Mean, Median, Mode and Range (B) Worksheet | PDF Printable ... - Free Printable

Let’s solve each part step by step.

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Section A – Row 1: Numbers are 5, 7, 8, 4, 7

First, sort them: 4, 5, 7, 7, 8

- Median: Middle number → 7
- Mode: Most frequent → 7 (appears twice)
- Mean: Add all: 4+5+7+7+8 = 31 → divide by 5 → 31 ÷ 5 = 6.2
- Range: Biggest - smallest → 8 - 4 = 4

✔ Answers: Median=7, Mode=7, Mean=6.2, Range=4

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Section A – Row 2: Numbers are 3, 6, 11, 4, 8, 0, and one missing card. Given: Mean = 5

There are 7 numbers total (including the blank). Let’s call the missing number x.

Sum of known numbers: 3 + 6 + 11 + 4 + 8 + 0 = 32

Total sum for mean of 5 over 7 numbers: 5 × 7 = 35

So, x = 35 - 32 = 3

Now full list: 3, 6, 11, 4, 8, 0, 3 → sort: 0, 3, 3, 4, 6, 8, 11

- Median: Middle (4th) → 4
- Mode: Most frequent → 3 (appears twice)
- Range: 11 - 0 = 11

✔ Answers: Missing number=3, Median=4, Range=11, Mode=3

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Section A – Row 3: Numbers are 7, 4, 2, and one missing. Given: Range = 7

Current numbers: 7, 4, 2 → min=2, max=7 → current range = 5

We need range = 7. So either:

- New number is bigger than 7 → then new max = x, so x - 2 = 7 → x = 9
OR
- New number is smaller than 2 → then new min = x, so 7 - x = 7 → x = 0

Check both possibilities.

Try x = 9: List = 2, 4, 7, 9 → sorted → median = (4+7)/2 = 5.5, mode? no repeats → none, mean = (2+4+7+9)/4 = 22/4 = 5.5

Try x = 0: List = 0, 2, 4, 7 → sorted → median = (2+4)/2 = 3, mode? none, mean = (0+2+4+7)/4 = 13/4 = 3.25

But we don’t have more info — wait! The problem gives “Range: 7” as a given, not something to find. So we must pick the value that makes range = 7.

Actually, let’s re-read: It says “Range: 7” is already filled in — meaning it’s given, and we have to find the missing number that makes range = 7.

Current min = 2, max = 7 → range = 5. To make range = 7, we need to extend by 2 on one end.

If we add 9, max becomes 9 → 9 - 2 = 7 ✔️
If we add 0, min becomes 0 → 7 - 0 = 7 ✔️

Which one? We need to see which fits with other stats? But none are given except range.

Wait — actually, looking at the layout, the red box is the fourth card — so we’re to fill in the missing number such that range = 7.

But there are two possible answers? That can’t be. Maybe we assume the numbers are positive integers? Or maybe we look at context.

Actually, let’s check if the problem expects us to choose based on typical worksheet patterns. Often they expect the higher number.

But let’s think differently — perhaps the range is given as 7, and we must find the missing number that makes range exactly 7.

Since current max-min = 5, we need to increase range by 2.

So adding 9 or 0 both work.

But let’s calculate mean and median for both and see if any match implied values? No other values are given.

Wait — actually, in the image, only “Range: 7” is given — others are blank. So we just need to find the missing number that makes range = 7.

But since two options exist, perhaps we missed something.

Look again: The cards are shown as 7, 4, 2, [blank]. If we add 0, list is 0,2,4,7 → range=7. If we add 9, list is 2,4,7,9 → range=7.

Both valid. But maybe the worksheet assumes non-negative? Both are non-negative.

Perhaps we should consider that in Section A Row 2, they used 0, so 0 is allowed.

But let’s see the next rows — maybe we can come back.

Actually, let’s proceed and note that both 0 and 9 are mathematically correct, but perhaps the intended answer is 9 because 0 might make mean too low? Not sure.

Wait — let’s calculate what the mean would be if we had to guess — but no mean is given.

Another idea: Perhaps the range is defined as max - min, and with existing numbers 2 and 7, to get range 7, the new number must be either 2 - 7 = -5? No, that would make min=-5, max=7, range=12 — too big.

No: if we add x, then new min = min(2,x), new max = max(7,x)

Set max - min = 7.

Case 1: x ≥ 7 → then max=x, min=2 → x - 2 = 7 → x=9

Case 2: x ≤ 2 → then max=7, min=x → 7 - x = 7 → x=0

Case 3: 2 < x < 7 → then max=7, min=2 → range=5 ≠7 → invalid

So only x=0 or x=9.

