Mean, Median, and Mode + Free Worksheet - Free Printable

Let's solve each problem step by step.

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Problem 1: Petrol Usage (Grouped Data)

Given:

| Volume of petrol (in litres) | Number of days |
|-------------------------------|----------------|
| 1 - 5 | 2 |
| 5 - 10 | 4 |
| 10 - 15 | 10 |
| 15 - 20 | 4 |

Total number of days = 2 + 4 + 10 + 4 = 20

We will find:
- Mean
- Median
- Mode

---

#### Step 1: Mean

For grouped data, we use the midpoint of each class interval and multiply by frequency.

| Class | Midpoint (x) | Frequency (f) | f × x |
|-------------|--------------|---------------|--------|
| 1 - 5 | (1+5)/2 = 3 | 2 | 6 |
| 5 - 10 | (5+10)/2 = 7.5 | 4 | 30 |
| 10 - 15 | (10+15)/2 = 12.5 | 10 | 125 |
| 15 - 20 | (15+20)/2 = 17.5 | 4 | 70 |
| Total | | 20 | 231|

$$
\text{Mean} = \frac{\sum f \times x}{\sum f} = \frac{231}{20} = 11.55 \text{ litres}
$$

✔ Mean = 11.55 litres

---

#### Step 2: Median

For grouped data, median is found using:

$$
\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h
$$

Where:
- $ L $ = lower boundary of median class
- $ N $ = total frequency = 20 → $ N/2 = 10 $
- $ F $ = cumulative frequency before median class
- $ f $ = frequency of median class
- $ h $ = class width

Cumulative frequency table:

| Class | Frequency | Cumulative Frequency |
|-------------|-----------|----------------------|
| 1 - 5 | 2 | 2 |
| 5 - 10 | 4 | 6 |
| 10 - 15 | 10 | 16 |
| 15 - 20 | 4 | 20 |

Since $ N/2 = 10 $, and cumulative frequency just exceeds 10 in the 10–15 class, this is the median class.

- $ L = 10 $
- $ F = 6 $ (cumulative before 10–15)
- $ f = 10 $
- $ h = 5 $

$$
\text{Median} = 10 + \left( \frac{10 - 6}{10} \right) \times 5 = 10 + \left( \frac{4}{10} \right) \times 5 = 10 + 2 = 12
$$

✔ Median = 12 litres

---

#### Step 3: Mode

For grouped data, mode is estimated using:

$$
\text{Mode} = L + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h
$$

Where:
- $ f_m $ = frequency of modal class (highest frequency)
- $ f_1 $ = frequency before modal class
- $ f_2 $ = frequency after modal class
- $ L $ = lower boundary of modal class
- $ h $ = class width

Highest frequency = 10 → class 10–15

- $ L = 10 $
- $ f_m = 10 $
- $ f_1 = 4 $ (from 5–10)
- $ f_2 = 4 $ (from 15–20)
- $ h = 5 $

$$
\text{Mode} = 10 + \left( \frac{10 - 4}{2(10) - 4 - 4} \right) \times 5 = 10 + \left( \frac{6}{12} \right) \times 5 = 10 + 2.5 = 12.5
$$

✔ Mode = 12.5 litres

---

✔ Answer for Problem 1:

- Mean: 11.55 litres
- Median: 12 litres
- Mode: 12.5 litres

---

Problem 2: Frequency Table (Discrete Data)

Given:

| x | 10 | 20 | 25 | 30 | 37 | 55 |
|----|----|----|----|----|----|----|
| f | 5 | 12 | 14 | 15 | 10 | 4 |

Total frequency $ N = 5 + 12 + 14 + 15 + 10 + 4 = 60 $

---

#### Step 1: Mean

$$
\text{Mean} = \frac{\sum f \cdot x}{\sum f}
$$

Compute $ f \cdot x $:

| x | f | f×x |
|----|----|-----|
| 10 | 5 | 50 |
| 20 | 12 | 240 |
| 25 | 14 | 350 |
| 30 | 15 | 450 |
| 37 | 10 | 370 |
| 55 | 4 | 220 |
| | | 1740 |

$$
\text{Mean} = \frac{1740}{60} = 29
$$

✔ Mean = 29

---

#### Step 2: Median

Since data is discrete and ordered, we can use cumulative frequency.

| x | f | Cumulative Frequency |
|----|----|----------------------|
| 10 | 5 | 5 |
| 20 | 12 | 17 |
| 25 | 14 | 31 |
| 30 | 15 | 46 |
| 37 | 10 | 56 |
| 55 | 4 | 60 |

$ N = 60 $, so $ N/2 = 30 $

Find the value where cumulative frequency first exceeds 30 → it's at x = 30 (cumulative = 46)

So, median = 30

✔ Median = 30

---

#### Step 3: Mode

The mode is the value with highest frequency.

