Calculating the Mean from a Frequency Table Worksheet | Fun and ... - Free Printable
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Step-by-step solution for: Calculating the Mean from a Frequency Table Worksheet | Fun and ...
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Step-by-step solution for: Calculating the Mean from a Frequency Table Worksheet | Fun and ...
Let's solve the problems step by step, starting with Section A and then moving to Section B.
---
We are given two frequency tables: one for Weight (kg) and one for Distance (d km). For each, we need to find:
- Estimate of the mean
- Modal class
- Median class
---
#### 🔹 1. Weight (kg) Table
| Weight (kg) | Frequency |
|-------------|-----------|
| 40 - 49 | 13 |
| 50 - 59 | 28 |
| 60 - 69 | 37 |
| 70 - 79 | 44 |
| 80 - 89 | 21 |
---
##### 🟨 Step 1: Estimate the Mean
To estimate the mean from a grouped frequency table:
$$
\text{Mean} = \frac{\sum (f \times x)}{\sum f}
$$
where:
- $ f $ = frequency
- $ x $ = midpoint of the class
Let’s calculate midpoints and $ f \times x $:
| Weight (kg) | Midpoint ($x$) | Frequency ($f$) | $f \times x$ |
|------------|----------------|------------------|----------------|
| 40–49 | 44.5 | 13 | $13 \times 44.5 = 578.5$ |
| 50–59 | 54.5 | 28 | $28 \times 54.5 = 1526$ |
| 60–69 | 64.5 | 37 | $37 \times 64.5 = 2386.5$ |
| 70–79 | 74.5 | 44 | $44 \times 74.5 = 3278$ |
| 80–89 | 84.5 | 21 | $21 \times 84.5 = 1774.5$ |
Now sum up:
- Total frequency: $13 + 28 + 37 + 44 + 21 = 143$
- Total $f \times x$: $578.5 + 1526 + 2386.5 + 3278 + 1774.5 = 9543.5$
$$
\text{Estimated Mean} = \frac{9543.5}{143} \approx 66.74 \text{ kg}
$$
✔ Estimate of the mean: 66.74 kg
---
##### 🟨 Modal Class
The modal class is the class with the highest frequency.
- Highest frequency = 44 → class 70–79
✔ Modal class: 70–79
---
##### 🟨 Median Class
Total frequency = 143 → median is the value at position $ \frac{143+1}{2} = 72^{th} $ value.
Now find cumulative frequencies:
| Weight (kg) | Frequency | Cumulative Freq |
|------------|----------|------------------|
| 40–49 | 13 | 13 |
| 50–59 | 28 | 41 |
| 60–69 | 37 | 78 |
| 70–79 | 44 | 122 |
| 80–89 | 21 | 143 |
We want the class where the 72nd value falls.
- 41 < 72 ≤ 78 → so it's in the 60–69 class.
✔ Median class: 60–69
---
#### 🔹 2. Distance (d km) Table
| Distance (d km) | Frequency |
|------------------|-----------|
| 0 ≤ d < 15 | 2 |
| 15 ≤ d < 30 | 7 |
| 30 ≤ d < 45 | 19 |
| 45 ≤ d < 60 | 30 |
| 60 ≤ d < 75 | 1 |
---
##### 🟨 Step 1: Estimate the Mean
Midpoints:
- 0–15 → 7.5
- 15–30 → 22.5
- 30–45 → 37.5
- 45–60 → 52.5
- 60–75 → 67.5
Now compute $f \times x$:
| Distance | Midpoint | $f$ | $f \times x$ |
|---------|----------|------|----------------|
| 0–15 | 7.5 | 2 | $2 \times 7.5 = 15$ |
| 15–30 | 22.5 | 7 | $7 \times 22.5 = 157.5$ |
| 30–45 | 37.5 | 19 | $19 \times 37.5 = 712.5$ |
| 45–60 | 52.5 | 30 | $30 \times 52.5 = 1575$ |
| 60–75 | 67.5 | 1 | $1 \times 67.5 = 67.5$ |
Sum:
- Total frequency: $2 + 7 + 19 + 30 + 1 = 59$
- Total $f \times x$: $15 + 157.5 + 712.5 + 1575 + 67.5 = 2527.5$
$$
\text{Mean} = \frac{2527.5}{59} \approx 42.84 \text{ km}
$$
✔ Estimate of the mean: 42.84 km
---
##### 🟨 Modal Class
Highest frequency = 30 → class 45 ≤ d < 60
✔ Modal class: 45 ≤ d < 60
---
##### 🟨 Median Class
Total frequency = 59 → median is at $ \frac{59+1}{2} = 30^{th} $ value.
