41 siblings. Don't just average the values themselves — you must weight them by how often they occur.

Which measure of central tendency is best for data with outliers?+

The median is typically best when you have outliers or skewed data, because it represents the middle value and isn't affected by extreme values. In this problem, the median of 1 sibling better represents typical families than the mean of 1.41, which is pulled up by the few families with many siblings.

What does it mean when a dataset has multiple modes?+

When two or more values tie for the highest frequency, the dataset is bimodal or multimodal. Here, 0 siblings appears 6 times, which is the highest frequency, making 0 the single mode. If another value also appeared 6 times, we'd have two modes, suggesting the data has two distinct peaks or typical values.

NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-07-31