Math worksheet for calculating mean, median, and mode with ten number sets.

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

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Problem 1: 32, 33, 22, 85, 58

Step 1: Mean
Add all numbers: 32 + 33 + 22 + 85 + 58 = 230
Divide by count (5): 230 ÷ 5 = 46

Step 2: Median
Order the numbers: 22, 32, 33, 58, 85
Middle number is 33

Step 3: Mode
No number repeats → No mode

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Problem 2: 18, 18, 26, 36, 64

Mean: 18+18+26+36+64 = 162 → 162 ÷ 5 = 32.4

Median: Ordered: 18, 18, 26, 36, 64 → middle is 26

Mode: 18 appears twice → 18

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Problem 3: 58, 84, 90, 90, 97

Mean: 58+84+90+90+97 = 419 → 419 ÷ 5 = 83.8

Median: Ordered: 58, 84, 90, 90, 97 → middle is 90

Mode: 90 appears twice → 90

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Problem 4: 9, 15, 28, 10, 8

Mean: 9+15+28+10+8 = 70 → 70 ÷ 5 = 14

Median: Order: 8, 9, 10, 15, 28 → middle is 10

Mode: No repeats → No mode

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Problem 5: 21, 15, 16, 25, 13, 18

Mean: 21+15+16+25+13+18 = 108 → 108 ÷ 6 = 18

Median: Order: 13, 15, 16, 18, 21, 25 → two middle numbers: 16 and 18 → average = (16+18)/2 = 17

Mode: No repeats → No mode

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Problem 6: 53, 44, 10, 45, 59, 97, 77

Mean: 53+44+10+45+59+97+77 = 385 → 385 ÷ 7 = 55

Median: Order: 10, 44, 45, 53, 59, 77, 97 → middle is 53

Mode: No repeats → No mode

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Problem 7: 5, 3, 2, 6, 5, 2, 5

Mean: 5+3+2+6+5+2+5 = 28 → 28 ÷ 7 = 4

Median: Order: 2, 2, 3, 5, 5, 5, 6 → middle is 5

Mode: 5 appears 3 times → 5

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Problem 8: 6, 6, 6, 7, 7, 5, 4

Mean: 6+6+6+7+7+5+4 = 41 → 41 ÷ 7 ≈ 5.857 (we can write as fraction or decimal — let’s use decimal rounded to one place: 5.9)

Wait — better to keep exact: 41 ÷ 7 = 5.857... but since it's a worksheet, maybe they expect fraction? But usually decimal is fine. Let’s check: 41/7 = 5 and 6/7 → but we’ll go with 5.9 if rounding, but actually in math worksheets, sometimes they want exact. Let me recalculate sum: 6+6+6=18, 7+7=14, 5+4=9 → 18+14=32+9=41 → yes. So mean = 41/7 ≈ 5.86 — but perhaps leave as mixed number? Actually, for this level, decimal is okay. I’ll use 5.9 (rounded to nearest tenth).

But wait — let’s be precise. The problem doesn’t specify, so I’ll use exact value: 41/7 or approximately 5.86. But looking at other problems, they used decimals like 32.4, so we’ll do same: 5.9 (if rounding to tenths). Actually 41÷7 = 5.857... → rounds to 5.9

But let’s double-check: 7 × 5.857 = 41? Yes. So for consistency, I’ll write 5.9

Actually — better to calculate exactly: 41 ÷ 7 = 5.857142... → but in school, often round to one decimal: 5.9

Alternatively, maybe they expect fraction? But no, previous answers used decimals. So I’ll go with 5.9

Wait — problem 2 had 32.4 which is exact (162/5=32.4), so here 41/7 is not terminating. Hmm. Maybe I should leave as fraction? But the worksheet has blanks — probably expects decimal. Let me see: 41 ÷ 7 = 5.857... — perhaps write as 5.86? But to match style, let’s use one decimal: 5.9

Actually, let’s check calculation again: 6+6+6+7+7+5+4 = 41, yes. 41/7 = 5.857... — I think for accuracy, I’ll write 5.86 (two decimals) but problem 2 had one decimal. Wait, 162/5=32.4 — that’s exact. Here it’s not. Perhaps the worksheet allows fractions? But no indication. I’ll use 5.9 as rounded.

But to be safe, let’s compute exactly: 41 ÷ 7 = 5 and 6/7 — but I think for this context, decimal is expected. I’ll put 5.9

Actually, let’s look at problem 3: 419/5=83.8 — exact. Problem 4: 70/5=14 — exact. Problem 5: 108/6=18 — exact. Problem 6: 385/7=55 — exact. Problem 7: 28/7=4 — exact. Only problem 2 and now 8 are non-integer. Problem 2 was 32.4 — exact decimal. For 8, 41/7 is repeating. So perhaps write as fraction? But the blank says “Mean: ___” — likely expects decimal. I’ll write 5.9 (rounded to nearest tenth).

