Mean, Median, Mode and Range Worksheets - Free Printable

Let’s solve each problem step by step. We’ll find the mean, median, mode, and sometimes range for each set of numbers.

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Problem 1:

Data: 7, 12, 16, 12, 6, 4, 1

First, sort the data:
→ 1, 4, 6, 7, 12, 12, 16

- Mean: Add all numbers → 1+4+6+7+12+12+16 = 58
Divide by count (7) → 58 ÷ 7 ≈ 8.3

- Median: Middle number in sorted list → 4th number is 7

- Mode: Most frequent number → 12 appears twice, others once → 12

- Range: Max - Min → 16 - 1 = 15

✔ Answer: Mean ≈ 8.3, Median = 7, Mode = 12, Range = 15

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Problem 2:

Data: 4, 3, 8, 3, 6, 2, 9

Sort: 2, 3, 3, 4, 6, 8, 9

- Mean: 2+3+3+4+6+8+9 = 35 → 35 ÷ 7 = 5

- Median: 4th number → 4

- Mode: 3 appears twice → 3

- Range: 9 - 2 = 7

✔ Answer: Mean = 5, Median = 4, Mode = 3, Range = 7

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Problem 3:

Data: 6, 1, 3, 1, 6

Sort: 1, 1, 3, 6, 6

- Mean: 1+1+3+6+6 = 17 → 17 ÷ 5 = 3.4

- Median: 3rd number → 3

- Mode: Both 1 and 6 appear twice → bimodal: 1 and 6

- Range: 6 - 1 = 5

✔ Answer: Mean = 3.4, Median = 3, Mode = 1 and 6, Range = 5

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Problem 4:

What must be the missing piece so mean = 10?
Data: 14, ___, 9, 12, 2, 8, 2, 1 → That’s 8 numbers total.

Let missing number = x

Sum of known numbers: 14 + 9 + 12 + 2 + 8 + 2 + 1 = 48
Total sum needed for mean 10: 10 × 8 = 80
So: 48 + x = 80 → x = 80 - 48 = 32

✔ Answer: 32

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Problem 5:

What must be the missing piece so mode = 4?
Data: 6, 2, 3, 3, 2, 4, ___

Current counts:
- 2 → twice
- 3 → twice
- 4 → once
- 6 → once

To make 4 the mode, it must appear more than any other → needs to appear at least 3 times? But we only have one blank.

Wait — if we add another 4, then 4 appears twice, same as 2 and 3 → still not unique mode.

But maybe the question allows multiple modes? Or perhaps they want 4 to be *a* mode?

Actually, let’s check: current max frequency is 2 (for 2 and 3). If we add a 4, then 4 also has frequency 2 → so modes are 2, 3, and 4 → but that’s not helpful.

Wait — maybe I miscounted.

Original data: 6, 2, 3, 3, 2, 4, ___ → that’s 7 numbers.

If we add a 4 → now 4 appears twice, same as 2 and 3 → still three modes.

But if we add a 2 → then 2 appears 3 times → mode becomes 2.

If we add a 3 → 3 appears 3 times → mode becomes 3.

We want mode = 4 → so we need 4 to appear MORE than any other.

Currently, 2 and 3 appear twice. So we need 4 to appear 3 times → but we only have one blank → impossible unless we change existing data? No.

Wait — maybe the data is written wrong? Let me re-read:

“6, 2, 3, 3, 2, 4, ___”

That’s six numbers plus blank → seven total.

Perhaps the intention is that adding one number makes 4 the sole mode.

But with current data, no single number can become mode unless it already has highest frequency.

Unless... maybe “mode = 4” means 4 should be one of the modes? Then adding 4 would make it tied.

But typically, “the mode” implies unique.

Wait — let’s look again: maybe I missed something.

Another thought: perhaps the data is: 6, 2, 3, 3, 2, 4, and we add one number.

If we add 4 → frequencies: 2:2, 3:2, 4:2, 6:1 → modes are 2,3,4 → not good.

If we add 2 → 2:3, 3:2, 4:1, 6:1 → mode=2

If we add 3 → 3:3, 2:2, 4:1, 6:1 → mode=3

If we add 6 → 6:2, 2:2, 3:2, 4:1 → modes 2,3,6

None give mode=4 alone.

Unless... wait — what if the original data has only one 2 and one 3? Let me double-check the problem.

Problem says: “6, 2, 3, 3, 2, 4, ___”

Yes, two 2s and two 3s.

This seems impossible to make 4 the unique mode with one addition.

