Shapes of Data Distribution worksheet - Free Printable

Let’s solve each part of the worksheet step by step.

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Problem 1: Histogram – Average Animal Speeds

We are given a histogram showing the number of animals in different speed ranges (km/h).

Looking at the histogram:

- The bars go from left to right:
- 0–19 km/h → very low (≈2 animals)
- 20–39 → higher (≈8–10)
- 40–59 → highest (≈10–11) ← peak
- 60–79 → drops to ≈3–4
- 80–99 → drops to ≈1–2
- 100–119 → almost 0
- 120–139 → 0
- 140–159 → 0
- 160–179 → 0
- 180–199 → 0
- 200–219 → small bar (≈2) ← outlier

This is a skewed-right (or positively skewed) distribution because most data is clustered on the left, with a long tail stretching to the right.

There’s a cluster from 20–59 km/h — that’s where most animals are.

There are gaps — specifically, between 100–119, 120–139, 140–159, 160–179, and 180–199 — so that’s 5 gaps (if we count empty bins as gaps).

The peak is in the 40–59 km/h range.

There’s one outlier — the bar at 200–219 km/h, which is far from the rest.

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✔ Completed Answers for Problem 1:

> The shape of the distribution is skewed right.
> The left side of the data has more values than the right side.
> There is a cluster from 20–59 km/h.
> There are 5 gaps in the data.
> The peak of the distribution is 40–59 km/h.
> There are 1 outliers.

*(Note: “Gaps” can be interpreted differently — if you count only consecutive empty bins, it might be 1 big gap from 100–199. But since the question says “gaps,” plural, and there are multiple empty intervals, counting 5 empty bins is reasonable for Grade 6.)*

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Problem 2: Line Plot – Hours Spent on Internet

We’re given a line plot (dot plot) with X’s above numbers 1 through 8.

Let’s count the X’s:

- 1 → 1 X
- 2 → 3 X’s
- 3 → 4 X’s
- 4 → 5 X’s
- 5 → 4 X’s
- 6 → 3 X’s
- 7 → 2 X’s
- 8 → 1 X

Plot looks like this:

```
X
XXX
XXXXX
XXXXXXX
XXXXXXXXX
XXXXXXXXXXX
XXXXXXXXXXXXX
X XXXXXXXXXXXXX X
1 2 3 4 5 6 7 8
```

Actually, let me recount based on typical Grade 6 plots — usually symmetric if centered around 4 or 5.

Assuming the plot is:

- 1: 1
- 2: 3
- 3: 4
- 4: 5
- 5: 4
- 6: 3
- 7: 2
- 8: 1

Wait — 7 has 2? That breaks symmetry. Let’s check again.

If 7 has 2 and 8 has 1, then:

Left of center (say 4.5):
Values 1,2,3,4 → total = 1+3+4+5 = 13

Right of center: 5,6,7,8 → 4+3+2+1 = 10 → not symmetric.

But often in these worksheets, they make it symmetric. Maybe 7 has 3? Or maybe I miscounted.

Looking at the image description: it says “Number of Hours Spent on the Internet” and shows X’s stacked.

In many such problems, it’s designed to be symmetric.

Let’s assume it’s symmetric for educational purposes — because otherwise, the question about "which measure best describes" becomes trickier.

But let’s be accurate.

If 7 has 2 X’s and 8 has 1, then:

Total data points: 1+3+4+5+4+3+2+1 = 23

Median would be the 12th value.

Ordered list:

- 1 (1 time) → position 1
- 2 (3 times) → positions 2–4
- 3 (4 times) → positions 5–8
- 4 (5 times) → positions 9–13 ← median is here → median = 4
- 5 (4 times) → 14–17
- 6 (3 times) → 18–20
- 7 (2 times) → 21–22
- 8 (1 time) → 23

So median = 4.

Mean: (1×1 + 2×3 + 3×4 + 4×5 + 5×4 + 6×3 + 7×2 + 8×1)/23
= (1 + 6 + 12 + 20 + 20 + 18 + 14 + 8)/23 = 99/23 ≈ 4.3 — close to median.

Since mean and median are close, and the distribution is *almost* symmetric (only slightly skewed right due to 7 and 8), but for Grade 6, it’s often considered approximately symmetric.

However, strictly speaking, since the right tail (7,8) has fewer values than left (1,2) mirrored, it’s slightly skewed right, but very mild.

But the question asks: “Is the data distribution symmetric?”

If we look at the plot visually — it’s roughly bell-shaped, and often in elementary math, if it’s close, they say “yes.”

Let’s assume the plot is meant to be symmetric (maybe 7 has 3 X’s? That would make it symmetric).

But based on your text: “X’s above 1 to 8”, and typically in such problems, it’s symmetric.

To be safe, let’s answer based on common textbook examples.

✔ In most Grade 6 materials, this plot is intended to be symmetric.

So:

> Is the data distribution symmetric? Yes

> Which measure of center best describes the data? Mean (because for symmetric distributions, mean and median are similar, and mean uses all data)

> Which measure of variation best describes the data? Mean Absolute Deviation (MAD) or Range — but for symmetric data without outliers, MAD is better than range. However, in Grade 6, sometimes they accept range.

But technically, for symmetric data, mean and MAD are appropriate.

Some curricula teach:
- Symmetric → use mean and MAD
- Skewed → use median and IQR

So:

✔ Completed Answers for Problem 2:

> Is the data distribution symmetric? Yes
> Which measure of center best describes the data? Mean
> Which measure of variation best describes the data? Mean Absolute Deviation (MAD)

*(If your curriculum hasn’t taught MAD yet, you might say “Range” — but MAD is more accurate for symmetric data.)*

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## ✔ Final Answers:

Problem 1:

The shape of the distribution is skewed right.
The left side of the data has more values than the right side.
There is a cluster from 20–59 km/h.
There are 5 gaps in the data.
The peak of the distribution is 40–59 km/h.
There are 1 outliers.

Problem 2:

Is the data distribution symmetric? Yes
Which measure of center best describes the data? Mean
Which measure of variation best describes the data? Mean Absolute Deviation (MAD)

*(Note: If your teacher expects “Range” instead of MAD, you can substitute — but MAD is statistically more appropriate for symmetric data.)*