Answer :
Alright, let's solve this problem step by step:
1. Understanding the Given Data:
- Arithmetic mean (A) of the distribution is 33.
- Assumed mean (A') is 35.
- Step deviations (d) are: -3, -2, -1, 0, 1, 2.
- Frequency (f) corresponding to these step deviations are: 5, 10, 25, 30, 20, 10.
2. Calculate Total Number of Observations (N):
The total number of observations is the sum of all frequencies.
[tex]\[ N = 5 + 10 + 25 + 30 + 20 + 10 = 100 \][/tex]
3. Calculate Summation of Step Deviations Times Frequencies (Σf_i d_i):
We need to calculate the product of each step deviation with its corresponding frequency, and then sum these products.
[tex]\[ \Sigma f_i d_i = (5 \times -3) + (10 \times -2) + (25 \times -1) + (30 \times 0) + (20 \times 1) + (10 \times 2) \][/tex]
Breaking it down:
[tex]\[ \Sigma f_i d_i = (5 \times -3) + (10 \times -2) + (25 \times -1) + (30 \times 0) + (20 \times 1) + (10 \times 2) \][/tex]
[tex]\[ \Sigma f_i d_i = -15 + (-20) + (-25) + 0 + 20 + 20 = -20 \][/tex]
4. Calculate Mean Deviation:
The mean deviation is given by the formula:
[tex]\[ \text{Mean Deviation} = \frac{\Sigma f_i d_i}{N} \][/tex]
Substituting the values:
[tex]\[ \text{Mean Deviation} = \frac{-20}{100} = -0.2 \][/tex]
5. Relating Arithmetic Mean, Assumed Mean, and Mean Deviation:
The arithmetic mean can be related to the assumed mean through the following formula:
[tex]\[ \text{Arithmetic Mean} (A) = \text{Assumed Mean} (A') + (\text{Mean Deviation} \times \text{Class Interval}) \][/tex]
We know:
[tex]\( A = 33 \)[/tex]
[tex]\( A' = 35 \)[/tex]
[tex]\( \text{Mean Deviation} = -0.2 \)[/tex]
Let [tex]\( C \)[/tex] be the class interval.
Substituting the known values into the equation:
[tex]\[ 33 = 35 + (-0.2 \times C) \][/tex]
6. Solve for [tex]\( C \)[/tex] (Class Interval):
[tex]\[ 33 = 35 - 0.2C \][/tex]
Subtract 35 from both sides:
[tex]\[ 33 - 35 = -0.2C \][/tex]
[tex]\[ -2 = -0.2C \][/tex]
Divide both sides by -0.2:
[tex]\[ C = \frac{-2}{-0.2} = 10 \][/tex]
So, the class interval [tex]\(C\)[/tex] is 10.
1. Understanding the Given Data:
- Arithmetic mean (A) of the distribution is 33.
- Assumed mean (A') is 35.
- Step deviations (d) are: -3, -2, -1, 0, 1, 2.
- Frequency (f) corresponding to these step deviations are: 5, 10, 25, 30, 20, 10.
2. Calculate Total Number of Observations (N):
The total number of observations is the sum of all frequencies.
[tex]\[ N = 5 + 10 + 25 + 30 + 20 + 10 = 100 \][/tex]
3. Calculate Summation of Step Deviations Times Frequencies (Σf_i d_i):
We need to calculate the product of each step deviation with its corresponding frequency, and then sum these products.
[tex]\[ \Sigma f_i d_i = (5 \times -3) + (10 \times -2) + (25 \times -1) + (30 \times 0) + (20 \times 1) + (10 \times 2) \][/tex]
Breaking it down:
[tex]\[ \Sigma f_i d_i = (5 \times -3) + (10 \times -2) + (25 \times -1) + (30 \times 0) + (20 \times 1) + (10 \times 2) \][/tex]
[tex]\[ \Sigma f_i d_i = -15 + (-20) + (-25) + 0 + 20 + 20 = -20 \][/tex]
4. Calculate Mean Deviation:
The mean deviation is given by the formula:
[tex]\[ \text{Mean Deviation} = \frac{\Sigma f_i d_i}{N} \][/tex]
Substituting the values:
[tex]\[ \text{Mean Deviation} = \frac{-20}{100} = -0.2 \][/tex]
5. Relating Arithmetic Mean, Assumed Mean, and Mean Deviation:
The arithmetic mean can be related to the assumed mean through the following formula:
[tex]\[ \text{Arithmetic Mean} (A) = \text{Assumed Mean} (A') + (\text{Mean Deviation} \times \text{Class Interval}) \][/tex]
We know:
[tex]\( A = 33 \)[/tex]
[tex]\( A' = 35 \)[/tex]
[tex]\( \text{Mean Deviation} = -0.2 \)[/tex]
Let [tex]\( C \)[/tex] be the class interval.
Substituting the known values into the equation:
[tex]\[ 33 = 35 + (-0.2 \times C) \][/tex]
6. Solve for [tex]\( C \)[/tex] (Class Interval):
[tex]\[ 33 = 35 - 0.2C \][/tex]
Subtract 35 from both sides:
[tex]\[ 33 - 35 = -0.2C \][/tex]
[tex]\[ -2 = -0.2C \][/tex]
Divide both sides by -0.2:
[tex]\[ C = \frac{-2}{-0.2} = 10 \][/tex]
So, the class interval [tex]\(C\)[/tex] is 10.
