Printable math worksheet for practicing 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: 62, 10, 24, 24, 10, 62, 24, 10, 62, 62

First, sort them to make it easier:
→ 10, 10, 10, 24, 24, 24, 62, 62, 62, 62

Mean: Add all numbers → divide by count (10 numbers)
Sum = 10+10+10+24+24+24+62+62+62+62
= (10×3) + (24×3) + (62×4) = 30 + 72 + 248 = 350
Mean = 350 ÷ 10 = 35

Median: Middle value(s). Since 10 numbers (even), average of 5th and 6th.
Sorted: positions 5 and 6 are both 24 → Median = (24 + 24)/2 = 24

Mode: Most frequent number → 62 appears 4 times → 62

Range: Max - Min = 62 - 10 = 52

✔ Problem 1: Mean=35, Median=24, Mode=62, Range=52

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Problem 2:

Numbers: 3, 17, 17, 11, 8, 13, 5, 18

Sort: 3, 5, 8, 11, 13, 17, 17, 18 → 8 numbers

Mean: Sum = 3+5+8+11+13+17+17+18 = let’s add:
3+5=8; 8+8=16; 16+11=27; 27+13=40; 40+17=57; 57+17=74; 74+18=92
Mean = 92 ÷ 8 = 11.5

Median: Average of 4th and 5th → 11 and 13 → (11+13)/2 = 12

Mode: 17 appears twice, others once → 17

Range: 18 - 3 = 15

✔ Problem 2: Mean=11.5, Median=12, Mode=17, Range=15

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Problem 3:

Numbers: 13, 6, 24, 18, 33, 5, 13, 48, 9, 11, 36, 28, 15, 6, 13

Count: 15 numbers → odd → median is middle one (8th after sorting)

Sort:
5, 6, 6, 9, 11, 13, 13, 13, 15, 18, 24, 28, 33, 36, 48

Mean: Add all:
5+6=11; +6=17; +9=26; +11=37; +13=50; +13=63; +13=76; +15=91; +18=109; +24=133; +28=161; +33=194; +36=230; +48=278
Sum = 278
Mean = 278 ÷ 15 ≈ 18.533... → Let’s keep as fraction or decimal? Usually round to 1 or 2 decimals. But since others were exact, maybe check again.

Wait — let me recalculate sum carefully:

List: 5, 6, 6, 9, 11, 13, 13, 13, 15, 18, 24, 28, 33, 36, 48

Group:
(5+6+6) = 17
(9+11) = 20 → total 37
(13×3)=39 → total 76
15+18=33 → total 109
24+28=52 → total 161
33+36=69 → total 230
+48 = 278 → correct.

278 ÷ 15 = 18.533... → we can write as 18.53 (rounded to two decimals) or leave as fraction? In school worksheets, often they expect exact if possible, but here it’s not. Let’s use 18.53 unless specified otherwise. Actually, let’s see if I made a mistake — no, 15 × 18 = 270, remainder 8 → so 18 and 8/15 ≈ 18.533. OK.

But wait — maybe the worksheet expects whole numbers? Let me double-check the original list:

Original: 13,6,24,18,33,5,13,48,9,11,36,28,15,6,13 → yes, 15 numbers.

Actually, perhaps I should present mean as 18.5 if rounding to one decimal? But let’s be precise. Since other problems had .5, maybe this is fine. Alternatively, maybe I miscalculated?

Wait — let me add in pairs:

5 + 48 = 53
6 + 36 = 42
6 + 33 = 39
9 + 28 = 37
11 + 24 = 35
13 + 18 = 31
13 + 15 = 28
13 left → now add these: 53+42=95; +39=134; +37=171; +35=206; +31=237; +28=265; +13=278 → same.

So mean = 278 / 15 = 18.533... → I'll go with 18.53 (but actually, in many curricula, they might accept 18.5 or even 18.53). To be safe, let's use 18.53.

But wait — looking at problem 2, they used 11.5, so decimals are allowed. So we’ll use 18.53? Actually, 278 ÷ 15 = 18.5333... which rounds to 18.53 if two decimals, or 18.5 if one. Since problem 2 used one decimal, let’s use 18.5? No — 11.5 was exact. Here it’s repeating. Maybe better to write as fraction? But worksheet has boxes — probably expects decimal.

Alternatively, perhaps I should check if the sum is wrong? Let me try adding in order:

Start over:

5
+6 = 11
+6 = 17
+9 = 26
+11 = 37
+13 = 50
+13 = 63
+13 = 76
+15 = 91
+18 = 109
+24 = 133
+28 = 161
+33 = 194
+36 = 230
+48 = 278 → confirmed.

278 ÷ 15 = 18.533... → I think for consistency, since problem 2 had .5, and this is common, I’ll report 18.53 but actually, let’s see what the standard is. Perhaps the worksheet expects exact values only when possible. But here it’s not. I’ll proceed with 18.53 for now, but note that sometimes they want rounded to nearest tenth: 18.5.

Wait — let’s look ahead. Problem 4 and 5 may have similar issues. For accuracy, I’ll calculate exactly and then decide.

Actually, let’s do all first and then finalize.

Median: 15 numbers → 8th in sorted list:
Sorted: pos1:5, 2:6, 3:6, 4:9, 5:11, 6:13, 7:13, 8:13, 9:15, ... → 8th is 13

Mode: 13 appears three times, 6 appears twice → 13

Range: 48 - 5 = 43

So for Problem 3: Mean≈18.53, Median=13, Mode=13, Range=43

But to match format, perhaps write mean as 18.5? Or 18.53? I think 18.53 is more accurate. However, let’s check online or standard practice — in many cases, they round to one decimal if needed. Since 278/15 = 18.533..., rounding to one decimal is 18.5. And problem 2 used 11.5, which is exact half. So perhaps 18.5 is acceptable.

