Worksheet on Mean Absolute Deviation (MAD) with exercises on interpreting MAD values, using graphs, and analyzing a dot plot of siblings in a class.

Explanation:
Let’s solve each part step by step.

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Part 1: Interpreting the MAD (Mean Absolute Deviation)
This is a fill-in-the-blank concept question. We need to recall what MAD tells us:

- MAD measures the *average distance* of each data point from the mean.
- A low MAD means data points are close to the mean → low variability → numbers are close together (or consistent).
- A high MAD means data points are far from the mean → high variability → numbers are spread out (or scattered).

So the completed sentences should be:

- *If the MAD value is small, then the data set has a low variability. That means the numbers are close together.*
- *If the MAD value is large, then the data set has a high variability. That means the numbers are spread out.*

(“Small” and “large” are more student-friendly than “low/high” for the first blank; but since the sentence says “MAD value is ______”, “small” and “large” fit best.)

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Part 2: Using Graphs — Bar graph of reading minutes

The bar graph shows:
- 2 students read 5 minutes
- 4 students read 10 minutes
- 3 students read 15 minutes
- 1 student read 20 minutes

So total number of students = 2 + 4 + 3 + 1 = 10 students

Let’s list all the data values (each student’s minutes):
5, 5, 10, 10, 10, 10, 15, 15, 15, 20

Step 1: Find the mean
Sum = (2×5) + (4×10) + (3×15) + (1×20) = 10 + 40 + 45 + 20 = 115
Mean = 115 ÷ 10 = 11.5

Step 2: Find absolute deviations from the mean (|value − 11.5|):

- For 5: |5 − 11.5| = 6.5 → two of these → 2 × 6.5 = 13
- For 10: |10 − 11.5| = 1.5 → four of these → 4 × 1.5 = 6
- For 15: |15 − 11.5| = 3.5 → three of these → 3 × 3.5 = 10.5
- For 20: |20 − 11.5| = 8.5 → one of these → 8.5

Total absolute deviation = 13 + 6 + 10.5 + 8.5 = 38

Step 3: MAD = total absolute deviation ÷ number of data points = 38 ÷ 10 = 3.8

So MAD = 3.8 minutes

Variability: Since MAD is 3.8, and the range is 20 − 5 = 15, this is moderate variability — not extremely tight, not extremely spread out.

But the question only asks to *find the MAD* and *determine the variability*. Since MAD = 3.8, we can say the variability is moderate (or just report the MAD as the measure of variability). In many school contexts, stating the MAD *is* stating the variability.

So answer: MAD = 3.8; variability is moderate (or simply “the MAD is 3.8” suffices per instruction).

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Part 3: You try — Dot plot of number of siblings

Dot plot shows:
- 0 siblings: 1 dot
- 1 sibling: 2 dots
- 2 siblings: 3 dots
- 3 siblings: 4 dots
- 4 siblings: 4 dots
- 5 siblings: 0 dots
- 6 siblings: 1 dot

Let’s list all data points (each student’s number of siblings):

0,
1, 1,
2, 2, 2,
3, 3, 3, 3,
4, 4, 4, 4,
6

Count total students: 1 + 2 + 3 + 4 + 4 + 0 + 1 = 15 students

Step 1: Find the mean
Sum =
0×1 = 0
1×2 = 2
2×3 = 6
3×4 = 12
4×4 = 16
6×1 = 6
Total = 0 + 2 + 6 + 12 + 16 + 6 = 42

Mean = 42 ÷ 15 = 2.8

Step 2: Find absolute deviations from 2.8:

- 0: |0 − 2.8| = 2.8 → 1 time → 2.8
- 1: |1 − 2.8| = 1.8 → 2 times → 3.6
- 2: |2 − 2.8| = 0.8 → 3 times → 2.4
- 3: |3 − 2.8| = 0.2 → 4 times → 0.8
- 4: |4 − 2.8| = 1.2 → 4 times → 4.8
- 6: |6 − 2.8| = 3.2 → 1 time → 3.2

Add them up:
2.8 + 3.6 = 6.4
6.4 + 2.4 = 8.8
8.8 + 0.8 = 9.6
9.6 + 4.8 = 14.4
14.4 + 3.2 = 17.6

Step 3: MAD = 17.6 ÷ 15 = 1.1733… ≈ 1.17 (if rounding to nearest hundredth)
Or as a fraction: 17.6 / 15 = 176/150 = 88/75 ≈ 1.173

In many middle-school contexts, they may keep it as a decimal rounded to nearest tenth: 1.2

Let’s double-check sum:
Data: [0,1,1,2,2,2,3,3,3,3,4,4,4,4,6]
Mean = 42/15 = 2.8 ✔
Deviations:
0 → 2.8
1 → 1.8 (×2 = 3.6)
2 → 0.8 (×3 = 2.4)
3 → 0.2 (×4 = 0.8)
4 → 1.2 (×4 = 4.8)
6 → 3.2
Total = 2.8+3.6+2.4+0.8+4.8+3.2 = let's add in pairs:
2.8 + 3.2 = 6
3.6 + 4.8 = 8.4
2.4 + 0.8 = 3.2
6 + 8.4 = 14.4; 14.4 + 3.2 = 17.6 ✔

17.6 ÷ 15 = 1.173... So MAD ≈ 1.2 (to nearest tenth) is appropriate for this level.

Question 2: What is the outlier? Do you think that changes the MAD?

Outlier = value far from others. Most students have 0–4 siblings. One student has 6 siblings — that’s the outlier.

Does it change the MAD? Yes — because MAD uses *all* data points, and the outlier (6) is 3.2 away from the mean (2.8), which contributes significantly (3.2) to the total deviation. Without the outlier, the MAD would be smaller.

Let’s quickly test: Remove the 6 → 14 students, sum = 42 − 6 = 36, mean = 36/14 ≈ 2.571
Deviations without 6: total was 17.6 − 3.2 = 14.4
MAD = 14.4 / 14 ≈ 1.03 — smaller than 1.17. So yes, the outlier increases the MAD.

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Now compile final answers clearly:

1. Interpreting MAD blanks:
- small, variability, close together
- large, variability, spread out

2. Graph problem: MAD = 3.8

3. Dot plot:
- MAD ≈ 1.2 (or 1.17)
- Outlier is 6 siblings; yes, it increases the MAD.

Since the instructions say: *Only provide the final answer* after reasoning, and no sections are requested (ExplainAnswer = false, etc.), we output only the final answers in a clean format matching the worksheet structure.

But the user didn’t specify how to format the final answer — however, based on typical expectations for such worksheets, we give the filled-in blanks and computed values.

Given the worksheet has blanks, we’ll fill them precisely:

For “Interpreting the MAD”:
- If the MAD value is small, then the data set has a low variability. That means the numbers are close together.
- If the MAD value is large, then the data set has a high variability. That means the numbers are spread out.

For “Using Graphs”:
MAD = 3.8
Variability: moderate (or just state MAD = 3.8)

For “You try”:
1. MAD = 1.2 (rounded to nearest tenth)
2. Outlier = 6; yes, it increases the MAD.

Since the instruction says: *Final Answer: [Provide only the final correct answer]* — and there are multiple parts, we’ll list them concisely.

Final Answer:
Interpreting MAD: small, variability, close together; large, variability, spread out.
Graph MAD: 3.8.
Dot plot MAD: 1.2; outlier is 6 siblings, and yes, it increases the MAD.