Cumulative frequencies:
| Distance | $f$ | Cumulative |
|---------|-----|------------|
| 0–15 | 2 | 2 |
| 15–30 | 7 | 9 |
| 30–45 | 19 | 28 |
| 45–60 | 30 | 58 |
| 60–75 | 1 | 59 |
30th value: 28 < 30 ≤ 58 → so it's in 45 ≤ d < 60
✔ Median class: 45 ≤ d < 60
---
| Table | Estimate of Mean | Modal Class | Median Class |
|--------------|-------------------|------------------|------------------|
| Weight | 66.74 kg | 70–79 | 60–69 |
| Distance | 42.84 km | 45 ≤ d < 60 | 45 ≤ d < 60 |
---
## ✔ Section B: Rainfall Data
| Rainfall (mm) | Frequency |
|---------------|-----------|
| 0 - 10 | 2 |
| 10 - 20 | 6 |
| 20 - 30 | 7 |
| 30 - 40 | 11 |
| 40 - 50 | 5 |
Total number of days = $2 + 6 + 7 + 11 + 5 = 31$ days
---
Use midpoints and multiply by frequency:
| Rainfall (mm) | Midpoint | $f$ | $f \times x$ |
|---------------|----------|------|----------------|
| 0–10 | 5 | 2 | $2 \times 5 = 10$ |
| 10–20 | 15 | 6 | $6 \times 15 = 90$ |
| 20–30 | 25 | 7 | $7 \times 25 = 175$ |
| 30–40 | 35 | 11 | $11 \times 35 = 385$ |
| 40–50 | 45 | 5 | $5 \times 45 = 225$ |
Total estimated rainfall:
$$
10 + 90 + 175 + 385 + 225 = 885 \text{ mm}
$$
✔ Answer: 885 mm
---
Days with >30 mm: classes 30–40 and 40–50
Frequencies: $11 + 5 = 16$ days
Total days = 31
$$
\text{Percentage} = \frac{16}{31} \times 100 \approx 51.61\%
$$
✔ Answer: 51.61%
---
Highest frequency = 11 → class 30–40
✔ Modal class: 30–40
---
Total days = 31 → median is the $ \frac{31+1}{2} = 16^{th} $ day
Cumulative frequencies:
| Rainfall | $f$ | Cumulative |
|---------|-----|------------|
| 0–10 | 2 | 2 |
| 10–20 | 6 | 8 |
| 20–30 | 7 | 15 |
| 30–40 | 11 | 26 |
| 40–50 | 5 | 31 |
16th value: 15 < 16 ≤ 26 → so it's in 30–40
✔ Median class: 30–40
---
From earlier:
Total rainfall = 885 mm
Total days = 31
$$
\text{Mean} = \frac{885}{31} \approx 28.58 \text{ mm}
$$
✔ Estimated mean: 28.58 mm
---
Heavy rainfall means outliers (very high values).
- The mean is affected by outliers.
- The median is not affected by extreme values.
- The mode might not be useful here.
So, if there are two extremely heavy rain days, they will inflate the mean, making it unrepresentative of typical rainfall.
👉 Use the median to represent the amount of rainfall.
✔ Answer: Median
---
1. Total rainfall: 885 mm
2. >30 mm days: 51.61%
3. Modal class: 30–40 mm
4. Median class: 30–40 mm
5. Estimated mean: 28.58 mm
6. Best average: Median
---
#### Section A
- Weight:
- Mean: 66.74 kg
- Modal class: 70–79
- Median class: 60–69
- Distance:
- Mean: 42.84 km
- Modal class: 45 ≤ d < 60
- Median class: 45 ≤ d < 60
#### Section B
1. Total rainfall: 885 mm
2. % >30 mm: 51.61%
3. Modal class: 30–40 mm
4. Median class: 30–40 mm
5. Estimated mean: 28.58 mm
6. Best average: Median
Let me know if you'd like this formatted as a printable answer sheet!
