Mean Median Mode Range Worksheets - Math Monks - Free Printable

Problem Analysis

The task involves finding the mean, median, and mode for grouped data in various scenarios. Let's solve each part step by step.

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Part 1: Volume of Petrol Used

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

#### Step 1: Calculate the Mean
The formula for the mean of grouped data is:
\[
\text{Mean} = \frac{\sum (f \cdot m)}{\sum f}
\]
where:
- \( f \) is the frequency of each class.
- \( m \) is the midpoint of each class.

##### Midpoints Calculation:
- For \( 1 - 5 \): \( m = \frac{1 + 5}{2} = 3 \)
- For \( 5 - 10 \): \( m = \frac{5 + 10}{2} = 7.5 \)
- For \( 10 - 15 \): \( m = \frac{10 + 15}{2} = 12.5 \)
- For \( 15 - 20 \): \( m = \frac{15 + 20}{2} = 17.5 \)

##### Frequency (\( f \)) and Midpoint (\( m \)) Table:
| Class Interval | Frequency (\( f \)) | Midpoint (\( m \)) | \( f \cdot m \) |
|----------------|----------------------|---------------------|------------------|
| 1 - 5 | 2 | 3 | \( 2 \cdot 3 = 6 \) |
| 5 - 10 | 4 | 7.5 | \( 4 \cdot 7.5 = 30 \) |
| 10 - 15 | 10 | 12.5 | \( 10 \cdot 12.5 = 125 \) |
| 15 - 20 | 4 | 17.5 | \( 4 \cdot 17.5 = 70 \) |

##### Summation:
\[
\sum f = 2 + 4 + 10 + 4 = 20
\]
\[
\sum (f \cdot m) = 6 + 30 + 125 + 70 = 231
\]

##### Mean Calculation:
\[
\text{Mean} = \frac{\sum (f \cdot m)}{\sum f} = \frac{231}{20} = 11.55
\]

#### Step 2: Calculate the Median
The formula for the median of grouped data is:
\[
\text{Median} = l + \left( \frac{\frac{N}{2} - F}{f} \right) \cdot h
\]
where:
- \( l \) is the lower limit of the median class.
- \( N \) is the total frequency.
- \( F \) is the cumulative frequency of the class preceding the median class.
- \( f \) is the frequency of the median class.
- \( h \) is the class width.

##### Cumulative Frequency Table:
| Class Interval | Frequency (\( f \)) | Cumulative Frequency (\( F \)) |
|----------------|----------------------|---------------------------------|
| 1 - 5 | 2 | 2 |
| 5 - 10 | 4 | 6 |
| 10 - 15 | 10 | 16 |
| 15 - 20 | 4 | 20 |

##### Median Class Identification:
- Total frequency, \( N = 20 \).
- Median position: \( \frac{N}{2} = \frac{20}{2} = 10 \).
- The cumulative frequency just greater than 10 is 16, so the median class is \( 10 - 15 \).

##### Median Calculation:
- \( l = 10 \) (lower limit of the median class),
- \( N = 20 \),
- \( F = 6 \) (cumulative frequency of the class preceding the median class),
- \( f = 10 \) (frequency of the median class),
- \( h = 5 \) (class width).

\[
\text{Median} = 10 + \left( \frac{10 - 6}{10} \right) \cdot 5 = 10 + \left( \frac{4}{10} \right) \cdot 5 = 10 + 2 = 12
\]

#### Step 3: Calculate the Mode
The mode for grouped data is given by:
\[
\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \cdot h
\]
where:
- \( l \) is the lower limit of the modal class.
- \( f_1 \) is the frequency of the modal class.
- \( f_0 \) is the frequency of the class preceding the modal class.
- \( f_2 \) is the frequency of the class succeeding the modal class.
- \( h \) is the class width.

##### Modal Class Identification:
- The highest frequency is 10, so the modal class is \( 10 - 15 \).

##### Mode Calculation:
- \( l = 10 \) (lower limit of the modal class),
- \( f_1 = 10 \) (frequency of the modal class),
- \( f_0 = 4 \) (frequency of the class preceding the modal class),
- \( f_2 = 4 \) (frequency of the class succeeding the modal class),
- \( h = 5 \) (class width).

\[
\text{Mode} = 10 + \left( \frac{10 - 4}{2 \cdot 10 - 4 - 4} \right) \cdot 5 = 10 + \left( \frac{6}{12} \right) \cdot 5 = 10 + 0.5 \cdot 5 = 10 + 2.5 = 12.5
\]

#### Final Answers for Part 1:
\[
\boxed{11.55, 12, 12.5}
\]

---

Part 2: Frequency Table

#### Given Data:
| \( x \) | 10 | 20 | 25 | 30 | 37 | 55 |
|---------|----|----|----|----|----|----|
| \( f \) | 5 | 12 | 14 | 15 | 10 | 4 |

