Mean Median Mode Range Worksheets - Math Monks - Free Printable

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

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🔢 Problem 1:

Numbers: 3, 0, 0, 2, 0, 3, 0, 2, 2, 2, 3, 3

First, sort them to make it easier:
→ 0, 0, 0, 0, 2, 2, 2, 2, 3, 3, 3, 3

There are 12 numbers.

#### Mean:
Add all numbers:
0+0+0+0 = 0
2+2+2+2 = 8
3+3+3+3 = 12
Total = 0 + 8 + 12 = 20
Mean = 20 ÷ 12 ≈ 1.67 (we’ll keep as fraction or decimal? Let’s use exact: 20/12 = 5/3 ≈ 1.67)

But let’s check again:
Actually, count carefully:

Original list: 3, 0, 0, 2, 0, 3, 0, 2, 2, 2, 3, 3 → that’s 12 numbers.

Zeros: positions 2,3,5,7 → four 0s → 0×4 = 0
Twos: positions 4,8,9,10 → four 2s → 2×4 = 8
Threes: positions 1,6,11,12 → four 3s → 3×4 = 12
Total = 0 + 8 + 12 = 20
Mean = 20 ÷ 12 = 5/3 ≈ 1.67

But maybe they want decimal rounded? Or exact? Since other problems may have whole numbers, let’s hold off — actually, 20÷12 = 1.666... so we can write 1.67 if rounding to two decimals. But let’s see what’s expected. In school, sometimes they accept fractions. But since others might be whole, perhaps I made a mistake?

Wait — let me recount the original list:

“3, 0, 0, 2, 0, 3, 0, 2, 2, 2, 3, 3” — yes, 12 numbers.

Count of 0s: indices 2,3,5,7 → 4 zeros
Count of 2s: indices 4,8,9,10 → 4 twos
Count of 3s: indices 1,6,11,12 → 4 threes
Yes, 4 each → total sum = 4*(0+2+3) = 4*5 = 20 → mean = 20/12 = 5/3 ≈ 1.67

But let’s proceed with exact value for now. Maybe later we’ll adjust.

#### Median:
Sorted: 0,0,0,0,2,2,2,2,3,3,3,3 → 12 numbers → median is average of 6th and 7th.

6th number: 2
7th number: 2
Median = (2+2)/2 = 2

#### Mode:
Which appears most? All appear 4 times → no unique mode? But wait — in some curricula, if multiple values tie for highest frequency, you list all. But here, 0,2,3 all appear 4 times → so modes are 0,2,3? But that’s unusual. Wait — did I miscount?

Original: 3,0,0,2,0,3,0,2,2,2,3,3

List them:

Position:
1: 3
2: 0
3: 0
4: 2
5: 0
6: 3
7: 0
8: 2
9: 2
10: 2
11: 3
12: 3

So:

0 appears at 2,3,5,7 → 4 times
2 appears at 4,8,9,10 → 4 times
3 appears at 1,6,11,12 → 4 times

Yes, all three appear 4 times → so multimodal: 0, 2, 3

But in many elementary worksheets, they might expect “no mode” or list all. However, looking at the format, probably they expect one answer. Did I misread the problem?

Wait — let me double-check the original input: “3, 0, 0, 2, 0, 3, 0, 2, 2, 2, 3, 3”

That’s correct. So technically, mode is 0, 2, and 3. But perhaps the worksheet expects us to say “all” or something. Alternatively, maybe I should report all.

But let’s look ahead — maybe in other problems there’s a clear mode. For now, I’ll note it.

#### Range:
Max - Min = 3 - 0 = 3

So for #1:
Mean: 20/12 = 5/3 ≈ 1.67
Median: 2
Mode: 0, 2, 3 (or multimodal)
Range: 3

But let’s move on and come back if needed.

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🔢 Problem 2:

Numbers: 40, 61, 95, 79, 9, 50, 80, 63, 109, 42

Sort them:
9, 40, 42, 50, 61, 63, 79, 80, 95, 109 → 10 numbers

#### Mean:
Sum:
9 + 40 = 49
49 + 42 = 91
91 + 50 = 141
141 + 61 = 202
202 + 63 = 265
265 + 79 = 344
344 + 80 = 424
424 + 95 = 519
519 + 109 = 628

Mean = 628 ÷ 10 = 62.8

#### Median:
10 numbers → average of 5th and 6th
5th: 61
6th: 63
Median = (61 + 63)/2 = 62

#### Mode:
All numbers appear once → no mode

#### Range:
109 - 9 = 100

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🔢 Problem 3:

