Answer :
Certainly! Let's go through the step-by-step solution to calculate the mean, standard deviation, and coefficient of variation for the given data:
### Step 1: Data Interpretation
We are given the class intervals and the corresponding number of donors (frequencies). The midpoint values (masses) for each class interval are calculated as follows:
- Class Interval: \(60-62\):
- Midpoint: \(\frac{60 + 62}{2} = 61\)
- Class Interval: \(62-65\):
- Midpoint: \(\frac{62 + 65}{2} = 63.5\)
- Class Interval: \(69-71\):
- Midpoint: \(\frac{69 + 71}{2} = 70\)
- Class Interval: \(72-76\):
- Midpoint: \(\frac{72 + 76}{2} = 74\)
### Step 2: Given Data Summary
- Masses (midpoints): \([61, 63.5, 70, 74]\)
- Frequencies: \([5, 18, 43, 8]\)
- Assumed Mean (\(A\)): \(67 \, \text{kg}\)
### Step 3: Mean Calculation
1. Calculate the Deviations:
[tex]\[ d_i = x_i - A \][/tex]
Here, \(x_i\) are the midpoints, and \(A\) is the assumed mean.
- For \(61 - 67 = -6\)
- For \(63.5 - 67 = -3.5\)
- For \(70 - 67 = 3\)
- For \(74 - 67 = 7\)
Deviations: \([-6, -3.5, 3, 7]\)
2. Weighted Deviations:
[tex]\[ f_id_i = f_i \times d_i \][/tex]
Multiply each deviation by its corresponding frequency.
- \(5 \times -6 = -30\)
- \(18 \times -3.5 = -63\)
- \(43 \times 3 = 129\)
- \(8 \times 7 = 56\)
Weighted Deviations: \([-30, -63, 129, 56]\)
3. Sum of Weighted Deviations:
[tex]\[ \sum (f_id_i) = -30 - 63 + 129 + 56 = 92 \][/tex]
4. Sum of Frequencies:
[tex]\[ \sum f_i = 5 + 18 + 43 + 8 = 74 \][/tex]
5. Apply the Formula for Mean:
[tex]\[ \bar{x} = A + \frac{\sum (f_id_i)}{\sum f_i} = 67 + \frac{92}{74} = 68.24 \, \text{kg \, (approx)} \][/tex]
### Step 4: Standard Deviation Calculation
1. Square of the Deviations:
[tex]\[ d_i^2 \][/tex]
- \((-6)^2 = 36\)
- \((-3.5)^2 = 12.25\)
- \(3^2 = 9\)
- \(7^2 = 49\)
Squared Deviations: \([36, 12.25, 9, 49]\)
2. Weighted Squared Deviations:
[tex]\[ f_i \times d_i^2 \][/tex]
- \(5 \times 36 = 180\)
- \(18 \times 12.25 = 220.5\)
- \(43 \times 9 = 387\)
- \(8 \times 49 = 392\)
Weighted Squared Deviations: \([180, 220.5, 387, 392]\)
3. Sum of Weighted Squared Deviations:
[tex]\[ \sum(f_i d_i^2) = 180 + 220.5 + 387 + 392 = 1179.5 \][/tex]
4. Variance:
[tex]\[ \sigma^2 = \frac{\sum(f_i d_i^2)}{\sum f_i} = \frac{1179.5}{74} = 15.937 \, (\text{approx}) \][/tex]
5. Standard Deviation:
[tex]\[ \sigma = \sqrt{15.937} = 3.99 \, \text{kg \, (approx)} \][/tex]
### Step 5: Coefficient of Variation
1. Coefficient of Variation:
[tex]\[ \text{CV} = \left(\frac{\sigma}{\bar{x}}\right) \times 100 \][/tex]
[tex]\[ \text{CV} = \left(\frac{3.99}{68.24}\right) \times 100 = 5.85\% \, (\text{approx}) \][/tex]
### Final Summarized Results
- Mean (\(\bar{x}\)): \(68.24 \, \text{kg\ (approx)}\)
- Standard Deviation (\(\sigma\)): \(3.99 \, \text{kg\ (approx)}\)
- Coefficient of Variation (CV): \(5.85\% \, \text{(approx)}\)
Thus, the mean is approximately [tex]\(68.24 \, \text{kg}\)[/tex], the standard deviation is approximately [tex]\(3.99 \, \text{kg}\)[/tex], and the coefficient of variation is approximately [tex]\(5.85\%\)[/tex].
