Answer :
To calculate the arithmetic mean of the given data, we first need to find the frequencies for each class interval and then proceed with the calculations. Here is a step-by-step process:
### Step 1: Calculate the Frequencies
The class intervals are given in a cumulative frequency form. Let's translate them into actual frequencies for each class interval:
- Frequency for the 0-10 class ([tex]\(f_1\)[/tex]):
[tex]\[ f_1 = \text{More than 0} - \text{More than 10} = 50 - 46 = 4 \][/tex]
- Frequency for the 10-20 class ([tex]\(f_2\)[/tex]):
[tex]\[ f_2 = \text{More than 10} - \text{More than 20} = 46 - 40 = 6 \][/tex]
- Frequency for the 20-30 class ([tex]\(f_3\)[/tex]):
[tex]\[ f_3 = \text{More than 20} - \text{More than 30} = 40 - 20 = 20 \][/tex]
- Frequency for the 30-40 class ([tex]\(f_4\)[/tex]):
[tex]\[ f_4 = \text{More than 30} - \text{More than 40} = 20 - 10 = 10 \][/tex]
- Frequency for the 40-50 class ([tex]\(f_5\)[/tex]):
[tex]\[ f_5 = \text{More than 40} - \text{More than 50} = 10 - 3 = 7 \][/tex]
- Frequency for the class above 50 ([tex]\(f_6\)[/tex]):
[tex]\[ f_6 = \text{More than 50} = 3 \][/tex]
### Step 2: Determine the Midpoints
For calculating the mean, we need the midpoint of each class interval:
- Midpoint of 0-10 class interval ([tex]\(m_1\)[/tex]):
[tex]\[ m_1 = 5 \][/tex]
- Midpoint of 10-20 class interval ([tex]\(m_2\)[/tex]):
[tex]\[ m_2 = 15 \][/tex]
- Midpoint of 20-30 class interval ([tex]\(m_3\)[/tex]):
[tex]\[ m_3 = 25 \][/tex]
- Midpoint of 30-40 class interval ([tex]\(m_4\)[/tex]):
[tex]\[ m_4 = 35 \][/tex]
- Midpoint of 40-50 class interval ([tex]\(m_5\)[/tex]):
[tex]\[ m_5 = 45 \][/tex]
- Midpoint of class above 50 ([tex]\(m_6\)[/tex]):
[tex]\[ m_6 = 55 \][/tex]
### Step 3: Sum of Frequencies
Calculate the total frequency:
[tex]\[ \text{Total Frequency } (N) = f_1 + f_2 + f_3 + f_4 + f_5 + f_6 = 4 + 6 + 20 + 10 + 7 + 3 = 50 \][/tex]
### Step 4: Sum of Frequency-Midpoint Products
Calculate the sum of the frequency multiplied by the midpoint for each class interval:
[tex]\[ \sum f_i m_i = (4 \cdot 5) + (6 \cdot 15) + (20 \cdot 25) + (10 \cdot 35) + (7 \cdot 45) + (3 \cdot 55) \][/tex]
Calculating each product:
- [tex]\(4 \cdot 5 = 20\)[/tex]
- [tex]\(6 \cdot 15 = 90\)[/tex]
- [tex]\(20 \cdot 25 = 500\)[/tex]
- [tex]\(10 \cdot 35 = 350\)[/tex]
- [tex]\(7 \cdot 45 = 315\)[/tex]
- [tex]\(3 \cdot 55 = 165\)[/tex]
Sum of these products:
[tex]\[ \sum f_i m_i = 20 + 90 + 500 + 350 + 315 + 165 = 1440 \][/tex]
### Step 5: Calculate the Arithmetic Mean
The arithmetic mean is calculated by dividing the sum of the frequency-midpoint products by the total frequency:
[tex]\[ \text{Arithmetic Mean} = \frac{\sum f_i m_i}{N} = \frac{1440}{50} = 28.8 \][/tex]
### Final Answer
The arithmetic mean of the given data is 28.8.
