Practice worksheet for calculating mean, median, mode, and range with four data sets.

Let’s solve each set of numbers step by step for Mean, Median, Mode, and Range.

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First Set: 9, 4, 2, 8, 2

Step 1: Sort the numbers → 2, 2, 4, 8, 9

- Mean: Add all numbers → 2 + 2 + 4 + 8 + 9 = 25
Divide by count (5) → 25 ÷ 5 = 5

- Median: Middle number in sorted list → 4th? Wait — 5 numbers → middle is 3rd → 4

- Mode: Most frequent number → 2 appears twice, others once → 2

- Range: Max - Min → 9 - 2 = 7

✔ First set done.

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Second Set: 9, 3, 9, 4, 4, 2, 4

Sort them → 2, 3, 4, 4, 4, 9, 9

Count = 7 numbers

- Mean: 2+3+4+4+4+9+9 = let’s add:
2+3=5; 5+4=9; 9+4=13; 13+4=17; 17+9=26; 26+9=35
35 ÷ 7 = 5

- Median: 7 numbers → middle is 4th → look at sorted:
1st:2, 2nd:3, 3rd:4, 4th:4 → 4

- Mode: 4 appears three times, 9 appears twice → 4

- Range: 9 - 2 = 7

✔ Second set done.

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Third Set: 6, 9, 10, 9, 3, 5

Sort → 3, 5, 6, 9, 9, 10

Count = 6

- Mean: 3+5+6+9+9+10 =
3+5=8; 8+6=14; 14+9=23; 23+9=32; 32+10=42
42 ÷ 6 = 7

- Median: Even count → average of 3rd and 4th → 6 and 9 → (6+9)/2 = 15/2 = 7.5

- Mode: 9 appears twice, others once → 9

- Range: 10 - 3 = 7

✔ Third set done.

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Fourth Set: 6, 9, 5, 2, 6, 5, 2

Sort → 2, 2, 5, 5, 6, 6, 9

Count = 7

- Mean: 2+2+5+5+6+6+9 =
2+2=4; 4+5=9; 9+5=14; 14+6=20; 20+6=26; 26+9=35
35 ÷ 7 = 5

- Median: 7 numbers → 4th → sorted: 1st:2, 2nd:2, 3rd:5, 4th:5 → 5

- Mode: 2, 5, and 6 each appear twice → so there are three modes: 2, 5, 6
But sometimes worksheets expect you to list all or say “no single mode” — but since multiple values tie for most frequent, we list them all. However, looking at context, maybe they want all modes listed. Let’s check:
Actually, in many school contexts, if more than one value ties, you list all. So → 2, 5, 6

Wait — let me double-check frequencies:
- 2: appears twice
- 5: appears twice
- 6: appears twice
- 9: once

Yes — multimodal. But some teachers might accept “none” or “multiple”. Since the worksheet doesn’t specify, I’ll go with listing all modes that appear most frequently.

But wait — actually, in standard definition, mode is the value(s) that appear most often. Here, 2, 5, and 6 all appear twice — which is the highest frequency. So yes, three modes.

However, looking back at previous sets, when there was a tie (like second set had 4 appearing 3x, 9 appearing 2x — no tie), here it’s a 3-way tie.

But let me check common practice: In elementary/middle school, sometimes they say “if more than one, list them” or “if all same frequency, no mode” — but here not all same — 9 appears only once.

Actually, 2,5,6 each appear twice — that’s the max. So modes are 2, 5, 6.

But perhaps the worksheet expects just one? Let me see the data again.

Alternatively, maybe I miscounted.

Original: 6,9,5,2,6,5,2 → that’s seven numbers.

Sorted: 2,2,5,5,6,6,9 → yes.

Frequencies:
- 2: 2 times
- 5: 2 times
- 6: 2 times
- 9: 1 time

So yes — three modes.

But in many basic worksheets, they might not expect multiple modes. However, mathematically correct is to report all.

Looking at the first set: 2 appeared twice, others once → mode=2.

Second set: 4 appeared three times → mode=4.

Third set: 9 appeared twice → mode=9.

Fourth set: three values tied at two appearances → so modes are 2,5,6.

I think we should write: 2, 5, 6

But let me confirm range:

- Range: 9 - 2 = 7

Okay.

Wait — actually, let me recalculate mean for fourth set to be sure:

Numbers: 6,9,5,2,6,5,2

Add: 6+9=15; 15+5=20; 20+2=22; 22+6=28; 28+5=33; 33+2=35 → yes, 35÷7=5.

Median: 4th in sorted 2,2,5,5,6,6,9 → 4th is 5 → correct.

Mode: as above.

Range: 9-2=7.

Now, for mode — since the problem likely expects a single answer per blank, but mathematically there are three modes, I need to decide.

In many standardized tests or worksheets, if there is a tie, they may say "no mode" or list all. But looking at the structure, each has one blank for mode.

Perhaps I made a mistake? Let me count again:

Set: 6,9,5,2,6,5,2

List:

Position 1: 6

2:9

3:5

4:2

5:6

6:5

7:2

So values: 6 (twice), 9 (once), 5 (twice), 2 (twice)

Yes — 2,5,6 each twice.

But perhaps in this context, since it's a worksheet for beginners, they might consider only if one value is strictly most frequent. But here, no single value is more frequent than others — all three are tied.

Standard definition: mode is the value that appears most frequently. If multiple values tie for most frequent, then there are multiple modes.

However, some curricula teach that if there is a tie, there is no mode — but that’s incorrect. The correct term is "multimodal".

Given that, and since the worksheet has only one line for "Mode =", I think the expected answer might be to list all, or perhaps they expect us to recognize it's multimodal.

But looking at similar problems online, often they list all modes separated by commas.

For example, if modes are 2,5,6, write "2, 5, 6"

I think that’s acceptable.

To be safe, let me see if any other interpretation.

Another thought: perhaps I should check if the problem intends for us to pick one — but no, mathematically it’s clear.

I’ll go with listing all.

But let me double-check the entire thing.

Perhaps for consistency, since in other cases there was a unique mode, here it’s different.

I recall that in some systems, if all values have same frequency, no mode, but here 9 has lower frequency.

No — 2,5,6 have higher frequency than 9.

So modes are 2,5,6.

Final decision: for mode in fourth set, answer is 2, 5, 6

But let me write it as "2, 5, 6" in the answer.

Now, compiling all answers.

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Final Answer:

First Set (9, 4, 2, 8, 2):
Mean = 5
Median = 4
Mode = 2
Range = 7

Second Set (9, 3, 9, 4, 4, 2, 4):
Mean = 5
Median = 4
Mode = 4
Range = 7

Third Set (6, 9, 10, 9, 3, 5):
Mean = 7
Median = 7.5
Mode = 9
Range = 7

Fourth Set (6, 9, 5, 2, 6, 5, 2):
Mean = 5
Median = 5
Mode = 2, 5, 6
Range = 7