Now, perhaps the worksheet intends for us to use the fact that in the first row, numbers were positive, but 0 was used in row 2.

Maybe we can look at the answer format — but since this is text, I’ll go with 9 as it's more common in such problems to add a larger number.

But let’s hold on and do other rows first.

Actually, let’s skip and come back — or better, let’s assume x=9 for now and verify later if needed.

So for now, missing number = 9

List: 2,4,7,9

- Mean = (2+4+7+9)/4 = 22/4 = 5.5
- Median = (4+7)/2 = 5.5
- Mode = none (all unique) — but usually we say "no mode" or leave blank? In worksheets, sometimes they write "none"

But in the answer boxes, it’s empty — so perhaps we write "none" or leave it? But the instruction is to solve, so we’ll state it.

Actually, looking at the image, for mode, if no repeat, it’s often left blank or written as "none". But since the box is there, perhaps we put "none".

But let’s see row 4 — it has mode given as 15, etc.

For consistency, if no mode, we can write "none".

But let’s continue.

I think I made a mistake — in the original problem, for row 3, it says "Range: 7" is given, and we need to find the missing number. But also, after finding it, we need to find mean, mode, median.

With x=9: mean=5.5, median=5.5, mode=none

With x=0: mean=3.25, median=3, mode=none

Neither seems special.

Perhaps the problem has a typo, or I misread.

Another thought: maybe the range is 7, and the numbers are 7,4,2, and the missing one, and perhaps the missing one is such that when added, the range is 7, but maybe they expect the number that makes the set have a mode or something — but no.

Let’s look at the next row for clue.

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Section A – Row 4: Numbers are 10,8,15,19,4,15,15, and one missing. Given: Median = 12.5

There are 8 numbers total (7 given + 1 missing).

Sort the known numbers: 4,8,10,15,15,15,19

Add missing number x.

Median for 8 numbers is average of 4th and 5th when sorted.

Given median = 12.5, so (4th + 5th)/2 = 12.5 → 4th + 5th = 25

Current sorted without x: positions 1 to 7: 4,8,10,15,15,15,19

When we add x, it will insert somewhere.

Let’s denote the sorted list including x.

The 4th and 5th numbers must sum to 25.

Possible pairs that sum to 25: 10 and 15, 12 and 13, etc., but our numbers are integers, so likely 10 and 15.

10 + 15 = 25, yes.

So if 4th is 10 and 5th is 15, or vice versa, but since sorted, 4th ≤ 5th, so 4th=10, 5th=15.

In the current list without x, the 4th number is 15 (since 4,8,10,15,... so index 4 is 15).

After adding x, we need the 4th to be 10 and 5th to be 15.

That means x must be less than or equal to 10, so that it pushes the 15s to later positions.

Current list: pos1:4, pos2:8, pos3:10, pos4:15, pos5:15, pos6:15, pos7:19

If we add x ≤ 10, say x=10, then sorted: 4,8,10,10,15,15,15,19 → 4th=10, 5th=15 → sum=25, median=12.5 ✔️

If x=9: 4,8,9,10,15,15,15,19 → 4th=10, 5th=15 → same

If x=11: 4,8,10,11,15,15,15,19 → 4th=11, 5th=15 → sum=26, median=13 ≠12.5

If x=5: 4,5,8,10,15,15,15,19 → 4th=10, 5th=15 → good

So x can be any number ≤10, but probably integer, and likely they expect a specific value.

But we have more information? No, only median is given.

But we need to find mean, range, mode as well, but they are not given, so perhaps x is determined solely by median condition.

But multiple x satisfy it.

Unless... perhaps the missing number is such that it doesn't change the mode or something, but mode is 15 already.

Another idea: perhaps the range or mean is to be calculated, but no values are given for them.

Looking back at the image, for this row, only "Median: 12.5" is given, others are blank, so we need to find the missing number that makes median 12.5, then compute the rest.

But as above, many x work.

Perhaps in the context, the missing number is an integer, and we can choose one, but that doesn't make sense for a worksheet.

Unless I miscalculated.

Let's list the positions.

With 8 numbers, median is average of 4th and 5th in sorted order.

Current numbers: 4,8,10,15,15,15,19 — that's 7 numbers.

Add x.

The sorted list will have 8 numbers.

The 4th and 5th must average to 12.5, so sum to 25.

Possible combinations from the numbers available.

The numbers are mostly 15s, so likely the 4th and 5th are 10 and 15.

To have 4th=10 and 5th=15, we need at least three numbers ≤10, and at least four numbers ≥15? Let's think.

In sorted order, position 4 is 10, position 5 is 15.

That means there are exactly 3 numbers less than or equal to 10, and the 4th is 10, then 5th is 15.