### Step 1: Calculate the Frequencies
The class intervals are given in a cumulative frequency form. Let's translate them into actual frequencies for each class interval:
- Frequency for the 0-10 class ([tex]\(f_1\)[/tex]):
[tex]\[ f_1 = \text{More than 0} - \text{More than 10} = 50 - 46 = 4 \][/tex]
- Frequency for the 10-20 class ([tex]\(f_2\)[/tex]):
[tex]\[ f_2 = \text{More than 10} - \text{More than 20} = 46 - 40 = 6 \][/tex]
- Frequency for the 20-30 class ([tex]\(f_3\)[/tex]):
[tex]\[ f_3 = \text{More than 20} - \text{More than 30} = 40 - 20 = 20 \][/tex]
- Frequency for the 30-40 class ([tex]\(f_4\)[/tex]):
[tex]\[ f_4 = \text{More than 30} - \text{More than 40} = 20 - 10 = 10 \][/tex]
- Frequency for the 40-50 class ([tex]\(f_5\)[/tex]):
[tex]\[ f_5 = \text{More than 40} - \text{More than 50} = 10 - 3 = 7 \][/tex]
- Frequency for the class above 50 ([tex]\(f_6\)[/tex]):
[tex]\[ f_6 = \text{More than 50} = 3 \][/tex]
### Step 2: Determine the Midpoints
For calculating the mean, we need the midpoint of each class interval:
- Midpoint of 0-10 class interval ([tex]\(m_1\)[/tex]):
[tex]\[ m_1 = 5 \][/tex]
- Midpoint of 10-20 class interval ([tex]\(m_2\)[/tex]):
[tex]\[ m_2 = 15 \][/tex]
- Midpoint of 20-30 class interval ([tex]\(m_3\)[/tex]):
[tex]\[ m_3 = 25 \][/tex]
- Midpoint of 30-40 class interval ([tex]\(m_4\)[/tex]):
[tex]\[ m_4 = 35 \][/tex]
- Midpoint of 40-50 class interval ([tex]\(m_5\)[/tex]):
[tex]\[ m_5 = 45 \][/tex]
- Midpoint of class above 50 ([tex]\(m_6\)[/tex]):
[tex]\[ m_6 = 55 \][/tex]
### Step 3: Sum of Frequencies
Calculate the total frequency:
[tex]\[ \text{Total Frequency } (N) = f_1 + f_2 + f_3 + f_4 + f_5 + f_6 = 4 + 6 + 20 + 10 + 7 + 3 = 50 \][/tex]
### Step 4: Sum of Frequency-Midpoint Products
Calculate the sum of the frequency multiplied by the midpoint for each class interval:
[tex]\[ \sum f_i m_i = (4 \cdot 5) + (6 \cdot 15) + (20 \cdot 25) + (10 \cdot 35) + (7 \cdot 45) + (3 \cdot 55) \][/tex]
Calculating each product:
- [tex]\(4 \cdot 5 = 20\)[/tex]
- [tex]\(6 \cdot 15 = 90\)[/tex]
- [tex]\(20 \cdot 25 = 500\)[/tex]
- [tex]\(10 \cdot 35 = 350\)[/tex]
- [tex]\(7 \cdot 45 = 315\)[/tex]
- [tex]\(3 \cdot 55 = 165\)[/tex]
Sum of these products:
[tex]\[ \sum f_i m_i = 20 + 90 + 500 + 350 + 315 + 165 = 1440 \][/tex]
### Step 5: Calculate the Arithmetic Mean
The arithmetic mean is calculated by dividing the sum of the frequency-midpoint products by the total frequency:
[tex]\[ \text{Arithmetic Mean} = \frac{\sum f_i m_i}{N} = \frac{1440}{50} = 28.8 \][/tex]
### Final Answer
The arithmetic mean of the given data is 28.8.