#### Step 1: Calculate the Mean
The formula for the mean is:
\[
\text{Mean} = \frac{\sum (f \cdot x)}{\sum f}
\]

##### Frequency (\( f \)) and Value (\( x \)) Table:
| \( x \) | \( f \) | \( f \cdot x \) |
|---------|---------|----------------|
| 10 | 5 | \( 5 \cdot 10 = 50 \) |
| 20 | 12 | \( 12 \cdot 20 = 240 \) |
| 25 | 14 | \( 14 \cdot 25 = 350 \) |
| 30 | 15 | \( 15 \cdot 30 = 450 \) |
| 37 | 10 | \( 10 \cdot 37 = 370 \) |
| 55 | 4 | \( 4 \cdot 55 = 220 \) |

##### Summation:
\[
\sum f = 5 + 12 + 14 + 15 + 10 + 4 = 60
\]
\[
\sum (f \cdot x) = 50 + 240 + 350 + 450 + 370 + 220 = 1680
\]

##### Mean Calculation:
\[
\text{Mean} = \frac{\sum (f \cdot x)}{\sum f} = \frac{1680}{60} = 28
\]

#### Step 2: Calculate the Median
The formula for the median is:
\[
\text{Median} = l + \left( \frac{\frac{N}{2} - F}{f} \right) \cdot h
\]

##### Cumulative Frequency Table:
| \( x \) | \( f \) | Cumulative Frequency (\( F \)) |
|---------|---------|---------------------------------|
| 10 | 5 | 5 |
| 20 | 12 | 17 |
| 25 | 14 | 31 |
| 30 | 15 | 46 |
| 37 | 10 | 56 |
| 55 | 4 | 60 |

##### Median Class Identification:
- Total frequency, \( N = 60 \).
- Median position: \( \frac{N}{2} = \frac{60}{2} = 30 \).
- The cumulative frequency just greater than 30 is 31, so the median class is \( 25 \).

##### Median Calculation:
- \( l = 25 \) (lower limit of the median class),
- \( N = 60 \),
- \( F = 17 \) (cumulative frequency of the class preceding the median class),
- \( f = 14 \) (frequency of the median class),
- \( h = 5 \) (class width).

\[
\text{Median} = 25 + \left( \frac{30 - 17}{14} \right) \cdot 5 = 25 + \left( \frac{13}{14} \right) \cdot 5 = 25 + 4.64 \approx 29.64
\]

#### Step 3: Calculate the Mode
The mode is the value with the highest frequency. Here, the highest frequency is 15, corresponding to \( x = 30 \).

#### Final Answers for Part 2:
\[
\boxed{28, 29.64, 30}
\]

---

Part 3: Grouped Data

#### Subpart (a): Ages of 100 People
#### Given Data:
| Age Range | Frequency (\( f \)) |
|-----------|----------------------|
| 8 - 10 | 12 |
| 11 - 13 | 25 |
| 14 - 16 | 37 |
| 17 - 19 | 26 |

##### Step 1: Calculate the Mean
The formula for the mean is:
\[
\text{Mean} = \frac{\sum (f \cdot m)}{\sum f}
\]

##### Midpoints Calculation:
- For \( 8 - 10 \): \( m = \frac{8 + 10}{2} = 9 \)
- For \( 11 - 13 \): \( m = \frac{11 + 13}{2} = 12 \)
- For \( 14 - 16 \): \( m = \frac{14 + 16}{2} = 15 \)
- For \( 17 - 19 \): \( m = \frac{17 + 19}{2} = 18 \)

##### Frequency (\( f \)) and Midpoint (\( m \)) Table:
| Age Range | Frequency (\( f \)) | Midpoint (\( m \)) | \( f \cdot m \) |
|-----------|----------------------|---------------------|------------------|
| 8 - 10 | 12 | 9 | \( 12 \cdot 9 = 108 \) |
| 11 - 13 | 25 | 12 | \( 25 \cdot 12 = 300 \) |
| 14 - 16 | 37 | 15 | \( 37 \cdot 15 = 555 \) |
| 17 - 19 | 26 | 18 | \( 26 \cdot 18 = 468 \) |

##### Summation:
\[
\sum f = 12 + 25 + 37 + 26 = 100
\]
\[
\sum (f \cdot m) = 108 + 300 + 555 + 468 = 1431
\]

##### Mean Calculation:
\[
\text{Mean} = \frac{\sum (f \cdot m)}{\sum f} = \frac{1431}{100} = 14.31
\]

##### Step 2: Calculate the Median
The formula for the median is:
\[
\text{Median} = l + \left( \frac{\frac{N}{2} - F}{f} \right) \cdot h
\]

##### Cumulative Frequency Table:
| Age Range | Frequency (\( f \)) | Cumulative Frequency (\( F \)) |
|-----------|----------------------|---------------------------------|
| 8 - 10 | 12 | 12 |
| 11 - 13 | 25 | 37 |
| 14 - 16 | 37 | 74 |
| 17 - 19 | 26 | 100 |

##### Median Class Identification:
- Total frequency, \( N = 100 \).
- Median position: \( \frac{N}{2} = \frac{100}{2} = 50 \).
- The cumulative frequency just greater than 50 is 74, so the median class is \( 14 - 16 \).

