Worksheet for calculating mean, median, mode, and range of given number sets.

Let’s solve each problem step by step. For each set of numbers, we need to find:

- Mean = average (add all numbers, divide by how many there are)
- Median = middle number when sorted from smallest to largest
- Mode = number that appears most often
- Range = biggest number minus smallest number

We’ll do this for each of the 5 sets.

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Problem 1: {93,47,47,47,10,83,14,44,27,91,44,84,72,60,47,87,84,23,15,38}

Step 1: Sort the numbers
Sorted: 10, 14, 15, 23, 27, 38, 44, 44, 47, 47, 47, 47, 60, 72, 83, 84, 84, 87, 91, 93

Count: 20 numbers → even count → median is average of 10th and 11th numbers

10th = 47, 11th = 47 → Median = (47 + 47)/2 = 47

Mode: Which number appears most?
47 appears 4 times → more than any other → Mode = 47

Range: Max - Min = 93 - 10 = 83

Mean: Add all numbers:

Let’s add carefully:

Group them to make it easier:

(10+14+15+23+27) = 89
(38+44+44+47+47) = 220
(47+47+60+72+83) = 309
(84+84+87+91+93) = 439

Total = 89 + 220 = 309; 309 + 309 = 618; 618 + 439 = 1057

Wait — let me double-check addition:

Better to list and sum:

10
14 → 24
15 → 39
23 → 62
27 → 89
38 → 127
44 → 171
44 → 215
47 → 262
47 → 309
47 → 356
47 → 403
60 → 463
72 → 535
83 → 618
84 → 702
84 → 786
87 → 873
91 → 964
93 → 1057

Yes, total = 1057
Number of values = 20
Mean = 1057 ÷ 20 = 52.85

✔ So for #1:
Mean = 52.85, Median = 47, Mode = 47, Range = 83

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Problem 2: {60,29,26,86,29,79,31,25,95,84,20,84,36,79,13,69,84,26,89,24}

Sort:
13, 20, 24, 25, 26, 26, 29, 29, 31, 36, 60, 69, 79, 79, 84, 84, 84, 86, 89, 95

Count = 20 → median = average of 10th and 11th

10th = 36, 11th = 60 → Median = (36 + 60)/2 = 48

Mode: Look for most frequent
84 appears 3 times → others less → Mode = 84

Range: 95 - 13 = 82

Mean: Add all

List:

13
20 → 33
24 → 57
25 → 82
26 → 108
26 → 134
29 → 163
29 → 192
31 → 223
36 → 259
60 → 319
69 → 388
79 → 467
79 → 546
84 → 630
84 → 714
84 → 798
86 → 884
89 → 973
95 → 1068

Total = 1068
Mean = 1068 ÷ 20 = 53.4

✔ #2: Mean=53.4, Median=48, Mode=84, Range=82

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Problem 3: {26,74,49,26,86,55,29,26,74,18,94,27,73,48,19,84,62,19,32,95}

Sort:
18, 19, 19, 26, 26, 26, 27, 29, 32, 48, 49, 55, 62, 73, 74, 74, 84, 86, 94, 95

Count = 20 → median = avg of 10th and 11th

10th = 48, 11th = 49 → Median = (48+49)/2 = 48.5

Mode: 26 appears 3 times → Mode = 26

Range: 95 - 18 = 77

Mean: Add all

18
19 → 37
19 → 56
26 → 82
26 → 108
26 → 134
27 → 161
29 → 190
32 → 222
48 → 270
49 → 319
55 → 374
62 → 436
73 → 509
74 → 583
74 → 657
84 → 741
86 → 827
94 → 921
95 → 1016

Total = 1016
Mean = 1016 ÷ 20 = 50.8

✔ #3: Mean=50.8, Median=48.5, Mode=26, Range=77

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Problem 4: {20,43,38,86,18,63,49,80,58,11,48,72,38,95,10,40,63,20,97,18}

Sort:
10, 11, 18, 18, 20, 20, 38, 38, 40, 43, 48, 49, 58, 63, 63, 72, 80, 86, 95, 97

Count = 20 → median = avg of 10th and 11th

10th = 43, 11th = 48 → Median = (43+48)/2 = 45.5

Mode: 18, 20, 38, 63 each appear twice → multiple modes? But usually we pick one or say “no mode” if tie. However, in school math, sometimes they expect you to list all or pick first. Let’s check frequency:

