Math worksheet for calculating mean, median, mode, and range.

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, let’s 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 = Total ÷ Count = 20 ÷ 12 ≈ 1.67 (but let’s keep as fraction: 20/12 = 5/3 ≈ 1.67)

Wait — actually, let me recalculate sum carefully:

List: 3,0,0,2,0,3,0,2,2,2,3,3
Count: 12 numbers

Sum:
3+0=3; +0=3; +2=5; +0=5; +3=8; +0=8; +2=10; +2=12; +2=14; +3=17; +3=20 → Yes, 20

Mean = 20 ÷ 12 = 5/3 ≈ 1.67

But maybe they want decimal? Let’s use 1.67 or exact fraction? Since others may be whole, perhaps we should check if I miscounted.

Wait — let’s list again with counts:

Zeros: four 0s → 0×4 = 0
Twos: four 2s → 2×4 = 8
Threes: four 3s → 3×4 = 12
Total = 0+8+12 = 20 → correct.

So mean = 20/12 = 5/3 ≈ 1.67

But in school problems, sometimes they expect rounded decimals. Let’s hold that.

#### 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 number appears most often?
0 appears 4 times
2 appears 4 times
3 appears 4 times
→ All appear same frequency → No mode? But wait — usually if multiple modes, we say “multimodal” but in basic math, sometimes they say “no mode” or list all.

Actually, looking at original list:
Original: 3,0,0,2,0,3,0,2,2,2,3,3
Let’s count again:

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

Yes, all three appear 4 times → so technically, modes are 0, 2, and 3

But some curricula say if all appear equally, there’s no mode. However, standard definition: mode is value(s) that appear most frequently. If tie, all are modes.

But let’s see what the worksheet expects. Maybe I made a mistake?

Wait — original list has 12 numbers:
Index:
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: indices 2,3,5,7 → 4
2: 4,8,9,10 → 4
3: 1,6,11,12 → 4

Yes. So multimodal.

But perhaps in this context, since it's elementary, they might accept "none" or list all. Let’s note that.

Actually, let me double-check the problem statement — it says “find the mean, median, mode and range”. It doesn’t specify how to handle ties. In many textbooks, if multiple values tie for highest frequency, you list them all as modes.

But let’s proceed and see other problems — maybe pattern emerges.

Alternatively, perhaps I misread the list? Let me write it again:

Problem ①: 3, 0, 0, 2, 0, 3, 0, 2, 2, 2, 3, 3

That’s 12 numbers. Counts:

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

Yes. So mode = 0, 2, 3

But maybe the worksheet expects only one mode? Or perhaps I should report all.

To be safe, I'll calculate range first.

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

Now back to mode — since all have same frequency, perhaps the answer is "no unique mode", but I think for accuracy, we should state all modes.

However, let’s look ahead — maybe in other problems, modes are clear. Perhaps for consistency, I’ll assume they want all modes listed.

But let’s move on and come back.

Actually, let me check online or standard practice — in K-8 math, if multiple modes, they often say “the modes are...” or if all equal, “no mode”. But here, since 0,2,3 all appear 4 times, and no number appears more, they are all modes.

I think it’s fine to list them.

But let’s do the calculation for mean again — 20/12 = 1.666... which is 1.67 if rounded to two decimals.

Perhaps they want fraction? Unlikely. Let’s keep as decimal.

Wait — maybe I can leave mean as 5/3, but probably not.

Another thought: perhaps I miscounted the numbers? Let me count the commas.

"3, 0, 0, 2, 0, 3, 0, 2, 2, 2, 3, 3" — that’s 12 numbers. Yes.

Okay, let’s proceed with:

Mean: 20/12 = 5/3 ≈ 1.67
Median: 2
Mode: 0, 2, 3 (all appear 4 times)
Range: 3

But for the box, they might expect single values. Perhaps in this case, since it's multimodal, but let's see problem ②.

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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

Count: 10 numbers

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

Sum = 628
Mean = 628 ÷ 10 = 62.8

#### Median:
10 numbers → average of 5th and 6th
Sorted: 1:9, 2:40, 3:42, 4:50, 5:61, 6:63, 7:79, 8:80, 9:95, 10:109
5th: 61, 6th: 63
Median = (61+63)/2 = 124/2 = 62

#### Mode:
Check frequencies:
9:1, 40:1, 42:1, 50:1, 61:1, 63:1, 79:1, 80:1, 95:1, 109:1 → all appear once → no mode

In such cases, we say "no mode" or "none".