##### Median Calculation:
- \( l = 14 \) (lower limit of the median class),
- \( N = 100 \),
- \( F = 37 \) (cumulative frequency of the class preceding the median class),
- \( f = 37 \) (frequency of the median class),
- \( h = 3 \) (class width).

\[
\text{Median} = 14 + \left( \frac{50 - 37}{37} \right) \cdot 3 = 14 + \left( \frac{13}{37} \right) \cdot 3 \approx 14 + 1.05 = 15.05
\]

##### Step 3: Calculate the Mode
The mode is the value with the highest frequency. Here, the highest frequency is 37, corresponding to the age range \( 14 - 16 \).

#### Final Answers for Subpart (a):
\[
\boxed{14.31, 15.05, 14 - 16}
\]

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#### Subpart (b): Prices of 150 Stamps
#### Given Data:
| Price Range | Frequency (\( f \)) |
|-------------|----------------------|
| 10 - 20 | 24 |
| 20 - 30 | 38 |
| 30 - 40 | 52 |
| 40 - 50 | 36 |

##### Step 1: Calculate the Mean
The formula for the mean is:
\[
\text{Mean} = \frac{\sum (f \cdot m)}{\sum f}
\]

##### Midpoints Calculation:
- For \( 10 - 20 \): \( m = \frac{10 + 20}{2} = 15 \)
- For \( 20 - 30 \): \( m = \frac{20 + 30}{2} = 25 \)
- For \( 30 - 40 \): \( m = \frac{30 + 40}{2} = 35 \)
- For \( 40 - 50 \): \( m = \frac{40 + 50}{2} = 45 \)

##### Frequency (\( f \)) and Midpoint (\( m \)) Table:
| Price Range | Frequency (\( f \)) | Midpoint (\( m \)) | \( f \cdot m \) |
|-------------|----------------------|---------------------|------------------|
| 10 - 20 | 24 | 15 | \( 24 \cdot 15 = 360 \) |
| 20 - 30 | 38 | 25 | \( 38 \cdot 25 = 950 \) |
| 30 - 40 | 52 | 35 | \( 52 \cdot 35 = 1820 \) |
| 40 - 50 | 36 | 45 | \( 36 \cdot 45 = 1620 \) |

##### Summation:
\[
\sum f = 24 + 38 + 52 + 36 = 150
\]
\[
\sum (f \cdot m) = 360 + 950 + 1820 + 1620 = 4750
\]

##### Mean Calculation:
\[
\text{Mean} = \frac{\sum (f \cdot m)}{\sum f} = \frac{4750}{150} \approx 31.67
\]

##### Step 2: Calculate the Median
The formula for the median is:
\[
\text{Median} = l + \left( \frac{\frac{N}{2} - F}{f} \right) \cdot h
\]

##### Cumulative Frequency Table:
| Price Range | Frequency (\( f \)) | Cumulative Frequency (\( F \)) |
|-------------|----------------------|---------------------------------|
| 10 - 20 | 24 | 24 |
| 20 - 30 | 38 | 62 |
| 30 - 40 | 52 | 114 |
| 40 - 50 | 36 | 150 |

##### Median Class Identification:
- Total frequency, \( N = 150 \).
- Median position: \( \frac{N}{2} = \frac{150}{2} = 75 \).
- The cumulative frequency just greater than 75 is 114, so the median class is \( 30 - 40 \).

##### Median Calculation:
- \( l = 30 \) (lower limit of the median class),
- \( N = 150 \),
- \( F = 62 \) (cumulative frequency of the class preceding the median class),
- \( f = 52 \) (frequency of the median class),
- \( h = 10 \) (class width).

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

##### Step 3: Calculate the Mode
The mode is the value with the highest frequency. Here, the highest frequency is 52, corresponding to the price range \( 30 - 40 \).

#### Final Answers for Subpart (b):
\[
\boxed{31.67, 32.5, 30 - 40}
\]

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Final Summary of All Parts

1. Part 1: \(\boxed{11.55, 12, 12.5}\)
2. Part 2: \(\boxed{28, 29.64, 30}\)
3. Subpart (a): \(\boxed{14.31, 15.05, 14 - 16}\)
4. Subpart (b): \(\boxed{31.67, 32.5, 30 - 40}\)