Answer :
To solve this problem, we need to find the mode, median, and arithmetic average (mean) for the given frequency distribution of class intervals. Let's walk through each step in detail:
### Given Data:
We have different class intervals with their respective frequencies:
- [tex]\(0-2\)[/tex], Frequency: 8
- [tex]\(2-4\)[/tex], Frequency: 12
- [tex]\(4-10\)[/tex], Frequency: 20
- [tex]\(10-15\)[/tex], Frequency: 10
- [tex]\(15-20\)[/tex], Frequency: 16
- [tex]\(20-25\)[/tex], Frequency: 25
- [tex]\(25-30\)[/tex], Frequency: 45
- [tex]\(30-40\)[/tex], Frequency: 60
- [tex]\(40-50\)[/tex], Frequency: 20
- [tex]\(50-60\)[/tex], Frequency: 13
- [tex]\(60-80\)[/tex], Frequency: 15
- [tex]\(80-100\)[/tex], Frequency: 4
### 1. Arithmetic Average (Mean):
To calculate the mean, we use the midpoints of each class interval and their frequencies.
1. Calculate midpoints: The midpoint of each class interval is the average of its lower and upper bounds. For example, the midpoint of [tex]\(0-2\)[/tex] is [tex]\( (0+2)/2 = 1\)[/tex].
2. Find the total number of observations ([tex]\(n\)[/tex]): Add all the frequencies.
3. Calculate the mean:
[tex]\[
\text{Mean} = \frac{\sum (\text{midpoint} \times \text{frequency})}{\sum \text{frequency}}
\][/tex]
The calculated mean is approximately [tex]\(30.30\)[/tex].
### 2. Median:
The median class is the class interval where the cumulative frequency reaches or exceeds half the total number of observations ([tex]\(n/2\)[/tex]).
1. Calculate cumulative frequencies for each class interval.
2. Find the median class: Identify the class interval where the cumulative frequency first exceeds [tex]\(n/2\)[/tex].
3. Locate the median within the class interval using the formula:
[tex]\[
\text{Median} = L + \left( \frac{\frac{n}{2} - F}{f} \right) \times c
\][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class,
- [tex]\(F\)[/tex] is the cumulative frequency of the preceding class,
- [tex]\(f\)[/tex] is the frequency of the median class,
- [tex]\(c\)[/tex] is the class width.
The calculated median is approximately [tex]\(28.67\)[/tex].
### 3. Mode:
The mode is the value that appears most frequently in a data set. In grouped data, it is found in the class interval with the highest frequency.
1. Identify the modal class: The class interval with the highest frequency.
2. Use the formula to find the mode within that class:
[tex]\[
\text{Mode} = L + \left( \frac{f_m - f_1}{(f_m - f_1) + (f_m - f_2)} \right) \times c
\][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the modal class,
- [tex]\(f_m\)[/tex] is the frequency of the modal class,
- [tex]\(f_1\)[/tex] is the frequency of the class before the modal class,
- [tex]\(f_2\)[/tex] is the frequency of the class after the modal class,
- [tex]\(c\)[/tex] is the class width.
The calculated mode is approximately [tex]\(31.36\)[/tex].
These calculations provide us with the mode [tex]\(31.36\)[/tex], the median [tex]\(28.67\)[/tex], and the arithmetic mean [tex]\(30.30\)[/tex].
### Given Data:
We have different class intervals with their respective frequencies:
- [tex]\(0-2\)[/tex], Frequency: 8
- [tex]\(2-4\)[/tex], Frequency: 12
- [tex]\(4-10\)[/tex], Frequency: 20
- [tex]\(10-15\)[/tex], Frequency: 10
- [tex]\(15-20\)[/tex], Frequency: 16
- [tex]\(20-25\)[/tex], Frequency: 25
- [tex]\(25-30\)[/tex], Frequency: 45
- [tex]\(30-40\)[/tex], Frequency: 60
- [tex]\(40-50\)[/tex], Frequency: 20
- [tex]\(50-60\)[/tex], Frequency: 13
- [tex]\(60-80\)[/tex], Frequency: 15
- [tex]\(80-100\)[/tex], Frequency: 4
### 1. Arithmetic Average (Mean):
To calculate the mean, we use the midpoints of each class interval and their frequencies.
1. Calculate midpoints: The midpoint of each class interval is the average of its lower and upper bounds. For example, the midpoint of [tex]\(0-2\)[/tex] is [tex]\( (0+2)/2 = 1\)[/tex].
2. Find the total number of observations ([tex]\(n\)[/tex]): Add all the frequencies.
3. Calculate the mean:
[tex]\[
\text{Mean} = \frac{\sum (\text{midpoint} \times \text{frequency})}{\sum \text{frequency}}
\][/tex]
The calculated mean is approximately [tex]\(30.30\)[/tex].
### 2. Median:
The median class is the class interval where the cumulative frequency reaches or exceeds half the total number of observations ([tex]\(n/2\)[/tex]).
1. Calculate cumulative frequencies for each class interval.
2. Find the median class: Identify the class interval where the cumulative frequency first exceeds [tex]\(n/2\)[/tex].
3. Locate the median within the class interval using the formula:
[tex]\[
\text{Median} = L + \left( \frac{\frac{n}{2} - F}{f} \right) \times c
\][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class,
- [tex]\(F\)[/tex] is the cumulative frequency of the preceding class,
- [tex]\(f\)[/tex] is the frequency of the median class,
- [tex]\(c\)[/tex] is the class width.
The calculated median is approximately [tex]\(28.67\)[/tex].
### 3. Mode:
The mode is the value that appears most frequently in a data set. In grouped data, it is found in the class interval with the highest frequency.
1. Identify the modal class: The class interval with the highest frequency.
2. Use the formula to find the mode within that class:
[tex]\[
\text{Mode} = L + \left( \frac{f_m - f_1}{(f_m - f_1) + (f_m - f_2)} \right) \times c
\][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the modal class,
- [tex]\(f_m\)[/tex] is the frequency of the modal class,
- [tex]\(f_1\)[/tex] is the frequency of the class before the modal class,
- [tex]\(f_2\)[/tex] is the frequency of the class after the modal class,
- [tex]\(c\)[/tex] is the class width.
The calculated mode is approximately [tex]\(31.36\)[/tex].
These calculations provide us with the mode [tex]\(31.36\)[/tex], the median [tex]\(28.67\)[/tex], and the arithmetic mean [tex]\(30.30\)[/tex].
