Answer :
Sure! Let's go through the step-by-step process to find the mean, median, and mode for the given frequency distribution table.
### Step 1: Calculate the Midpoints of Each Class Interval
First, we need to find the midpoints of each class interval. The midpoint ([tex]\(x_i\)[/tex]) for a class interval [tex]\((a, b)\)[/tex] is calculated using the formula:
[tex]\[ x_i = \frac{a + b}{2} \][/tex]
For the given class intervals:
- [tex]\(10-20\)[/tex]: [tex]\( \frac{10 + 20}{2} = 15.0 \)[/tex]
- [tex]\(20-30\)[/tex]: [tex]\( \frac{20 + 30}{2} = 25.0 \)[/tex]
- [tex]\(30-40\)[/tex]: [tex]\( \frac{30 + 40}{2} = 35.0 \)[/tex]
- [tex]\(40-50\)[/tex]: [tex]\( \frac{40 + 50}{2} = 45.0 \)[/tex]
- [tex]\(50-60\)[/tex]: [tex]\( \frac{50 + 60}{2} = 55.0 \)[/tex]
So, the midpoints are [tex]\(15.0, 25.0, 35.0, 45.0, 55.0\)[/tex].
### Step 2: Calculate the Mean
The mean ([tex]\(\bar{x}\)[/tex]) is calculated using the formula:
[tex]\[ \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i} \][/tex]
Where:
- [tex]\(x_i\)[/tex] is the midpoint of each class interval
- [tex]\(f_i\)[/tex] is the frequency of each class interval
Substitute the values:
[tex]\[ \bar{x} = \frac{(15.0 \cdot 5) + (25.0 \cdot 8) + (35.0 \cdot 12) + (45.0 \cdot 6) + (55.0 \cdot 3)}{5 + 8 + 12 + 6 + 3} \][/tex]
[tex]\[ \bar{x} = \frac{75 + 200 + 420 + 270 + 165}{34} = \frac{1130}{34} \approx 33.24 \][/tex]
### Step 3: Calculate the Median
To find the median, we need to locate the class interval containing the median. Follow these steps:
1. Calculate the cumulative frequency:
- Cumulative frequency up to [tex]\(10-20\)[/tex]: [tex]\(5\)[/tex]
- Cumulative frequency up to [tex]\(20-30\)[/tex]: [tex]\(5 + 8 = 13\)[/tex]
- Cumulative frequency up to [tex]\(30-40\)[/tex]: [tex]\(13 + 12 = 25\)[/tex]
- Cumulative frequency up to [tex]\(40-50\)[/tex]: [tex]\(25 + 6 = 31\)[/tex]
- Cumulative frequency up to [tex]\(50-60\)[/tex]: [tex]\(31 + 3 = 34\)[/tex]
2. Identify the median class, which is the class where the cumulative frequency is greater than or equal to [tex]\(\frac{N}{2}\)[/tex]:
- [tex]\(N = 34\)[/tex]; [tex]\(\frac{N}{2} = 17\)[/tex]
- The cumulative frequency first exceeds [tex]\(17\)[/tex] in the class interval [tex]\(30-40\)[/tex].
3. Calculate the median using the formula:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f_m} \right) \cdot c \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class: [tex]\(30\)[/tex]
- [tex]\(N\)[/tex] is the total frequency: [tex]\(34\)[/tex]
- [tex]\(F\)[/tex] is the cumulative frequency before the median class: [tex]\(13\)[/tex]
- [tex]\(f_m\)[/tex] is the frequency of the median class: [tex]\(12\)[/tex]
- [tex]\(c\)[/tex] is the class interval width: [tex]\(10\)[/tex]
Substitute the values:
[tex]\[ \text{Median} = 30 + \left( \frac{17 - 13}{12} \right) \cdot 10 = 30 + \left( \frac{4}{12} \right) \cdot 10 = 30 + 3.33 \approx 33.33 \][/tex]
### Step 4: Calculate the Mode
To find the mode, identify the modal class, which is the class interval with the highest frequency. The frequency values are [tex]\(5, 8, 12, 6, 3\)[/tex]:
- The highest frequency is [tex]\(12\)[/tex] for the class interval [tex]\(30-40\)[/tex].
