Answer :
To find the mean of the given data using the step deviation method, let's follow these steps:
### Step 1: Identify the class intervals and frequencies
We have the following class intervals and their corresponding frequencies:
- [tex]\(10-20\)[/tex]: Frequency [tex]\(F = 2\)[/tex]
- [tex]\(20-30\)[/tex]: Frequency [tex]\(F = 3\)[/tex]
- [tex]\(30-40\)[/tex]: Frequency [tex]\(F = 5\)[/tex]
- [tex]\(40-50\)[/tex]: Frequency [tex]\(F = 7\)[/tex]
- [tex]\(50-60\)[/tex]: Frequency [tex]\(F = 5\)[/tex]
- [tex]\(60-70\)[/tex]: Frequency [tex]\(F = 3\)[/tex]
### Step 2: Calculate the midpoints of each class interval
The midpoint of a class interval is given by the formula [tex]\(\frac{\text{lower limit} + \text{upper limit}}{2}\)[/tex]. Let's calculate the midpoints for each class:
- [tex]\( \frac{10 + 20}{2} = 15.0 \)[/tex]
- [tex]\( \frac{20 + 30}{2} = 25.0 \)[/tex]
- [tex]\( \frac{30 + 40}{2} = 35.0 \)[/tex]
- [tex]\( \frac{40 + 50}{2} = 45.0 \)[/tex]
- [tex]\( \frac{50 + 60}{2} = 55.0 \)[/tex]
- [tex]\( \frac{60 + 70}{2} = 65.0 \)[/tex]
So, the midpoints are: [tex]\[ 15.0, 25.0, 35.0, 45.0, 55.0, 65.0 \][/tex]
### Step 3: Calculate the sum of the frequencies
Sum of all frequencies [tex]\(n = 2 + 3 + 5 + 7 + 5 + 3 = 25\)[/tex]
### Step 4: Choose an assumed mean [tex]\(A\)[/tex]
We can choose any midpoint as the assumed mean [tex]\(A\)[/tex]. Here, we will choose [tex]\(A = 45.0\)[/tex].
### Step 5: Calculate the deviations ([tex]\(d_i = \text{midpoint} - A\)[/tex])
Let's calculate the deviations for each midpoint:
- For [tex]\(15.0\)[/tex]: [tex]\(d = 15.0 - 45.0 = -30.0\)[/tex]
- For [tex]\(25.0\)[/tex]: [tex]\(d = 25.0 - 45.0 = -20.0\)[/tex]
- For [tex]\(35.0\)[/tex]: [tex]\(d = 35.0 - 45.0 = -10.0\)[/tex]
- For [tex]\(45.0\)[/tex]: [tex]\(d = 45.0 - 45.0 = 0.0\)[/tex]
- For [tex]\(55.0\)[/tex]: [tex]\(d = 55.0 - 45.0 = 10.0\)[/tex]
- For [tex]\(65.0\)[/tex]: [tex]\(d = 65.0 - 45.0 = 20.0\)[/tex]
So, the deviations are: [tex]\[ -30.0, -20.0, -10.0, 0.0, 10.0, 20.0 \][/tex]
### Step 6: Calculate [tex]\( \sum{F_i \cdot d_i} \)[/tex]
Now, we calculate the sum of the products of the frequencies and their respective deviations.
[tex]\[ Fd\_sum = (2 \times -30.0) + (3 \times -20.0) + (5 \times -10.0) + (7 \times 0.0) + (5 \times 10.0) + (3 \times 20.0) \][/tex]
[tex]\[ Fd\_sum = -60.0 + (-60.0) + (-50.0) + 0.0 + 50.0 + 60.0 \][/tex]
[tex]\[ Fd\_sum = -60.0 \][/tex]
### Step 7: Determine the class interval [tex]\( h \)[/tex]
The class interval [tex]\( h \)[/tex] is the difference between the upper and lower boundaries of any class. Here, [tex]\( h = 20 - 10 = 10 \)[/tex].
### Step 8: Calculate the mean using the step deviation formula
The mean [tex]\( \bar{x} \)[/tex] is given by:
[tex]\[ \bar{x} = A + \left( \frac{\sum{F_i \cdot d_i}}{n} \right) \times h \][/tex]
Substitute the values:
[tex]\[ \bar{x} = 45.0 + \left( \frac{-60.0}{25} \right) \times 10 \][/tex]
[tex]\[ \bar{x} = 45.0 + \left( -2.4 \right) \times 10 \][/tex]
[tex]\[ \bar{x} = 45.0 - 24.0 \][/tex]
[tex]\[ \bar{x} = 21.0 \][/tex]
### Conclusion
The mean of the given data using the step deviation method is [tex]\( 21.0 \)[/tex].