I’ll go with 18.5 for simplicity, as it’s commonly done in such worksheets.

Actually, let’s confirm with calculator: 278 ÷ 15 = 18.5333... → if we round to one decimal place, it’s 18.5. Yes.

So Mean = 18.5

✔ Problem 3: Mean=18.5, Median=13, Mode=13, Range=43

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Problem 4:

Numbers: 8, 9, 5, 8, 3, 7, 3, 9, 2

Sort: 2, 3, 3, 5, 7, 8, 8, 9, 9 → 9 numbers

Mean: Sum = 2+3+3+5+7+8+8+9+9
= (2) + (3+3)=6 → 8; +5=13; +7=20; +8+8=36; +9+9=54
Sum = 54
Mean = 54 ÷ 9 = 6

Median: 9 numbers → 5th is middle → sorted: pos5 is 7

Mode: 3,8,9 each appear twice → multimodal? But usually, if multiple modes, list all. However, in many school contexts, if there’s a tie, they might say “no mode” or list all. But here, three numbers tie for most frequent (each twice). The definition of mode is the most frequent value — if multiple, it’s multimodal. But in basic worksheets, sometimes they expect you to list all or say "none" if no single mode. Wait — actually, each appears twice, and no number appears more than twice, so all three are modes? That’s unusual.

Let me check counts:

2:1, 3:2, 5:1, 7:1, 8:2, 9:2 → so 3,8,9 each appear twice → so modes are 3,8,9.

But in some curricula, they might consider it as having no unique mode, but technically, it has multiple modes. However, looking at the worksheet, it has one box for mode — so perhaps they expect to list all or choose one? This is ambiguous.

In many elementary/middle school contexts, if there is a tie for highest frequency, they still call them all modes. But since the box is singular, maybe they expect to write all? Or perhaps I miscounted.

Wait — is there a number that appears more than twice? No. So strictly, modes are 3,8,9.

But let’s see the answer format — probably they expect to list them comma-separated or something. Since it’s a box, maybe write "3,8,9".

Alternatively, perhaps the problem intends for us to recognize multiple modes. I’ll go with listing them.

Range: 9 - 2 = 7

✔ Problem 4: Mean=6, Median=7, Mode=3,8,9, Range=7

But to fit the box, perhaps write "3, 8, 9"

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Problem 5:

Numbers: 6, 3, 9, 5, 7, 2, 5, 5, 3

Sort: 2, 3, 3, 5, 5, 5, 6, 7, 9 → 9 numbers

Mean: Sum = 2+3+3+5+5+5+6+7+9
= 2+3=5; +3=8; +5=13; +5=18; +5=23; +6=29; +7=36; +9=45
Sum = 45
Mean = 45 ÷ 9 = 5

Median: 5th number → sorted: pos5 is 5

Mode: 5 appears three times → 5

Range: 9 - 2 = 7

✔ Problem 5: Mean=5, Median=5, Mode=5, Range=7

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Problem 6:

Numbers: 86, 13, 60, 55, 61, 97, 30, 98, 79, 52, 18

Count: 11 numbers → odd → median is 6th after sorting

Sort:
13, 18, 30, 52, 55, 60, 61, 79, 86, 97, 98

Mean: Sum = 13+18+30+52+55+60+61+79+86+97+98

Add step by step:

13+18=31
31+30=61
61+52=113
113+55=168
168+60=228
228+61=289
289+79=368
368+86=454
454+97=551
551+98=649

Sum = 649
Mean = 649 ÷ 11 = ?
11 × 59 = 649 → because 11×60=660, minus 11=649 → so 59

Exactly 59.

Median: 11 numbers → 6th is 60

Mode: All numbers appear once → no mode? Or "none"? In some contexts, they say "no mode" if all frequencies are 1.

Check: 13,18,30,52,55,60,61,79,86,97,98 — all unique → so no mode

But the worksheet has a box — perhaps write "none" or leave blank? Typically, in such cases, we say "no mode".

Range: 98 - 13 = 85

✔ Problem 6: Mean=59, Median=60, Mode=no mode, Range=85

But how to write "no mode"? Perhaps "none" or just state it. Since it's a box, maybe "N/A" or "none". I think "none" is appropriate.

Now, compiling all answers.

For Problem 4 mode, since multiple, I’ll write "3, 8, 9"

For Problem 6 mode, "none"

Final Answers:

Problem 1: Mean=35, Median=24, Mode=62, Range=52
Problem 2: Mean=11.5, Median=12, Mode=17, Range=15
Problem 3: Mean=18.5, Median=13, Mode=13, Range=43
Problem 4: Mean=6, Median=7, Mode=3,8,9, Range=7
Problem 5: Mean=5, Median=5, Mode=5, Range=7
Problem 6: Mean=59, Median=60, Mode=none, Range=85

Note: For Problem 3 mean, I used 18.5 (rounded to one decimal) as 278/15=18.533...≈18.5
For Problem 4 mode, listed all modes.
For Problem 6 mode, "none" since no repeated values.

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Final Answer:
1. Mean: 35, Median: 24, Mode: 62, Range: 52
2. Mean: 11.5, Median: 12, Mode: 17, Range: 15
3. Mean: 18.5, Median: 13, Mode: 13, Range: 43
4. Mean: 6, Median: 7, Mode: 3, 8, 9, Range: 7
5. Mean: 5, Median: 5, Mode: 5, Range: 7
6. Mean: 59, Median: 60, Mode: none, Range: 85