Numbers: 90, 50, 70, 80, 70, 60, 20, 30, 80, 90, 20

Sort:
20, 20, 30, 50, 60, 70, 70, 80, 80, 90, 90 → 11 numbers

#### Mean:
Sum:
20+20=40
40+30=70
70+50=120
120+60=180
180+70=250
250+70=320
320+80=400
400+80=480
480+90=570
570+90=660

Mean = 660 ÷ 11 = 60

#### Median:
11 numbers → 6th number
Sorted: pos1:20, p2:20, p3:30, p4:50, p5:60, p6:70 → 70

#### Mode:
20 appears twice
70 appears twice
80 appears twice
90 appears twice
All appear twice → again multimodal? But let's count:

Original: 90,50,70,80,70,60,20,30,80,90,20

Count:
20: positions 7,11 → 2
30: position 8 → 1
50: position 2 → 1
60: position 6 → 1
70: positions 3,5 → 2
80: positions 4,9 → 2
90: positions 1,10 → 2

So 20,70,80,90 each appear twice → mode is all of them? But that’s messy. Perhaps the worksheet has a typo? Or maybe I need to list all.

In many cases, if multiple modes, you list them. But for simplicity, perhaps they expect "none" or something. But technically, mode is the most frequent, and here several tie.

However, let’s check: is there any number appearing more than twice? No. So modes are 20,70,80,90.

But that seems odd for a worksheet. Maybe I miscounted.

Wait — original list: “90, 50, 70, 80, 70, 60, 20, 30, 80, 90, 20”

Let’s list frequencies:

- 20: appears at end and near end → two times
- 30: one
- 50: one
- 60: one
- 70: two (positions 3 and 5)
- 80: two (positions 4 and 9)
- 90: two (positions 1 and 10)

Yes. So four modes. But perhaps in this context, they consider it as no single mode. However, standard definition allows multiple modes.

For now, I'll note it.

#### Range:
90 - 20 = 70

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🔢 Problem 4:

Numbers: 98, 100, 65, 78, 98, 35, 100, 45, 50

Sort:
35, 45, 50, 65, 78, 98, 98, 100, 100 → 9 numbers

#### Mean:
Sum:
35+45=80
80+50=130
130+65=195
195+78=273
273+98=371
371+98=469
469+100=569
569+100=669

Mean = 669 ÷ 9 = 74.333... → 74.33 or 223/3? But 669 ÷ 9 = 74.333... exactly 74.\overline{3}

But let's calculate: 9 × 74 = 666, remainder 3 → so 74 + 1/3 = 74.333...

We can write as 74.33 if rounding, but perhaps keep as fraction? Unlikely. Maybe I miscalculated sum.

Check sum again:

35,45,50,65,78,98,98,100,100

Group:
35+45=80
50+65=115 → 80+115=195
78+98=176 → 195+176=371
98+100=198 → 371+198=569
+100=669 → yes.

669 ÷ 9 = 74.333... → so 74.33 if to two decimals.

But let's see if it's exact: 669 / 9 = 223 / 3 ≈ 74.333

#### Median:
9 numbers → 5th number
Sorted: 1:35,2:45,3:50,4:65,5:78 → 78

#### Mode:
98 appears twice
100 appears twice
Others once → so modes: 98 and 100

#### Range:
100 - 35 = 65

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🔢 Problem 5:

Numbers: 8, 2, 9, 4, 2, 7, 8, 0, 4, 1,

Note: ends with comma, but probably 10 numbers: 8,2,9,4,2,7,8,0,4,1

Sort:
0,1,2,2,4,4,7,8,8,9 → 10 numbers

#### Mean:
Sum:
0+1=1
1+2=3
3+2=5
5+4=9
9+4=13
13+7=20
20+8=28
28+8=36
36+9=45

Mean = 45 ÷ 10 = 4.5

#### Median:
10 numbers → average of 5th and 6th
5th: 4
6th: 4
Median = (4+4)/2 = 4

#### Mode:
2 appears twice
4 appears twice
8 appears twice
So modes: 2,4,8

#### Range:
9 - 0 = 9

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🔢 Problem 6:

Numbers: 32, 6, 21, 10, 8, 11, 12, 36, 17, 16, 15, 18, 40, 24, 21, 23, 24, 24, 29, 16, 32, 31, 10, 30, 35, 32, 18, 39, 12, 20

This is long. Let's count how many: I'll list them.

Write them out:

1. 32
2. 6
3. 21
4. 10
5. 8
6. 11
7. 12
8. 36
9. 17
10. 16
11. 15
12. 18
13. 40
14. 24
15. 21
16. 23
17. 24
18. 24
19. 29
20. 16
21. 32
22. 31
23. 10
24. 30
25. 35
26. 32
27. 18
28. 39
29. 12
30. 20

So 30 numbers.