#### Range:
Max - Min = 109 - 9 = 100

So for ②:
Mean: 62.8
Median: 62
Mode: none
Range: 100

Back to ① — similarly, if all appear same frequency, perhaps "no mode"? But in ①, they appear 4 times each, while in ②, all appear once. The definition is: mode is the value that appears most frequently. If multiple values tie for most frequent, they are all modes. If all values appear same frequency, then all are modes? That seems odd.

Standard definition: if every value appears the same number of times, there is no mode. For example, in {1,2,3}, no mode. In {1,1,2,2}, modes are 1 and 2.

In ①, 0,2,3 each appear 4 times, and no other numbers, so they are all modes. But in some contexts, they might say "bimodal" or "trimodal", but for simplicity, perhaps list them.

To resolve, let's look at problem .

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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

Count: 11 numbers

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

Sum = 660
Mean = 660 ÷ 11 = 60

#### Median:
11 numbers → 6th number
Sorted: 1:20,2:20,3:30,4:50,5:60,6:70,7:70,8:80,9:80,10:90,11:90
6th: 70

#### Mode:
Frequencies:
20:2, 30:1, 50:1, 60:1, 70:2, 80:2, 90:2
So 20,70,80,90 each appear twice? Wait:

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

- 20: twice
- 30: once
- 50: once
- 60: once
- 70: twice
- 80: twice
- 90: twice

So modes are 20,70,80,90 — all appear twice, and that's the highest frequency.

Again, multimodal.

This suggests that for these worksheets, they might expect to list all modes or say "multiple modes". But perhaps in the answer boxes, they allow multiple entries.

Maybe I should just calculate and see.

For now, let's continue.

#### Range:
Max - Min = 90 - 20 = 70

So ③:
Mean: 60
Median: 70
Mode: 20,70,80,90
Range: 70

But this is getting messy. Perhaps for ①, since 0,2,3 all appear 4 times, mode is those three.

Let's do problem ④.

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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

Count: 9 numbers

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

Sum = 669
Mean = 669 ÷ 9 = 74.333... = 74.3 recurring, or 223/3 ≈ 74.33

Calculate: 9*74 = 666, 669-666=3, so 74 + 3/9 = 74 + 1/3 ≈ 74.33

#### Median:
9 numbers → 5th number
Sorted: 1:35,2:45,3:50,4:65,5:78,6:98,7:98,8:100,9:100
5th: 78

#### Mode:
Frequencies:
35:1, 45:1, 50:1, 65:1, 78:1, 98:2, 100:2
So 98 and 100 each appear twice → modes are 98 and 100

#### Range:
Max - Min = 100 - 35 = 65

So ④:
Mean: 74.33 (or 74.3)
Median: 78
Mode: 98, 100
Range: 65

Now problem ⑤.

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Problem ⑤:

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

Note: there's a comma after 1, but probably it's 10 numbers: 8,2,9,4,2,7,8,0,4,1

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

Count: 10 numbers

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

Sum = 45
Mean = 45 ÷ 10 = 4.5

#### Median:
10 numbers → average of 5th and 6th
Sorted: 1:0,2:1,3:2,4:2,5:4,6:4,7:7,8:8,9:8,10:9
5th: 4, 6th: 4
Median = (4+4)/2 = 4

#### Mode:
Frequencies:
0:1, 1:1, 2:2, 4:2, 7:1, 8:2, 9:1
So 2,4,8 each appear twice → modes are 2,4,8

#### Range:
Max - Min = 9 - 0 = 9

So ⑤:
Mean: 4.5
Median: 4
Mode: 2,4,8
Range: 9

Finally, problem ⑥.

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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

Let me count: that's 30 numbers? Let's list them:

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

Yes, 30 numbers.

Sort them to make it easier.

First, 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 count: from 6 to 40.

Write sorted:

Start with smallest:

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

Count: 30 numbers. Good.

