Averages from Frequency Tables Worksheet | Fun and Engaging ... - Free Printable

Let's solve each section of this worksheet step by step.

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Section A: Mode and Median from Frequency Tables

We'll find the mode (most frequent value) and median (middle value when ordered) for each table.

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#### 1. Height of Year 9 Students

| Heights | Frequency |
|--------|----------|
| 153 cm | 26 |
| 154 cm | 30 |
| 155 cm | 19 |
| 156 cm | 21 |

- Mode: The height with the highest frequency → 154 cm (frequency = 30)
- Median:
- Total number of students = 26 + 30 + 19 + 21 = 96
- Median is the average of the 48th and 49th values (since 96 is even).
- Cumulative frequencies:
- 153 cm: 26
- 154 cm: 26 + 30 = 56
- Both the 48th and 49th values fall in the 154 cm group.
- So, Median = 154 cm

✔ Mode: 154 cm, Median: 154 cm

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#### 2. Shoe Size of Year 3 Students

| Size | Frequency |
|--------|----------|
| 4½ | 12 |
| 5 | 10 |
| 5½ | 8 |
| 6 | 15 |

- Mode: Highest frequency → 6 (frequency = 15)
- Median:
- Total = 12 + 10 + 8 + 15 = 45
- Middle value is the 23rd (since 45 is odd).
- Cumulative:
- 4½: 12
- 5: 12 + 10 = 22
- 5½: 22 + 8 = 30
- 23rd value falls in 5½ group
- So, Median = 5½

✔ Mode: 6, Median: 5½

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#### 3. Goals Scored in 10 Football Games

| Goals | Frequency |
|-------|----------|
| 0 | 4 |
| 1 | 2 |
| 2 | 3 |
| 3 | 2 |

- Mode: Most frequent → 0 (frequency = 4)
- Median:
- Total games = 4 + 2 + 3 + 2 = 11
- Middle value is the 6th game.
- Cumulative:
- 0 goals: 4 games
- 1 goal: 4 + 2 = 6 games
- 6th game is the last one in "1 goal" → so median is 1

✔ Mode: 0, Median: 1

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Section B: Mean and Range from Frequency Tables

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#### 1. Age of Children at a Children’s Party

| Age | Frequency |
|-----|----------|
| 1 | 9 |
| 2 | 5 |
| 3 | 4 |
| 4 | 2 |

- Range = Max − Min = 4 − 1 = 3
- Mean:
$$
\text{Mean} = \frac{(1×9) + (2×5) + (3×4) + (4×2)}{9+5+4+2} = \frac{9 + 10 + 12 + 8}{20} = \frac{39}{20} = 1.95
$$

✔ Range: 3, Mean: 1.95

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#### 2. Total Number of Pets

| No. pets | Frequency |
|---------|----------|
| 0 | 12 |
| 1 | 7 |
| 2 | 4 |
| 3 | 1 |

- Range = Max − Min = 3 − 0 = 3
- Mean:
$$
\text{Mean} = \frac{(0×12) + (1×7) + (2×4) + (3×1)}{12+7+4+1} = \frac{0 + 7 + 8 + 3}{24} = \frac{18}{24} = 0.75
$$

✔ Range: 3, Mean: 0.75

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#### 3. Midnight Temperature

| Temp | Frequency |
|------|----------|
| -8°C | 5 |
| -7°C | 8 |
| -6°C | 6 |
| -5°C | 4 |

- Range = Max − Min = (-5) − (-8) = 3°C
- Mean:
$$
\text{Mean} = \frac{(-8×5) + (-7×8) + (-6×6) + (-5×4)}{5+8+6+4} = \frac{-40 -56 -36 -20}{23} = \frac{-152}{23} ≈ -6.61
$$

✔ Range: 3°C, Mean: -6.61°C

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Section C: Goldfish Lifespan

Given:

| Lifespan (years) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|------------------|---|---|---|---|---|---|---|---|
| Frequency | 4 | 5 | 9 |10 | 8 | X | 3 | 2 |

Average lifespan = 4.24 years

Let the missing frequency be x.

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#### 1) Find the missing data (value of x)

Use formula for mean:
$$
\text{Mean} = \frac{\sum (\text{value} × \text{frequency})}{\sum \text{frequency}}
$$

Total frequency = 4 + 5 + 9 + 10 + 8 + x + 3 + 2 = 41 + x

Sum of (value × frequency):
$$
(1×4) + (2×5) + (3×9) + (4×10) + (5×8) + (6×x) + (7×3) + (8×2) = 4 + 10 + 27 + 40 + 40 + 6x + 21 + 16 = 158 + 6x
$$

Now:
$$
\frac{158 + 6x}{41 + x} = 4.24
$$

Multiply both sides:
$$
158 + 6x = 4.24(41 + x)
$$
$$
158 + 6x = 173.84 + 4.24x
$$
$$
6x - 4.24x = 173.84 - 158
$$
$$
1.76x = 15.84
$$
$$
x = \frac{15.84}{1.76} = 9
$$

✔ Missing frequency (x) = 9

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#### 2) How many people did Michael ask?

Total frequency = 41 + x = 41 + 9 = 50

✔ Michael asked 50 people

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#### 3) Judy used median instead of mean. Whose results were more reliable? Why?

- Michael used the mean (average).
- Judy used the median.

The mean can be affected by outliers or extreme values. In this case, the lifespans go up to 8 years, but most are clustered around 3–5 years.

But since we have a skewed distribution (some very long-lived fish), the mean might be pulled higher by those few longer lifespans.

The median is less sensitive to outliers and gives a better central measure for skewed data.

So, Judy’s median is more reliable because it is not affected by extreme values (e.g., fish living 7 or 8 years).

✔ Answer: Judy’s results were more reliable because the median is less affected by outliers and gives a better representation of the typical lifespan.

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✔ Final Answers Summary

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#### Section A

1. Height of Year 9 students
- Mode: 154 cm
- Median: 154 cm

2. Shoe size of Year 3 students
- Mode: 6
- Median: 5½

3. Goals scored in football games
- Mode: 0
- Median: 1

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#### Section B

1. Age of children
- Range: 3
- Mean: 1.95

2. Number of pets
- Range: 3
- Mean: 0.75

3. Midnight temperature
- Range: 3°C
- Mean: -6.61°C

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#### Section C

1. Missing frequency (x) = 9
2. Number of people asked = 50
3. Judy’s median is more reliable because it is not influenced by extreme values (outliers), unlike the mean.

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