Maximum frequency = 15 → corresponds to x = 30

✔ Mode = 30

---

✔ Answer for Problem 2:

- Mean: 29
- Median: 30
- Mode: 30

---

Problem 3a: Ages of 100 People

Given:

| Age | 8 - 10 | 11 - 13 | 14 - 16 | 17 - 19 |
|-----------|--------|---------|---------|---------|
| Frequency | 12 | 25 | 37 | 26 |

Total $ N = 100 $

---

#### Step 1: Mean

Use midpoints:

| Class | Midpoint (x) | f | f×x |
|-----------|--------------|---|-----|
| 8 - 10 | 9 | 12 | 108 |
| 11 - 13 | 12 | 25 | 300 |
| 14 - 16 | 15 | 37 | 555 |
| 17 - 19 | 18 | 26 | 468 |
| | | 100 | 1431 |

$$
\text{Mean} = \frac{1431}{100} = 14.31
$$

✔ Mean = 14.31 years

---

#### Step 2: Median

$ N = 100 $, so $ N/2 = 50 $

Cumulative frequency:

| Class | f | Cumulative |
|-----------|----|------------|
| 8 - 10 | 12 | 12 |
| 11 - 13 | 25 | 37 |
| 14 - 16 | 37 | 74 |
| 17 - 19 | 26 | 100 |

Cumulative frequency reaches 74 in 14–16, which is > 50 → median class = 14–16

- $ L = 14 $
- $ F = 37 $ (before 14–16)
- $ f = 37 $
- $ h = 3 $ (class width)

Wait! Actually, the class 11–13 ends at 13, and 14–16 starts at 14 → no gap? But since it's continuous, we assume boundaries are inclusive.

But let's check:

- $ N/2 = 50 $
- Cumulative before 14–16 = 12 + 25 = 37
- So 50 – 37 = 13 people into the class of size 37

$$
\text{Median} = 14 + \left( \frac{50 - 37}{37} \right) \times 3 = 14 + \left( \frac{13}{37} \right) \times 3 \approx 14 + 1.054 = 15.054
$$

✔ Median ≈ 15.05 years

---

#### Step 3: Mode

Modal class: highest frequency = 37 → 14–16

- $ L = 14 $
- $ f_m = 37 $
- $ f_1 = 25 $ (previous class)
- $ f_2 = 26 $ (next class)
- $ h = 3 $

$$
\text{Mode} = 14 + \left( \frac{37 - 25}{2(37) - 25 - 26} \right) \times 3 = 14 + \left( \frac{12}{74 - 51} \right) \times 3 = 14 + \left( \frac{12}{23} \right) \times 3
$$
$$
= 14 + 1.565 \approx 15.565
$$

✔ Mode ≈ 15.57 years

---

✔ Answer for Problem 3a:

- Mean: 14.31 years
- Median: ~15.05 years
- Mode: ~15.57 years

---

Problem 3b: Prices of Stamps

| Price | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
|-----------|---------|---------|---------|---------|
| Frequency | 24 | 38 | 52 | 36 |

Total $ N = 24 + 38 + 52 + 36 = 150 $

---

#### Step 1: Mean

Midpoints:

| Class | Midpoint (x) | f | f×x |
|-----------|--------------|---|-----|
| 10 - 20 | 15 | 24 | 360 |
| 20 - 30 | 25 | 38 | 950 |
| 30 - 40 | 35 | 52 | 1820 |
| 40 - 50 | 45 | 36 | 1620 |
| | | 150 | 4750 |

$$
\text{Mean} = \frac{4750}{150} \approx 31.67
$$

✔ Mean ≈ 31.67

---

#### Step 2: Median

$ N = 150 $ → $ N/2 = 75 $

Cumulative frequency:

| Class | f | Cumulative |
|-----------|----|------------|
| 10 - 20 | 24 | 24 |
| 20 - 30 | 38 | 62 |
| 30 - 40 | 52 | 114 |
| 40 - 50 | 36 | 150 |

75 lies in 30–40 class → median class

- $ L = 30 $
- $ F = 62 $
- $ f = 52 $
- $ h = 10 $

$$
\text{Median} = 30 + \left( \frac{75 - 62}{52} \right) \times 10 = 30 + \left( \frac{13}{52} \right) \times 10 = 30 + 2.5 = 32.5
$$

✔ Median = 32.5

---

#### Step 3: Mode

Modal class: highest frequency = 52 → 30–40

- $ L = 30 $
- $ f_m = 52 $
- $ f_1 = 38 $
- $ f_2 = 36 $
- $ h = 10 $

$$
\text{Mode} = 30 + \left( \frac{52 - 38}{2(52) - 38 - 36} \right) \times 10 = 30 + \left( \frac{14}{104 - 74} \right) \times 10 = 30 + \left( \frac{14}{30} \right) \times 10
$$
$$
= 30 + 4.67 = 34.67
$$

✔ Mode ≈ 34.67

---

✔ Answer for Problem 3b:

- Mean: ~31.67
- Median: 32.5
- Mode: ~34.67

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✔ Final Summary:

#### Problem 1: Petrol Usage
- Mean: 11.55 litres
- Median: 12 litres
- Mode: 12.5 litres

#### Problem 2: Discrete Frequency Table
- Mean: 29
- Median: 30
- Mode: 30

#### Problem 3a: Ages
- Mean: 14.31 years
- Median: ~15.05 years
- Mode: ~15.57 years

#### Problem 3b: Stamp Prices
- Mean: ~31.67
- Median: 32.5
- Mode: ~34.67

Let me know if you'd like visual representations or further explanation!