#### Mean:
Sum all. This might take time, but let's group.

List with frequencies or add step by step.

Use pairing or calculator in mind.

Since it's large, let's sum in parts.

Group:

Small numbers: 6+8=14; +10=24; +10=34; +11=45; +12=57; +12=69; +15=84; +16=100; +16=116; +17=133; +18=151; +18=169; +20=189

Now medium: +21=210; +21=231; +23=254; +24=278; +24=302; +24=326; +29=355; +30=385; +31=416; +32=448; +32=480; +32=512; +35=547; +36=583; +39=622; +40=662

Sum = 662

Mean = 662 ÷ 30 = ?

30 * 22 = 660, so 662 / 30 = 22 + 2/30 = 22 + 1/15 ≈ 22.0667 or 22.07 if rounded.

Exactly 662/30 = 331/15 ≈ 22.0667

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

Sorted list:

Position:
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
18:24
19:24
20:24
21:29
22:30
23:31
24:32
25:32
26:32
27:35
28:36
29:39
30:40

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

#### Mode:
Find which number appears most frequently.

From sorted list:

6:1
8:1
10:2 (positions 3,4)
11:1
12:2 (6,7)
15:1
16:2 (9,10)
17:1
18:2 (12,13)
20:1
21:2 (15,16)
23:1
24:3 (18,19,20) ← ah! 24 appears three times
29:1
30:1
31:1
32:3 (24,25,26) ← also three times
35:1
36:1
39:1
40:1

So 24 appears 3 times, 32 appears 3 times. Others less.

Is there any with more? 24: positions 18,19,20 → three
32: 24,25,26 → three
Others: 10,12,16,18,21 each appear twice.

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

#### Range:
Max - Min = 40 - 6 = 34

So ⑥:
Mean: 662/30 ≈ 22.07
Median: 21
Mode: 24, 32
Range: 34

Now back to problem ①. Given that in other problems, when there are ties, we list all modes, for ①, modes are 0,2,3.

But let's confirm the sum for ① again.

Problem ①: 3,0,0,2,0,3,0,2,2,2,3,3

Sum: let's add sequentially:

Start: 3
+0=3
+0=3
+2=5
+0=5
+3=8
+0=8
+2=10
+2=12
+2=14
+3=17
+3=20 → yes, 20

Count: 12

Mean: 20/12 = 5/3 ≈ 1.6667 → typically rounded to 1.67

Median: sorted 0,0,0,0,2,2,2,2,3,3,3,3 → 6th and 7th are both 2 → median 2

Mode: 0,2,3 each appear 4 times → so modes are 0,2,3

Range: 3-0=3

Similarly for others.

Now, for the final answers, I need to present them clearly.

Also, for mean, in some cases like ④, 669/9=74.333..., which is 74.3 if rounded to one decimal, but 74.33 to two. Looking at ②, mean is 62.8, which is exact. ⑤ is 4.5, exact. ③ is 60, exact. ⑥ is approximately 22.07.

Perhaps for consistency, use two decimals where needed.

But in ①, 20/12 = 1.666... so 1.67

In ④, 669/9 = 74.333... so 74.33

In , 662/30 = 22.0666... so 22.07

Now, let's compile all answers.

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Final Answers:


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

For mode, in cases where there are multiple, I'll list them separated by commas.

In ②, since all appear once, "none" is appropriate.

Now, to box the final answer as per instruction.

The user wants the final answer section to contain only the final correct answer, after all reasoning.

So I'll write:

Final Answer:
① Mean: 1.67, Median: 2, Mode: 0,2,3, Range: 3
② Mean: 62.8, Median: 62, Mode: none, Range: 100
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
Mean: 22.07, Median: 21, Mode: 24,32, Range: 34

But to make it neat, perhaps format as:

For each problem, list the four values.

Since the worksheet has boxes, probably they expect numerical answers, and for mode, if multiple, list them.

I think this is accurate.

One thing: in problem ①, is the mean exactly 5/3? But 1.67 is fine.

Similarly, for ④, 74.33 is 74.3 recurring, but 74.33 is acceptable.

In ⑥, 22.07 is rounded.

To be precise, perhaps use fractions, but unlikely for this level.

I think decimals are fine.

So final answer.