But perhaps the question allows 4 to be a mode (not necessarily unique)? In that case, adding 4 would make it a mode along with 2 and 3.

But that’s not standard.

Wait — maybe I made a mistake in counting.

List: positions: 1:6, 2:2, 3:3, 4:3, 5:2, 6:4, 7:?

So values: 6,2,3,3,2,4 → so 2 appears at pos2 and pos5 → twice; 3 at pos3 and pos4 → twice; 4 once; 6 once.

Yes.

Perhaps the answer is 4, accepting that it becomes a mode (even if not unique).

Or maybe there's a typo in the problem.

But looking at similar problems, often they expect you to add the number to make it appear most frequently.

Here, since 2 and 3 are tied at 2, adding 4 won't help.

Unless... what if we add a number that is not 2,3,4,6? Say 5 — then all have freq 1 or 2, no mode? Worse.

I think there might be an error, but let's assume the intended answer is to add 4, making it appear twice, and perhaps the problem considers it acceptable.

But that doesn't make sense.

Another idea: perhaps "mode = 4" means that 4 should be the only mode, so we need to reduce others? But we can't.

Wait — let's read the problem again: "What must the missing piece of data be to have a mode of 4?"

In some contexts, if multiple modes, it's still called "modes", but here it says "a mode of 4", which might mean 4 is one of the modes.

In that case, adding 4 would work.

Perhaps the data is different.

Let me check online or think differently.

Maybe the data is: 6,2,3,3,2,4 and we add one number, and we want 4 to be the mode, so we need 4 to appear more than twice, but we can't.

Unless the blank is for two numbers? No, it says "piece", singular.

Perhaps I miscalculated the count.

List the numbers: given: 6,2,3,3,2,4 — that's six numbers. Blank is seventh.

Frequencies before adding: 2:2, 3:2, 4:1, 6:1.

After adding x:

If x=4, then 4:2, so modes are 2,3,4.

If the problem accepts that, then x=4.

Otherwise, it's impossible.

But let's look at the next problems for clue.

Perhaps in this context, "mode of 4" means 4 is a mode, not necessarily unique.

I think for school level, they might accept adding 4.

So I'll go with 4.

✔ Answer: 4

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Problem 6:

What must be the missing piece so mean = 7?
Data: 7, 12, ___, 3, 11 → 5 numbers total.

Let missing = x

Sum of known: 7+12+3+11 = 33

Total sum needed: 7 × 5 = 35

So 33 + x = 35 → x = 2

✔ Answer: 2

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Problem 7:

What must be the missing piece so mean = 40?
Data: 20, 30, 50, 40, ___ → 5 numbers.

Sum of known: 20+30+50+40 = 140

Total sum needed: 40 × 5 = 200

So 140 + x = 200 → x = 60

✔ Answer: 60

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Problem 8:

What must be the missing piece so mean = 27?
Data: 18, 11, 7, 15, 32, 27, ___ → 7 numbers.

Sum of known: 18+11+7+15+32+27 = let's calculate:
18+11=29, +7=36, +15=51, +32=83, +27=110

Total sum needed: 27 × 7 = 189

So 110 + x = 189 → x = 79

✔ Answer: 79

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Problem 9:

What must be the missing piece so mean = 37?
Data: 51, 52, ___, 23, 21, 17, 37, 45 → 8 numbers.

Sum of known: 51+52+23+21+17+37+45

Calculate step by step:
51+52=103
103+23=126
126+21=147
147+17=164
164+37=201
201+45=246

Total sum needed: 37 × 8 = 296

So 246 + x = 296 → x = 50

✔ Answer: 50

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Now, compiling all answers:

1. Mean ≈ 8.3, Median = 7, Mode = 12, Range = 15
2. Mean = 5, Median = 4, Mode = 3, Range = 7
3. Mean = 3.4, Median = 3, Mode = 1 and 6, Range = 5
4. Missing number for mean=10: 32
5. Missing number for mode=4: 4 (assuming it becomes a mode)
6. Missing number for mean=7: 2
7. Missing number for mean=40: 60
8. Missing number for mean=27: 79
9. Missing number for mean=37: 50

For problem 5, I'm a bit unsure, but based on common textbook problems, they likely expect 4.

Final Answer:
1. Mean: 8.3, Median: 7, Mode: 12, Range: 15
2. Mean: 5, Median: 4, Mode: 3, Range: 7
3. Mean: 3.4, Median: 3, Mode: 1 and 6, Range: 5
4. 32
5. 4
6. 2
7. 60
8. 79
9. 50