But let’s confirm: 7 * 5.9 = 41.3 — too big. 7*5.8=40.6, 7*5.85=40.95, 7*5.86=41.02 — close. Actually 41/7 = 5.857142... so to one decimal: 5.9, to two: 5.86. Since problem 2 used one decimal (32.4), I’ll use one decimal: 5.9

Okay.

Median: Order: 4, 5, 6, 6, 6, 7, 7 → middle is 6

Mode: 6 appears 3 times → 6

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Problem 9: 9, 19, 16, 9, 16, 2, 14, 8, 2

First, list: 9,19,16,9,16,2,14,8,2 → 9 numbers

Mean: Sum: 9+19=28, +16=44, +9=53, +16=69, +2=71, +14=85, +8=93, +2=95 → total 95
Count: 9 → 95 ÷ 9 ≈ 10.555... → round to 10.6 (one decimal)

Check: 9*10.555=95? 9*10=90, 9*0.555≈5, yes. So 10.6

Median: Order: 2,2,8,9,9,14,16,16,19 → 9 numbers, middle is 5th: 9

Mode: 2 appears twice, 9 appears twice, 16 appears twice → multiple modes? But typically, if more than one, we say "no mode" or list all? In many curricula, if multiple values tie for most frequent, there is no single mode, or sometimes called multimodal. But for this level, probably they expect to list all or say no mode? Looking at problem 7, only one mode. Here, three numbers appear twice. So technically, modes are 2,9,16. But the worksheet might expect "no mode" or list them. I think for simplicity, since it's elementary, they might consider no unique mode. But let's see standard definition: mode is the value that appears most frequently. If multiple, then multiple modes. But in some contexts, they say "no mode" if no single value appears more than others. Actually, in this case, 2,9,16 each appear twice, and others once, so they are all modes. But the blank is "Mode: ___" — singular. Probably, they expect to list all or say none. To be safe, I'll say 2, 9, 16 but that might not fit. Alternatively, in many school worksheets, if there is a tie, they say "no mode". Let me check online or recall: actually, common practice in elementary math is that if no number appears more than others, no mode; but here, several appear same max frequency. I think it's acceptable to say "no mode" or list. But to match the format, perhaps write "none" or list. Looking at problem 1,4,5,6 — they have "no mode". Here, since multiple, I'll write 2, 9, 16 but that might be messy. Another way: sometimes they say "bimodal" etc., but not here. I think for this worksheet, since it's basic, and blanks are small, probably they expect "no mode" if not unique. But let's calculate frequency:

Numbers: 2:2 times, 8:1, 9:2, 14:1, 16:2, 19:1 → so 2,9,16 each have highest frequency (2). So modes are 2,9,16. But perhaps the answer is to list them. I'll put 2, 9, 16

But to confirm, in problem 8, 6 appeared 3 times, others less, so clear mode. Here, tie. I think it's fine to list.

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Problem 10: 19, 5, 9, 12, 10, 5

Mean: 19+5+9+12+10+5 = 60 → 60 ÷ 6 = 10

Median: Order: 5,5,9,10,12,19 → two middle: 9 and 10 → average = (9+10)/2 = 9.5

Mode: 5 appears twice → 5

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Now, compiling all answers:

Problem 1: Mean=46, Median=33, Mode=no mode
Problem 2: Mean=32.4, Median=26, Mode=18
Problem 3: Mean=83.8, Median=90, Mode=90
Problem 4: Mean=14, Median=10, Mode=no mode
Problem 5: Mean=18, Median=17, Mode=no mode
Problem 6: Mean=55, Median=53, Mode=no mode
Problem 7: Mean=4, Median=5, Mode=5
Problem 8: Mean=5.9 (or 41/7), Median=6, Mode=6
Problem 9: Mean=10.6 (95/9≈10.555→10.6), Median=9, Mode=2,9,16
Problem 10: Mean=10, Median=9.5, Mode=5

For problem 8 mean, to be precise, 41÷7=5.857... — if we round to one decimal, it's 5.9, but 5.86 is more accurate. However, since problem 2 used 32.4 (which is exact), and here it's not, perhaps write as fraction? But I think for consistency, I'll use 5.9 as per rounding convention.

Similarly, problem 9 mean: 95÷9=10.555... → 10.6

For mode in problem 9, I'll list the modes.

Final Answer:

1. Mean: 46, Median: 33, Mode: no mode
2. Mean: 32.4, Median: 26, Mode: 18
3. Mean: 83.8, Median: 90, Mode: 90
4. Mean: 14, Median: 10, Mode: no mode
5. Mean: 18, Median: 17, Mode: no mode
6. Mean: 55, Median: 53, Mode: no mode
7. Mean: 4, Median: 5, Mode: 5
8. Mean: 5.9, Median: 6, Mode: 6
9. Mean: 10.6, Median: 9, Mode: 2, 9, 16
10. Mean: 10, Median: 9.5, Mode: 5

Note: For problem 8, if exact fraction is preferred, it's 41/7, but decimal is used in similar cases. For problem 9, modes are multiple.

I think this is correct.