Currently, without x, we have: 4,8,10 — that's three numbers ≤10, and then 15,15,15,19 — four numbers ≥15.

If we add x, if x ≤10, then we have four numbers ≤10: 4,8,10,x (sorted), then the next are 15,15,15,19.

So sorted list: say x=5: 4,5,8,10,15,15,15,19 — then 4th=10, 5th=15 — good.

If x=10: 4,8,10,10,15,15,15,19 — 4th=10, 5th=15 — good.

If x=11: 4,8,10,11,15,15,15,19 — 4th=11, 5th=15 — sum 26, median 13 — not good.

If x=14: same thing.

If x=15: 4,8,10,15,15,15,15,19 — 4th=15, 5th=15 — median 15 — not good.

So x must be ≤10.

But which one? Perhaps the worksheet has a specific number in mind, or perhaps we can choose x=10, as it's already in the list.

But let's see the mode — mode is 15, which is fine.

Range: if x=10, min=4, max=19, range=15

If x=0, min=0, max=19, range=19

etc.

But no constraint.

Perhaps in the image, the missing card is to be filled, and later we can see, but for now, let's assume x=10, as it's a nice number.

Or perhaps from the context of the worksheet, but I think for accuracy, we need to realize that any x≤10 works, but that can't be for a homework problem.

Another thought: perhaps the median is 12.5, and with the numbers, the only way is if the 4th and 5th are 10 and 15, and to have that, x must be such that it doesn't push the 10 to earlier, but it's already there.

Perhaps the missing number is 10, as it's symmetric or something.

Let's calculate the mean for x=10: sum = 4+8+10+15+15+15+19 +10 = let's calculate: 4+8=12, +10=22, +15=37, +15=52, +15=67, +19=86, +10=96 — mean = 96/8 = 12

But not given.

Perhaps we can leave it and move on.

I recall that in some worksheets, they might have a specific number, but here, let's look at the last row of Section A.

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Section A – Row 5: Six blank cards. Given: Median=2, Mode=1, Mean=3, Range=6

So we need to find six numbers that satisfy:

- Median = 2 → for 6 numbers, average of 3rd and 4th = 2, so 3rd + 4th = 4

- Mode = 1 → 1 appears most frequently, at least twice, and more than any other number

- Mean = 3 → sum = 3 * 6 = 18

- Range = 6 → max - min = 6

Let the numbers be a,b,c,d,e,f sorted: a≤b≤c≤d≤e≤f

Then c + d = 4 (since median=2)

Mode=1, so 1 must appear at least twice, and more than any other number.

Since c and d are at least b, and b≥a, and c+d=4, possible pairs for (c,d): (1,3), (2,2), (0,4), etc., but since mode is 1, likely 1 is involved.

If c=1, d=3, then since sorted, a≤b≤c=1, so a and b are ≤1, and since mode is 1, probably a and b are 1 or less.

But if a and b are 1, then we have at least three 1's (a,b,c), and d=3, then e and f.

Sum = a+b+c+d+e+f = 1+1+1+3+e+f = 6 + e+f = 18 → e+f=12

Range = f - a = f - 1 = 6 → f=7

Then e = 12 - f = 12-7=5

So numbers: 1,1,1,3,5,7

Check mode: 1 appears three times, others once — good.

Median: (c+d)/2 = (1+3)/2 = 2 — good.

Mean: (1+1+1+3+5+7)=18/6=3 — good.

Range: 7-1=6 — good.

Perfect.

So the six numbers are: 1,1,1,3,5,7

But the cards are blank, so we fill them with these numbers, probably in any order, but typically sorted or as per context.

In the image, the cards are empty, so we can list them as 1,1,1,3,5,7

Now, back to row 3.

For row 3: numbers 7,4,2, and missing x, range=7.

As before, x=0 or x=9.

With the successful solution for row 5, perhaps they allow 0.

Moreover, in row 2, they used 0.

So let's take x=0 for row 3.

Then list: 0,2,4,7

Sorted: 0,2,4,7

- Mean = (0+2+4+7)/4 = 13/4 = 3.25

- Median = (2+4)/2 = 3

- Mode = none (all unique)

- Range = 7-0=7 — good.

So missing number is 0.

Similarly, for row 4, with median=12.5, and we need to choose x≤10.

What if we choose x=10? Then list: 4,8,10,10,15,15,15,19

Sorted: 4,8,10,10,15,15,15,19

Median = (10+15)/2 = 12.5 — good.

Mean = (4+8+10+10+15+15+15+19) = let's calculate: 4+8=12, +10=22, +10=32, +15=47, +15=62, +15=77, +19=96 → 96/8=12

Mode = 15 (appears three times)

Range = 19-4=15

All good.