### Step 1: Identify the class intervals and frequencies
We have the following class intervals and their corresponding frequencies:
- [tex]\(10-20\)[/tex]: Frequency [tex]\(F = 2\)[/tex]
- [tex]\(20-30\)[/tex]: Frequency [tex]\(F = 3\)[/tex]
- [tex]\(30-40\)[/tex]: Frequency [tex]\(F = 5\)[/tex]
- [tex]\(40-50\)[/tex]: Frequency [tex]\(F = 7\)[/tex]
- [tex]\(50-60\)[/tex]: Frequency [tex]\(F = 5\)[/tex]
- [tex]\(60-70\)[/tex]: Frequency [tex]\(F = 3\)[/tex]
### Step 2: Calculate the midpoints of each class interval
The midpoint of a class interval is given by the formula [tex]\(\frac{\text{lower limit} + \text{upper limit}}{2}\)[/tex]. Let's calculate the midpoints for each class:
- [tex]\( \frac{10 + 20}{2} = 15.0 \)[/tex]
- [tex]\( \frac{20 + 30}{2} = 25.0 \)[/tex]
- [tex]\( \frac{30 + 40}{2} = 35.0 \)[/tex]
- [tex]\( \frac{40 + 50}{2} = 45.0 \)[/tex]
- [tex]\( \frac{50 + 60}{2} = 55.0 \)[/tex]
- [tex]\( \frac{60 + 70}{2} = 65.0 \)[/tex]
So, the midpoints are: [tex]\[ 15.0, 25.0, 35.0, 45.0, 55.0, 65.0 \][/tex]
### Step 3: Calculate the sum of the frequencies
Sum of all frequencies [tex]\(n = 2 + 3 + 5 + 7 + 5 + 3 = 25\)[/tex]
### Step 4: Choose an assumed mean [tex]\(A\)[/tex]
We can choose any midpoint as the assumed mean [tex]\(A\)[/tex]. Here, we will choose [tex]\(A = 45.0\)[/tex].
### Step 5: Calculate the deviations ([tex]\(d_i = \text{midpoint} - A\)[/tex])
Let's calculate the deviations for each midpoint:
- For [tex]\(15.0\)[/tex]: [tex]\(d = 15.0 - 45.0 = -30.0\)[/tex]
- For [tex]\(25.0\)[/tex]: [tex]\(d = 25.0 - 45.0 = -20.0\)[/tex]
- For [tex]\(35.0\)[/tex]: [tex]\(d = 35.0 - 45.0 = -10.0\)[/tex]
- For [tex]\(45.0\)[/tex]: [tex]\(d = 45.0 - 45.0 = 0.0\)[/tex]
- For [tex]\(55.0\)[/tex]: [tex]\(d = 55.0 - 45.0 = 10.0\)[/tex]
- For [tex]\(65.0\)[/tex]: [tex]\(d = 65.0 - 45.0 = 20.0\)[/tex]
So, the deviations are: [tex]\[ -30.0, -20.0, -10.0, 0.0, 10.0, 20.0 \][/tex]
### Step 6: Calculate [tex]\( \sum{F_i \cdot d_i} \)[/tex]
Now, we calculate the sum of the products of the frequencies and their respective deviations.
[tex]\[ Fd\_sum = (2 \times -30.0) + (3 \times -20.0) + (5 \times -10.0) + (7 \times 0.0) + (5 \times 10.0) + (3 \times 20.0) \][/tex]
[tex]\[ Fd\_sum = -60.0 + (-60.0) + (-50.0) + 0.0 + 50.0 + 60.0 \][/tex]
[tex]\[ Fd\_sum = -60.0 \][/tex]
### Step 7: Determine the class interval [tex]\( h \)[/tex]
The class interval [tex]\( h \)[/tex] is the difference between the upper and lower boundaries of any class. Here, [tex]\( h = 20 - 10 = 10 \)[/tex].
### Step 8: Calculate the mean using the step deviation formula
The mean [tex]\( \bar{x} \)[/tex] is given by:
[tex]\[ \bar{x} = A + \left( \frac{\sum{F_i \cdot d_i}}{n} \right) \times h \][/tex]
Substitute the values:
[tex]\[ \bar{x} = 45.0 + \left( \frac{-60.0}{25} \right) \times 10 \][/tex]
[tex]\[ \bar{x} = 45.0 + \left( -2.4 \right) \times 10 \][/tex]
[tex]\[ \bar{x} = 45.0 - 24.0 \][/tex]
[tex]\[ \bar{x} = 21.0 \][/tex]
### Conclusion
The mean of the given data using the step deviation method is [tex]\( 21.0 \)[/tex].