18: 2
20: 2
38: 2
63: 2
Others: once → so actually four numbers tied at 2 → technically multimodal. But since instructions don’t specify, and often in such worksheets they expect the most frequent even if tied — but here no single winner. Wait — let me recount original data:

Original: 20,43,38,86,18,63,49,80,58,11,48,72,38,95,10,40,63,20,97,18

Count occurrences:

10:1
11:1
18:2 (positions 5 and 20)
20:2 (1 and 18)
38:2 (3 and 13)
40:1
43:1
48:1
49:1
58:1
63:2 (6 and 17)
72:1
80:1
86:1
95:1
97:1

So yes — 18,20,38,63 each appear twice → no unique mode. But some curricula say "if no number repeats more than others, then no mode". Others accept multiple modes. Since this is likely elementary/middle school, and worksheet says “mode”, probably expects us to report if there's a clear one — but here there isn't. Hmm.

Wait — maybe I miscounted? Let me write frequencies:

Value : Count
10:1
11:1
18:2
20:2
38:2
40:1
43:1
48:1
49:1
58:1
63:2
72:1
80:1
86:1
95:1
97:1

Yes — four values with count 2. So strictly speaking, no single mode. But perhaps the worksheet expects us to leave blank or write “none”? Or maybe I made mistake?

Alternatively — perhaps in context, they consider the smallest or something? No — better to be accurate.

But looking back at problem 1, 47 appeared 4 times — clear mode. Here, no clear mode. In many standardized tests, if multiple modes exist, they might ask for all, but since instruction just says “mode”, and doesn’t specify, I think safest is to say no mode or list all. But let’s see what typical answer would be.

Actually, upon second thought — in many school contexts, if there is a tie for highest frequency, they still call those the modes. So we can say modes are 18, 20, 38, 63. But since the format probably expects one value, and this is ambiguous, let me check if I missed a repeat.

Wait — 63 appears twice? Yes: positions 6 and 17. 38: positions 3 and 13. 20: 1 and 18. 18: 5 and 20. All correct.

Perhaps the problem intends for us to recognize multiple modes. But to match expected output, maybe I should note that. However, for consistency with other problems where one mode exists, and since this is unusual, let me proceed and mark as multiple modes: 18, 20, 38, 63 — but since final answer needs concise, perhaps write “none” or list. I think for accuracy, I'll state it.

But let’s continue calculation.

Range: 97 - 10 = 87

Mean: Add all

10
11 → 21
18 → 39
18 → 57
20 → 77
20 → 97
38 → 135
38 → 173
40 → 213
43 → 256
48 → 304
49 → 353
58 → 411
63 → 474
63 → 537
72 → 609
80 → 689
86 → 775
95 → 870
97 → 967

Total = 967
Mean = 967 ÷ 20 = 48.35

For mode — since no single number appears more than others, and four numbers tie at 2, I will write no unique mode or perhaps the worksheet expects us to pick one? To avoid confusion, let me assume that in this context, if there is a tie, we report all, but since the answer format may not allow, I’ll note it. Alternatively, perhaps I made error — let me count again from original list:

Original set: 20,43,38,86,18,63,49,80,58,11,48,72,38,95,10,40,63,20,97,18

Write in order and count:

After sorting: 10,11,18,18,20,20,38,38,40,43,48,49,58,63,63,72,80,86,95,97 — same as before.

Frequencies: 18:2, 20:2, 38:2, 63:2 — yes.

I think for correctness, I'll state mode as 18, 20, 38, 63 but since the problem might expect a single answer, and to match style, perhaps it's acceptable to say "multiple modes" — but for now, I'll proceed and in final answer list them.

Actually, looking at problem 5, it has repeated numbers too — let’s finish and come back.

✔ #4: Mean=48.35, Median=45.5, Mode=18,20,38,63 (or none), Range=87

But to keep consistent, perhaps the worksheet assumes there is always a mode — maybe I should recheck. Another idea: perhaps "mode" means the most frequent, and if tie, any of them — but that’s not standard. I think best to be accurate.

For now, I'll write mode as multiple: 18, 20, 38, 63 but in final answer box, perhaps list as comma-separated.

Moving on.

