This worksheet provides five real-world scenarios for students to practice calculating statistical measures like mean, median, mode, and range.

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

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Problem 1:

Data: 2, 16, 18, 22, 28, 12, 24, 8

First, let’s sort the numbers in order:
→ 2, 8, 12, 16, 18, 22, 24, 28

There are 8 numbers (even count).

Mean: Add all numbers → 2+8+12+16+18+22+24+28 = 130
Divide by 8 → 130 ÷ 8 = 16.25

Median: Average of the two middle numbers (4th and 5th):
16 and 18 → (16 + 18) 2 = 17

Mode: No number repeats → No mode

Range: Largest - Smallest = 28 - 2 = 26

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Problem 2:

Phillip has fish food bags:
- 3 red
- 2 green
- 4 yellow
- 3 orange
- 13 black
- 15 blue

Total bags = 3+2+4+3+13+15 = 40

We treat this as a frequency list. To find mean/median/mode/range, we need to think of it as a data set with repeated values? But wait — these are *categories*, not numerical data. However, since the question asks for mean, median, etc., it likely wants us to treat the *counts* as the data set.

So data set is: [3, 2, 4, 3, 13, 15]

Sort them: 2, 3, 3, 4, 13, 15

Mean: (2+3+3+4+13+15) = 40 → 40 ÷ 6 ≈ 6.67

Median: Middle two numbers (3rd and 4th): 3 and 4 → (3+4)/2 = 3.5

Mode: 3 appears twice → 3

Range: 15 - 2 = 13

*(Note: If the problem meant something else, like total fish food per color, but no — it says “find mean, median, mode & range” for the counts given. So this interpretation is correct.)*

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Problem 3:

Ten students’ quiz scores:
15, 8, 7, 13, 9, 13, 8, 7, 15, 16

Sort them:
7, 7, 8, 8, 9, 13, 13, 15, 15, 16

Mean: Sum = 7+7+8+8+9+13+13+15+15+16 = let’s add:

7+7=14; 8+8=16; 9; 13+13=26; 15+15=30; 16
→ 14+16=30; +9=39; +26=65; +30=95; +16=111
Wait — that can’t be right. Let me recalculate:

Actually:
7 + 7 = 14
14 + 8 = 22
22 + 8 = 30
30 + 9 = 39
39 + 13 = 52
52 + 13 = 65
65 + 15 = 80
80 + 15 = 95
95 + 16 = 111

But there are 10 numbers → 111 ÷ 10 = 11.1

Wait — that seems high? Let me check again:

List: 7,7,8,8,9,13,13,15,15,16

Add pairwise:

(7+16)=23
(7+15)=22
(8+15)=23
(8+13)=21
(9+13)=22
Total: 23+22=45; +23=68; +21=89; +22=111 → yes, 111.

Mean = 111 / 10 = 11.1

Median: 5th and 6th numbers: 9 and 13 → (9+13)/2 = 11

Mode: 7, 8, 13, 15 each appear twice → so multiple modes? Usually we say “no unique mode” or list all. But often in school, if more than one, they might accept “multiple modes” or just pick one? Wait — actually, standard definition: mode is the most frequent value(s). Here, four values tie at frequency 2. So technically, modes are 7, 8, 13, 15. But sometimes problems expect you to say “no mode” if none is strictly higher — but here all have same max frequency. Hmm.

Looking back at problem 1: when no repeat, we said “no mode”. Here, multiple repeats — so we should report all modes? Or perhaps the problem expects us to note that there are multiple modes.

But let’s see what’s typical in middle school: if multiple values share highest frequency, they are all modes. So answer: 7, 8, 13, 15

However, some curricula say “if more than one, call it multimodal” — but for simplicity, maybe list them.

Alternatively, perhaps I miscounted? Let’s recount frequencies:

7: appears 2 times
8: 2 times
9: 1 time
13: 2 times
15: 2 times
16: 1 time

Yes — four values appear twice. So modes: 7, 8, 13, 15

But maybe the problem expects “no single mode”? Unlikely — usually they want you to list all.

Wait — looking at problem 4 and 5, they may have clear modes. For now, I’ll go with listing all.

Range: 16 - 7 = 9

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Problem 4:

Evan’s music album had:
13 rock songs
10 classical
12 hip hop
10 pop
15 R&B
13 gospel
17 rap songs

Again, treat the counts as data set: [13, 10, 12, 10, 15, 13, 17]

Sort: 10, 10, 12, 13, 13, 15, 17

Mean: Sum = 10+10+12+13+13+15+17 = let’s compute:

10+10=20; +12=32; +13=45; +13=58; +15=73; +17=90
→ 90 ÷ 7 ≈ 12.857 → round to 12.86? Or keep as fraction? Probably decimal is fine.

Actually, 90 ÷ 7 = 12.857... → typically rounded to nearest hundredth: 12.86

But maybe exact fraction? School level usually uses decimals.

Median: 7 numbers → 4th number is median → 13

Mode: 10 and 13 both appear twice → 10 and 13

Range: 17 - 10 = 7

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Problem 5:

Tamara recorded bird sightings over 30 days:
- 6 cardinals on 5 days
- 25 robins on 8 days
- 9 jays on 6 days
- 30 chickadees on 6 days
- 8 bluebirds on 5 days

This is grouped data. We need to reconstruct the full data set? Or use weighted averages?

Actually, for mean, we can do weighted average.

Total observations = 30 days.

To find mean: multiply each value by its frequency, sum, divide by total days.

So:

Cardinals: 6 × 5 = 30
Robins: 25 × 8 = 200
Jays: 9 × 6 = 54
Chickadees: 30 × 6 = 180
Bluebirds: 8 × 5 = 40

Sum = 30 + 200 = 230; +54=284; +180=464; +40=504

Mean = 504 30 = 16.8

Now, median: we need the middle value when all 30 daily counts are listed in order.

We have frequencies:

- 6 appears 5 times
- 8 appears 5 times
- 9 appears 6 times
- 25 appears 8 times
- 30 appears 6 times

Let’s build cumulative frequency to find the 15th and 16th values (since n=30, even → average of 15th and 16th).

Cumulative:

Up to 6: 5 days → positions 1-5: value 6
Up to 8: 5+5=10 → positions 6-10: value 8
Up to 9: 10+6=16 → positions 11-16: value 9
Up to 25: 16+8=24 → positions 17-24: value 25
Up to 30: 24+6=30 → positions 25-30: value 30

So 15th and 16th values are both 9 (since positions 11-16 are 9)

Thus, median = (9 + 9)/2 = 9

Mode: Which value appears most frequently? Look at frequencies:

6: 5
8: 5
9: 6
25: 8 ← highest!
30: 6

So 25 appears 8 times → mode = 25

Range: Max value - min value = 30 - 6 = 24

(Note: Range is based on the actual values observed, not frequencies.)

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Now, compiling final answers for each problem.

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Final Answer:

Problem 1:
Mean: 16.25
Median: 17
Mode: No mode
Range: 26

Problem 2:
Mean: 6.67
Median: 3.5
Mode: 3
Range: 13

Problem 3:
Mean: 11.1
Median: 11
Mode: 7, 8, 13, 15
Range: 9

Problem 4:
Mean: 12.86
Median: 13
Mode: 10, 13
Range: 7

Problem 5:
Mean: 16.8
Median: 9
Mode: 25
Range: 24