### Step 1: Data Interpretation
We are given the class intervals and the corresponding number of donors (frequencies). The midpoint values (masses) for each class interval are calculated as follows:
- Class Interval: \(60-62\):
- Midpoint: \(\frac{60 + 62}{2} = 61\)
- Class Interval: \(62-65\):
- Midpoint: \(\frac{62 + 65}{2} = 63.5\)
- Class Interval: \(69-71\):
- Midpoint: \(\frac{69 + 71}{2} = 70\)
- Class Interval: \(72-76\):
- Midpoint: \(\frac{72 + 76}{2} = 74\)
### Step 2: Given Data Summary
- Masses (midpoints): \([61, 63.5, 70, 74]\)
- Frequencies: \([5, 18, 43, 8]\)
- Assumed Mean (\(A\)): \(67 \, \text{kg}\)
### Step 3: Mean Calculation
1. Calculate the Deviations:
[tex]\[ d_i = x_i - A \][/tex]
Here, \(x_i\) are the midpoints, and \(A\) is the assumed mean.
- For \(61 - 67 = -6\)
- For \(63.5 - 67 = -3.5\)
- For \(70 - 67 = 3\)
- For \(74 - 67 = 7\)
Deviations: \([-6, -3.5, 3, 7]\)
2. Weighted Deviations:
[tex]\[ f_id_i = f_i \times d_i \][/tex]
Multiply each deviation by its corresponding frequency.
- \(5 \times -6 = -30\)
- \(18 \times -3.5 = -63\)
- \(43 \times 3 = 129\)
- \(8 \times 7 = 56\)
Weighted Deviations: \([-30, -63, 129, 56]\)
3. Sum of Weighted Deviations:
[tex]\[ \sum (f_id_i) = -30 - 63 + 129 + 56 = 92 \][/tex]
4. Sum of Frequencies:
[tex]\[ \sum f_i = 5 + 18 + 43 + 8 = 74 \][/tex]
5. Apply the Formula for Mean:
[tex]\[ \bar{x} = A + \frac{\sum (f_id_i)}{\sum f_i} = 67 + \frac{92}{74} = 68.24 \, \text{kg \, (approx)} \][/tex]
### Step 4: Standard Deviation Calculation
1. Square of the Deviations:
[tex]\[ d_i^2 \][/tex]
- \((-6)^2 = 36\)
- \((-3.5)^2 = 12.25\)
- \(3^2 = 9\)
- \(7^2 = 49\)
Squared Deviations: \([36, 12.25, 9, 49]\)
2. Weighted Squared Deviations:
[tex]\[ f_i \times d_i^2 \][/tex]
- \(5 \times 36 = 180\)
- \(18 \times 12.25 = 220.5\)
- \(43 \times 9 = 387\)
- \(8 \times 49 = 392\)
Weighted Squared Deviations: \([180, 220.5, 387, 392]\)
3. Sum of Weighted Squared Deviations:
[tex]\[ \sum(f_i d_i^2) = 180 + 220.5 + 387 + 392 = 1179.5 \][/tex]
4. Variance:
[tex]\[ \sigma^2 = \frac{\sum(f_i d_i^2)}{\sum f_i} = \frac{1179.5}{74} = 15.937 \, (\text{approx}) \][/tex]
5. Standard Deviation:
[tex]\[ \sigma = \sqrt{15.937} = 3.99 \, \text{kg \, (approx)} \][/tex]
### Step 5: Coefficient of Variation
1. Coefficient of Variation:
[tex]\[ \text{CV} = \left(\frac{\sigma}{\bar{x}}\right) \times 100 \][/tex]
[tex]\[ \text{CV} = \left(\frac{3.99}{68.24}\right) \times 100 = 5.85\% \, (\text{approx}) \][/tex]
### Final Summarized Results
- Mean (\(\bar{x}\)): \(68.24 \, \text{kg\ (approx)}\)
- Standard Deviation (\(\sigma\)): \(3.99 \, \text{kg\ (approx)}\)
- Coefficient of Variation (CV): \(5.85\% \, \text{(approx)}\)
Thus, the mean is approximately [tex]\(68.24 \, \text{kg}\)[/tex], the standard deviation is approximately [tex]\(3.99 \, \text{kg}\)[/tex], and the coefficient of variation is approximately [tex]\(5.85\%\)[/tex].