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Problem 5: {74,20,70,30,30,70,37,70,98,61,54,57,45,54,12,70,70,83,70,97}

Sort:
12, 20, 30, 30, 37, 45, 54, 54, 57, 61, 70, 70, 70, 70, 70, 70, 74, 83, 97, 98

Count = 20 → median = avg of 10th and 11th

10th = 61, 11th = 70 → Median = (61 + 70)/2 = 65.5

Mode: 70 appears how many times? Let’s count in sorted list: positions 11 to 16 → that’s 6 times!
Check original: 70 appears at indices: 3,6,8,16,17,19 — wait, let's list:

Original: 74,20,70,30,30,70,37,70,98,61,54,57,45,54,12,70,70,83,70,97

Positions with 70: 3rd,6th,8th,16th,17th,19th — that’s six times. Yes.

Other numbers: 30:2, 54:2, etc. So 70 appears 6 times → Mode = 70

Range: 98 - 12 = 86

Mean: Add all

12
20 → 32
30 → 62
30 → 92
37 → 129
45 → 174
54 → 228
54 → 282
57 → 339
61 → 400
70 → 470
70 → 540
70 → 610
70 → 680
70 → 750
70 → 820
74 → 894
83 → 977
97 → 1074
98 → 1172

Total = 1172
Mean = 1172 ÷ 20 = 58.6

✔ #5: Mean=58.6, Median=65.5, Mode=70, Range=86

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Now back to #4 mode issue. Since in #5, 70 clearly dominates, and in #1,47 does, but in #4, no single mode. Perhaps the worksheet allows for multiple modes, or maybe I should report the smallest one? But that’s arbitrary.

Upon checking common practice in such worksheets, sometimes they design problems to have a clear mode. Maybe I misread the set.

Let me double-check problem 4 original: {20,43,38,86,18,63,49,80,58,11,48,72,38,95,10,40,63,20,97,18}

Count 38: appears at position 3 and 13 → two times
20: position 1 and 18 → two
18: 5 and 20 → two
63: 6 and 17 → two

Yes. So no single mode. I think for accuracy, I'll state that there are multiple modes: 18, 20, 38, and 63. But since the answer format might expect a single value, and to avoid complication, perhaps in this context, we can say "no mode" — but technically, mode exists if there is repetition, and here there is, just not unique.

Another way: some definitions say mode is the value that appears most frequently, and if multiple, then multimodal. For school level, perhaps they expect us to list all or choose one. Given that, and to match the format, I'll write the mode as 18 (smallest) but that’s not correct. Better to be honest.

I recall that in some curricula, if there is a tie, they still call it mode and list them. So for final answer, I'll include all.

But let’s finalize all answers.

Summary:

1) Mean=52.85, Median=47, Mode=47, Range=83
2) Mean=53.4, Median=48, Mode=84, Range=82
3) Mean=50.8, Median=48.5, Mode=26, Range=77
4) Mean=48.35, Median=45.5, Mode=18,20,38,63, Range=87
5) Mean=58.6, Median=65.5, Mode=70, Range=86

For #4, since it's unusual, but mathematically correct, I'll keep it.

Now, for the final answer, I need to present only the answers, no explanation.

So:

Final Answer:
1) Mean: 52.85, Median: 47, Mode: 47, Range: 83
2) Mean: 53.4, Median: 48, Mode: 84, Range: 82
3) Mean: 50.8, Median: 48.5, Mode: 26, Range: 77
4) Mean: 48.35, Median: 45.5, Mode: 18, 20, 38, 63, Range: 87
5) Mean: 58.6, Median: 65.5, Mode: 70, Range: 86

But to make it neat, perhaps write as:

For each set:

1. Mean = 52.85, Median = 47, Mode = 47, Range = 83
2. Mean = 53.4, Median = 48, Mode = 84, Range = 82
3. Mean = 50.8, Median = 48.5, Mode = 26, Range = 77
4. Mean = 48.35, Median = 45.5, Mode = 18, 20, 38, 63, Range = 87
5. Mean = 58.6, Median = 65.5, Mode = 70, Range = 86

I think that's accurate.

Note: For #4, if the system expects a single mode, it might be an issue, but based on data, this is correct.

Perhaps in the original worksheet, they intended a different set, but based on given, this is right.

So final answer below.