If we chose x=5, mean would be different, but since no other constraints, x=10 is fine, and it's already in the list, so likely intended.

So for row 4, missing number is 10.

Now summarize Section A:

Row 1: 5,7,8,4,7 → sorted 4,5,7,7,8 → Median=7, Mode=7, Mean=6.2, Range=4

Row 2: 3,6,11,4,8,0, and missing=3 (since mean=5, sum=35, known sum=32, so x=3) → list 0,3,3,4,6,8,11 → Median=4, Mode=3, Range=11

Row 3: 7,4,2, missing=0 (to make range=7) → list 0,2,4,7 → Mean=3.25, Median=3, Mode=none

Row 4: 10,8,15,19,4,15,15, missing=10 → list 4,8,10,10,15,15,15,19 → Mean=12, Range=15, Mode=15

Row 5: six numbers: 1,1,1,3,5,7

Now Section B.

Section B Question 1:

The mean mass of twenty packages is 14.3 kg. What is the total mass?

Mean = total / number, so total = mean × number = 14.3 × 20

Calculate: 14.3 × 20 = 14.3 × 2 × 10 = 28.6 × 10 = 286 kg

Or: 14 × 20 = 280, 0.3 × 20 = 6, total 286 kg

✔ Total mass = 286 kg

Section B Question 2:

Perry has cards: 3,6,3,4

He takes another card. The mean goes up by 1. What is the number on his new card?

First, current mean of four cards: sum = 3+6+3+4 = 16, mean = 16/4 = 4

New mean = 4 + 1 = 5

Now he has five cards, so new sum = 5 × 5 = 25

Current sum = 16, so new card = 25 - 16 = 9

✔ New card is 9

Extension:

The average weight of 8 people increases by 2.5 kg when a new person comes in place of one of them weighing 45 kg. What is the weight of the new person?

Let the original average be A kg.

Original total weight = 8A

After replacement: one person of 45 kg is replaced by a new person of W kg.

New total weight = 8A - 45 + W

New average = A + 2.5

So new total = 8(A + 2.5) = 8A + 20

Set equal: 8A - 45 + W = 8A + 20

Subtract 8A from both sides: -45 + W = 20

So W = 20 + 45 = 65 kg

✔ Weight of new person is 65 kg

Now, to present the final answers clearly.

For Section A, since it's fill-in-the-blank, we'll list the answers for each row.

But the user asked to solve the problem, and the final answer should be provided.

Since there are multiple parts, perhaps list all answers.

But the instruction says "Final Answer: [Provide only the final correct answer]", implying one answer, but there are many.

Perhaps for the entire task, but that doesn't make sense.

Looking back, the user said "solve the problem accurately", and the image has multiple problems.

In such cases, we should provide all answers.

But the format says "Final Answer" singular.

Perhaps we need to box each answer, but the instruction is to have one Final Answer section.

Maybe for the extension or the main questions.

Another idea: perhaps the "problem" refers to the entire worksheet, but that's vague.

To comply, I'll provide the answers for Section B and Extension, as they are word problems, and for Section A, since it's fill-in, perhaps omit or include briefly.

But the user might expect all.

Let's see the response format: it says "Final Answer: [Provide only the final correct answer — this should come AFTER all explanations and reasoning.]"

And in the reasoning, I've done all.

Perhaps for the Final Answer, I'll put the answers to the numbered questions in Section B and the Extension, as they are distinct problems.

Section B has two questions, and Extension is separate.

So:

For Section B Q1: 286 kg

Q2: 9

Extension: 65 kg

And for Section A, since it's multiple blanks, perhaps not included in "final answer", but the user might want them.

To be safe, I'll include all in the reasoning, and for Final Answer, put the key results.

But let's check the initial request: "Solve the problem accurately." and the image is the task.

Perhaps the best is to list all answers clearly.

Since the assistant is to help with homework, and the image has several parts, I'll provide a comprehensive answer.

But to follow the format, I'll write the reasoning for all, and in Final Answer, put the answers for the non-fill-in parts.

Notice that in Section A, the last row has blanks to fill, but the given are median, mode, etc., so we found the numbers.

For the purpose, I'll summarize.

Let me write the Final Answer as the solutions to the explicit questions.

So:

- For Section B Q1: 286

- Section B Q2: 9

- Extension: 65

And for Section A, since it's practice, perhaps not required in final answer, but to be thorough, I'll mention in reasoning.

In the reasoning above, I have all.

So for Final Answer, I'll put:

The total mass is 286 kg, the new card is 9, and the new person's weight is 65 kg.

But that's three answers.

Perhaps box them separately.

Since the instruction is "the final correct answer", and there are multiple, I'll list them.

I think it's acceptable.

So let's proceed.