Averages from Frequency Tables Worksheet | Printable PDF Worksheets - Free Printable

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

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Section A: Mode and Median from Frequency Tables

#### 1. Height of Year 9 Students
| Heights | Frequency |
|--------|-----------|
| 153 cm | 26 |
| 154 cm | 30 |
| 155 cm | 19 |
| 156 cm | 21 |

- Mode: The value 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 when ordered.
- Cumulative frequencies:
- 153 cm: 26
- 154 cm: 26 + 30 = 56 → both 48th and 49th fall in this group
- So, median = 154 cm

✔ Mode: 154 cm, Median: 154 cm

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#### 2. Shoe Size of Year 3 Pupils
| Size | Frequency |
|--------|-----------|
| 4½ | 12 |
| 5 | 10 |
| 5½ | 8 |
| 6 | 15 |

- Mode: Highest frequency → 6 (frequency = 15)
- Median:
- Total = 12 + 10 + 8 + 15 = 45
- Median is the 23rd value (since (45+1)/2 = 23rd)
- Cumulative:
- 4½: 12
- 5: 12 + 10 = 22
- 5½: 22 + 8 = 30 → 23rd value is 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
- Median is the 6th value
- Cumulative:
- 0: 4
- 1: 4 + 2 = 6 → 6th value is 1
- So, median = 1

✔ Mode: 0, Median: 1

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Section B: Mean and Range from Frequency Tables

#### 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:
- Sum = (1×9) + (2×5) + (3×4) + (4×2) = 9 + 10 + 12 + 8 = 39
- Total children = 9 + 5 + 4 + 2 = 20
- Mean = 39 ÷ 20 = 1.95

✔ Range: 3, Mean: 1.95

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#### 2. Number of Pets
| No. pets | Frequency |
|----------|-----------|
| 0 | 12 |
| 1 | 7 |
| 2 | 4 |
| 3 | 1 |

- Range = 3 – 0 = 3
- Mean:
- Sum = (0×12) + (1×7) + (2×4) + (3×1) = 0 + 7 + 8 + 3 = 18
- Total pupils = 12 + 7 + 4 + 1 = 24
- Mean = 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 = Highest – Lowest = (-5) – (-8) = 3°C
- Mean:
- Sum = (-8×5) + (-7×8) + (-6×6) + (-5×4) = -40 -56 -36 -20 = -152
- Total cities = 5 + 8 + 6 + 4 = 23
- Mean = -152 ÷ 23 ≈ -6.61°C

✔ 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 (mean) lifespan = 4.24 years

We are to find:

#### 1) Missing frequency (X) for 6 years

Let total frequency be $ T $, and sum of lifespans be $ S $

We know:

$$
\text{Mean} = \frac{\text{Total sum}}{\text{Total frequency}} = 4.24
$$

So,

$$
\text{Sum} = \sum (\text{lifespan} \times \text{frequency}) = (1×4) + (2×5) + (3×9) + (4×10) + (5×8) + (6×X) + (7×3) + (8×2)
$$

Calculate known parts:

- 1×4 = 4
- 2×5 = 10
- 3×9 = 27
- 4×10 = 40
- 5×8 = 40
- 6×X = 6X
- 7×3 = 21
- 8×2 = 16

Total sum $ S = 4 + 10 + 27 + 40 + 40 + 6X + 21 + 16 = 158 + 6X $

Total frequency $ T = 4 + 5 + 9 + 10 + 8 + X + 3 + 2 = 41 + X $

Now use mean formula:

$$
\frac{158 + 6X}{41 + X} = 4.24
$$

Multiply both sides by $ 41 + X $:

$$
158 + 6X = 4.24(41 + X)
$$

Compute right-hand side:

$$
4.24 × 41 = 173.84 \\
4.24 × X = 4.24X
$$

So:

$$
158 + 6X = 173.84 + 4.24X
$$

Subtract 4.24X from both sides:

$$
158 + 1.76X = 173.84
$$

Subtract 158:

$$
1.76X = 15.84
$$

Divide:

$$
X = \frac{15.84}{1.76} = 9
$$

✔ 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 and why?

- Mean (used by Michael): Can be affected by extreme values (outliers). For example, if one goldfish lived 20 years, it would skew the mean upward.
- Median (used by Judy): The middle value; not affected by outliers.

In this case, since we have a range from 1 to 8 years, and most data clustered around 3–5 years, the median might be more reliable because it represents the "typical" lifespan better than the mean, especially if there are a few very long-lived fish.

But here, the mean is 4.24, which seems reasonable given the distribution. However, if there were outliers, the median would be more robust.

✔ Judy’s median result is more reliable because it is less affected by extreme values (outliers), giving a better representation of the typical lifespan.

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✔ Final Answers Summary:

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Section A

1. Height
- Mode: 154 cm
- Median: 154 cm

2. Shoe size
- Mode: 6
- Median: 5½

3. Goals
- Mode: 0
- Median: 1

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Section B

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

2. Pets
- Range: 3
- Mean: 0.75

3. Temperature
- Range: 3°C
- Mean: -6.61°C

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Section C

1. Missing frequency (6 years): 9
2. Number of people asked: 50
3. Whose result is more reliable?
➤ Judy's median is more reliable because it is not affected by extreme values (outliers), making it a better measure of central tendency for skewed data or data with anomalies.

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