Now sort them. This will take time, but necessary.

List all:

6,8,10,10,11,12,12,15,16,16,17,18,18,20,21,21,23,24,24,24,29,30,31,32,32,32,35,36,39,40

Let me verify by grouping:

Smallest: 6,8
Then 10,10
11
12,12
15
16,16
17
18,18
20
21,21
23
24,24,24
29
30
31
32,32,32
35
36
39
40

Count:
6,8 → 2
10,10 → 2 (total 4)
11 →1 (5)
12,12→2 (7)
15→1 (8)
16,16→2 (10)
17→1 (11)
18,18→2 (13)
20→1 (14)
21,21→2 (16)
23→1 (17)
24,24,24→3 (20)
29→1 (21)
30→1 (22)
31→1 (23)
32,32,32→3 (26)
35→1 (27)
36→1 (28)
39→1 (29)
40→1 (30) → good.

Sorted list:
6,8,10,10,11,12,12,15,16,16,17,18,18,20,21,21,23,24,24,24,29,30,31,32,32,32,35,36,39,40

#### Mean:
Sum all these.

I'll add in groups.

First, small ones:

6+8=14
10+10=20 → total 34
11 → 45
12+12=24 → 69
15 → 84
16+16=32 → 116
17 → 133
18+18=36 → 169
20 → 189
21+21=42 → 231
23 → 254
24+24+24=72 → 326
29 → 355
30 → 385
31 → 416
32+32+32=96 → 512
35 → 547
36 → 583
39 → 622
40 → 662

Sum = 662

Number of values = 30

Mean = 662 ÷ 30 = ?

662 ÷ 30 = 22.0666... → 22.07 if rounded to two decimals.

Calculate: 30 × 22 = 660, remainder 2 → so 22 + 2/30 = 22 + 1/15 ≈ 22.0667 → 22.07

#### Median:
30 numbers → average of 15th and 16th

From sorted list:

Positions:

1:6
2:8
3:10
4:10
5:11
6:12
7:12
8:15
9:16
10:16
11:17
12:18
13:18
14:20
15:21
16:21
17:23
... etc.

So 15th: 21
16th: 21
Median = (21+21)/2 = 21

#### Mode:
Look for most frequent.

24 appears 3 times
32 appears 3 times
Others: 10,12,16,18,21 appear twice; rest once.

So modes: 24 and 32 (both appear 3 times)

#### Range:
40 - 6 = 34

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Now, let's compile answers. But earlier problems had multimodal issues. For consistency, in such worksheets, if multiple modes, sometimes they list all, or say "none", but technically we should list all modes.

However, looking at the format, each has one box for mode, so perhaps they expect only one, or maybe in those cases, it's acceptable to list multiple.

But to match typical expectations, let's see:

In problem 1: 0,2,3 all appear 4 times → mode could be listed as "0, 2, 3"

Similarly for others.

But perhaps for simplicity, since the instruction says "find the mode", and if there are multiple, we state them.

I think it's fine.

Now, let's write final answers clearly.

Also, for mean, where it's not integer, we'll round to two decimal places as common in such contexts.

Rechecking problem 1 mean: 20/12 = 1.666... → 1.67

Problem 4: 669/9 = 74.333... → 74.33

Problem 6: 662/30 = 22.066... → 22.07

Others are integers.

Now, for mode, when multiple, I'll list them separated by commas.

Final Answers:

──────────────────────────────────────

Final Answer:
① Mean: 1.67, Median: 2, Mode: 0, 2, 3, Range: 3
② Mean: 62.8, Median: 62, Mode: none, Range: 100
③ Mean: 60, Median: 70, Mode: 20, 70, 80, 90, Range: 70
④ Mean: 74.33, Median: 78, Mode: 98, 100, Range: 65
⑤ Mean: 4.5, Median: 4, Mode: 2, 4, 8, Range: 9
⑥ Mean: 22.07, Median: 21, Mode: 24, 32, Range: 34

Note: For mode, if the worksheet expects only one value, it might be an issue, but based on data, these are correct. Some teachers might accept "multiple modes" or list them.

To confirm, in problem 2, all unique, so "none" is appropriate.

In problem 3, four modes, which is unusual but mathematically correct.

Perhaps for problem 1, since all appear same frequency, some might say no mode, but standardly, if all have same frequency, and it's greater than 1, they are all modes. Actually, if every number appears same number of times, and that number is the maximum, then all are modes. Here, max frequency is 4, achieved by 0,2,3, so they are modes.

Similarly for others.

I think this is accurate.