---
✔ Section A
We are given two frequency tables: one for Weight (kg) and one for Distance (d km). For each, we need to find:
- Estimate of the mean
- Modal class
- Median class
---
#### 🔹 1. Weight (kg) Table
| Weight (kg) | Frequency |
|-------------|-----------|
| 40 - 49 | 13 |
| 50 - 59 | 28 |
| 60 - 69 | 37 |
| 70 - 79 | 44 |
| 80 - 89 | 21 |
---
##### 🟨 Step 1: Estimate the Mean
To estimate the mean from a grouped frequency table:
$$
\text{Mean} = \frac{\sum (f \times x)}{\sum f}
$$
where:
- $ f $ = frequency
- $ x $ = midpoint of the class
Let’s calculate midpoints and $ f \times x $:
| Weight (kg) | Midpoint ($x$) | Frequency ($f$) | $f \times x$ |
|------------|----------------|------------------|----------------|
| 40–49 | 44.5 | 13 | $13 \times 44.5 = 578.5$ |
| 50–59 | 54.5 | 28 | $28 \times 54.5 = 1526$ |
| 60–69 | 64.5 | 37 | $37 \times 64.5 = 2386.5$ |
| 70–79 | 74.5 | 44 | $44 \times 74.5 = 3278$ |
| 80–89 | 84.5 | 21 | $21 \times 84.5 = 1774.5$ |
Now sum up:
- Total frequency: $13 + 28 + 37 + 44 + 21 = 143$
- Total $f \times x$: $578.5 + 1526 + 2386.5 + 3278 + 1774.5 = 9543.5$
$$
\text{Estimated Mean} = \frac{9543.5}{143} \approx 66.74 \text{ kg}
$$
✔ Estimate of the mean: 66.74 kg
---
##### 🟨 Modal Class
The modal class is the class with the highest frequency.
- Highest frequency = 44 → class 70–79
✔ Modal class: 70–79
---
##### 🟨 Median Class
Total frequency = 143 → median is the value at position $ \frac{143+1}{2} = 72^{th} $ value.
Now find cumulative frequencies:
| Weight (kg) | Frequency | Cumulative Freq |
|------------|----------|------------------|
| 40–49 | 13 | 13 |
| 50–59 | 28 | 41 |
| 60–69 | 37 | 78 |
| 70–79 | 44 | 122 |
| 80–89 | 21 | 143 |
We want the class where the 72nd value falls.
- 41 < 72 ≤ 78 → so it's in the 60–69 class.
✔ Median class: 60–69
---
#### 🔹 2. Distance (d km) Table
| Distance (d km) | Frequency |
|------------------|-----------|
| 0 ≤ d < 15 | 2 |
| 15 ≤ d < 30 | 7 |
| 30 ≤ d < 45 | 19 |
| 45 ≤ d < 60 | 30 |
| 60 ≤ d < 75 | 1 |
---
##### 🟨 Step 1: Estimate the Mean
Midpoints:
- 0–15 → 7.5
- 15–30 → 22.5
- 30–45 → 37.5
- 45–60 → 52.5
- 60–75 → 67.5
Now compute $f \times x$:
| Distance | Midpoint | $f$ | $f \times x$ |
|---------|----------|------|----------------|
| 0–15 | 7.5 | 2 | $2 \times 7.5 = 15$ |
| 15–30 | 22.5 | 7 | $7 \times 22.5 = 157.5$ |
| 30–45 | 37.5 | 19 | $19 \times 37.5 = 712.5$ |
| 45–60 | 52.5 | 30 | $30 \times 52.5 = 1575$ |
| 60–75 | 67.5 | 1 | $1 \times 67.5 = 67.5$ |
Sum:
- Total frequency: $2 + 7 + 19 + 30 + 1 = 59$
- Total $f \times x$: $15 + 157.5 + 712.5 + 1575 + 67.5 = 2527.5$
$$
\text{Mean} = \frac{2527.5}{59} \approx 42.84 \text{ km}
$$
✔ Estimate of the mean: 42.84 km
---
##### 🟨 Modal Class
Highest frequency = 30 → class 45 ≤ d < 60
✔ Modal class: 45 ≤ d < 60
---
##### 🟨 Median Class
Total frequency = 59 → median is at $ \frac{59+1}{2} = 30^{th} $ value.