Calculate the mode using the formula:
[tex]\[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \cdot c \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the modal class: [tex]\(30\)[/tex]
- [tex]\(f_1\)[/tex] is the frequency of the modal class: [tex]\(12\)[/tex]
- [tex]\(f_0\)[/tex] is the frequency of the class before the modal class: [tex]\(8\)[/tex]
- [tex]\(f_2\)[/tex] is the frequency of the class after the modal class: [tex]\(6\)[/tex]
- [tex]\(c\)[/tex] is the class interval width: [tex]\(10\)[/tex]
Substitute the values:
[tex]\[ \text{Mode} = 30 + \left( \frac{12 - 8}{2 \cdot 12 - 8 - 6} \right) \cdot 10 = 30 + \left( \frac{4}{24 - 14} \right) \cdot 10 = 30 + \left( \frac{4}{10} \right) \cdot 10 = 30 + 4 = 34.00 \][/tex]
### Summary
The ordered quantities (Mean, Median, Mode) for the given frequency distribution table are:
- Mean: 33.24
- Median: 33.33
- Mode: 34.00
### Step 1: Calculate the Midpoints of Each Class Interval
First, we need to find the midpoints of each class interval. The midpoint ([tex]\(x_i\)[/tex]) for a class interval [tex]\((a, b)\)[/tex] is calculated using the formula:
[tex]\[ x_i = \frac{a + b}{2} \][/tex]
For the given class intervals:
- [tex]\(10-20\)[/tex]: [tex]\( \frac{10 + 20}{2} = 15.0 \)[/tex]
- [tex]\(20-30\)[/tex]: [tex]\( \frac{20 + 30}{2} = 25.0 \)[/tex]
- [tex]\(30-40\)[/tex]: [tex]\( \frac{30 + 40}{2} = 35.0 \)[/tex]
- [tex]\(40-50\)[/tex]: [tex]\( \frac{40 + 50}{2} = 45.0 \)[/tex]
- [tex]\(50-60\)[/tex]: [tex]\( \frac{50 + 60}{2} = 55.0 \)[/tex]
So, the midpoints are [tex]\(15.0, 25.0, 35.0, 45.0, 55.0\)[/tex].
### Step 2: Calculate the Mean
The mean ([tex]\(\bar{x}\)[/tex]) is calculated using the formula:
[tex]\[ \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i} \][/tex]
Where:
- [tex]\(x_i\)[/tex] is the midpoint of each class interval
- [tex]\(f_i\)[/tex] is the frequency of each class interval
Substitute the values:
[tex]\[ \bar{x} = \frac{(15.0 \cdot 5) + (25.0 \cdot 8) + (35.0 \cdot 12) + (45.0 \cdot 6) + (55.0 \cdot 3)}{5 + 8 + 12 + 6 + 3} \][/tex]
[tex]\[ \bar{x} = \frac{75 + 200 + 420 + 270 + 165}{34} = \frac{1130}{34} \approx 33.24 \][/tex]
### Step 3: Calculate the Median
To find the median, we need to locate the class interval containing the median. Follow these steps:
1. Calculate the cumulative frequency:
- Cumulative frequency up to [tex]\(10-20\)[/tex]: [tex]\(5\)[/tex]
- Cumulative frequency up to [tex]\(20-30\)[/tex]: [tex]\(5 + 8 = 13\)[/tex]
- Cumulative frequency up to [tex]\(30-40\)[/tex]: [tex]\(13 + 12 = 25\)[/tex]
- Cumulative frequency up to [tex]\(40-50\)[/tex]: [tex]\(25 + 6 = 31\)[/tex]
- Cumulative frequency up to [tex]\(50-60\)[/tex]: [tex]\(31 + 3 = 34\)[/tex]
2. Identify the median class, which is the class where the cumulative frequency is greater than or equal to [tex]\(\frac{N}{2}\)[/tex]:
- [tex]\(N = 34\)[/tex]; [tex]\(\frac{N}{2} = 17\)[/tex]
- The cumulative frequency first exceeds [tex]\(17\)[/tex] in the class interval [tex]\(30-40\)[/tex].
3. Calculate the median using the formula:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f_m} \right) \cdot c \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class: [tex]\(30\)[/tex]
- [tex]\(N\)[/tex] is the total frequency: [tex]\(34\)[/tex]
- [tex]\(F\)[/tex] is the cumulative frequency before the median class: [tex]\(13\)[/tex]
- [tex]\(f_m\)[/tex] is the frequency of the median class: [tex]\(12\)[/tex]
- [tex]\(c\)[/tex] is the class interval width: [tex]\(10\)[/tex]
Substitute the values:
[tex]\[ \text{Median} = 30 + \left( \frac{17 - 13}{12} \right) \cdot 10 = 30 + \left( \frac{4}{12} \right) \cdot 10 = 30 + 3.33 \approx 33.33 \][/tex]
### Step 4: Calculate the Mode
To find the mode, identify the modal class, which is the class interval with the highest frequency. The frequency values are [tex]\(5, 8, 12, 6, 3\)[/tex]:
- The highest frequency is [tex]\(12\)[/tex] for the class interval [tex]\(30-40\)[/tex].
Calculate the mode using the formula:
[tex]\[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \cdot c \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the modal class: [tex]\(30\)[/tex]
- [tex]\(f_1\)[/tex] is the frequency of the modal class: [tex]\(12\)[/tex]
- [tex]\(f_0\)[/tex] is the frequency of the class before the modal class: [tex]\(8\)[/tex]
- [tex]\(f_2\)[/tex] is the frequency of the class after the modal class: [tex]\(6\)[/tex]
- [tex]\(c\)[/tex] is the class interval width: [tex]\(10\)[/tex]
Substitute the values:
[tex]\[ \text{Mode} = 30 + \left( \frac{12 - 8}{2 \cdot 12 - 8 - 6} \right) \cdot 10 = 30 + \left( \frac{4}{24 - 14} \right) \cdot 10 = 30 + \left( \frac{4}{10} \right) \cdot 10 = 30 + 4 = 34.00 \][/tex]
### Summary
The ordered quantities (Mean, Median, Mode) for the given frequency distribution table are:
- Mean: 33.24
- Median: 33.33
- Mode: 34.00