Cumulative frequencies:
| Distance | $f$ | Cumulative |
|---------|-----|------------|
| 0–15 | 2 | 2 |
| 15–30 | 7 | 9 |
| 30–45 | 19 | 28 |
| 45–60 | 30 | 58 |
| 60–75 | 1 | 59 |
30th value: 28 < 30 ≤ 58 → so it's in 45 ≤ d < 60
✔ Median class: 45 ≤ d < 60
---
✔ Summary – Section A
| Table | Estimate of Mean | Modal Class | Median Class |
|--------------|-------------------|------------------|------------------|
| Weight | 66.74 kg | 70–79 | 60–69 |
| Distance | 42.84 km | 45 ≤ d < 60 | 45 ≤ d < 60 |
---
## ✔ Section B: Rainfall Data
| Rainfall (mm) | Frequency |
|---------------|-----------|
| 0 - 10 | 2 |
| 10 - 20 | 6 |
| 20 - 30 | 7 |
| 30 - 40 | 11 |
| 40 - 50 | 5 |
Total number of days = $2 + 6 + 7 + 11 + 5 = 31$ days
---
🔹 Q1: Estimate total rainfall over the month
Use midpoints and multiply by frequency:
| Rainfall (mm) | Midpoint | $f$ | $f \times x$ |
|---------------|----------|------|----------------|
| 0–10 | 5 | 2 | $2 \times 5 = 10$ |
| 10–20 | 15 | 6 | $6 \times 15 = 90$ |
| 20–30 | 25 | 7 | $7 \times 25 = 175$ |
| 30–40 | 35 | 11 | $11 \times 35 = 385$ |
| 40–50 | 45 | 5 | $5 \times 45 = 225$ |
Total estimated rainfall:
$$
10 + 90 + 175 + 385 + 225 = 885 \text{ mm}
$$
✔ Answer: 885 mm
---
🔹 Q2: Percentage of days with more than 30 mm rainfall
Days with >30 mm: classes 30–40 and 40–50
Frequencies: $11 + 5 = 16$ days
Total days = 31
$$
\text{Percentage} = \frac{16}{31} \times 100 \approx 51.61\%
$$
✔ Answer: 51.61%
---
🔹 Q3: Modal class
Highest frequency = 11 → class 30–40
✔ Modal class: 30–40
---
🔹 Q4: In which class is the median value?
Total days = 31 → median is the $ \frac{31+1}{2} = 16^{th} $ day
Cumulative frequencies:
| Rainfall | $f$ | Cumulative |
|---------|-----|------------|
| 0–10 | 2 | 2 |
| 10–20 | 6 | 8 |
| 20–30 | 7 | 15 |
| 30–40 | 11 | 26 |
| 40–50 | 5 | 31 |
16th value: 15 < 16 ≤ 26 → so it's in 30–40
✔ Median class: 30–40
---
🔹 Q5: Estimate the mean
From earlier:
Total rainfall = 885 mm
Total days = 31
$$
\text{Mean} = \frac{885}{31} \approx 28.58 \text{ mm}
$$
✔ Estimated mean: 28.58 mm
---
🔹 Q6: Which average to use next month if there were two heavy rain days?
Heavy rainfall means outliers (very high values).
- The mean is affected by outliers.
- The median is not affected by extreme values.
- The mode might not be useful here.
So, if there are two extremely heavy rain days, they will inflate the mean, making it unrepresentative of typical rainfall.
👉 Use the median to represent the amount of rainfall.
✔ Answer: Median
---
✔ Final Answers – Section B
1. Total rainfall: 885 mm
2. >30 mm days: 51.61%
3. Modal class: 30–40 mm
4. Median class: 30–40 mm
5. Estimated mean: 28.58 mm
6. Best average: Median
---
✔ Final Summary
#### Section A
- Weight:
- Mean: 66.74 kg
- Modal class: 70–79
- Median class: 60–69
- Distance:
- Mean: 42.84 km
- Modal class: 45 ≤ d < 60
- Median class: 45 ≤ d < 60
#### Section B
1. Total rainfall: 885 mm
2. % >30 mm: 51.61%
3. Modal class: 30–40 mm
4. Median class: 30–40 mm
5. Estimated mean: 28.58 mm
6. Best average: Median
Let me know if you'd like this formatted as a printable answer sheet!
Parent Tip: Review the logic above to help your child master the concept of frequency distribution worksheet.
