Answer :
To find the standard deviation of the given distribution, follow these steps:
### Step 1: Determine the class midpoints
First, find the midpoint of each class interval. The midpoint (or class mark) is calculated as the average of the lower and upper boundaries of each interval.
For the intervals and calculations:
- For [tex]\(3-5\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{3 + 5}{2} = 4.0 \][/tex]
- For [tex]\(6-8\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{6 + 8}{2} = 7.0 \][/tex]
- For [tex]\(9-11\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{9 + 11}{2} = 10.0 \][/tex]
Thus, the midpoints are [tex]\(4.0\)[/tex], [tex]\(7.0\)[/tex], and [tex]\(10.0\)[/tex].
### Step 2: Calculate the mean of the distribution
The mean ([tex]\(\bar{x}\)[/tex]) is calculated by summing the product of each midpoint and its corresponding frequency, then dividing by the total frequency.
Given frequencies:
- [tex]\(\text{Frequency of } 3-5 = 2\)[/tex]
- [tex]\(\text{Frequency of } 6-8 = 2\)[/tex]
- [tex]\(\text{Frequency of } 9-11 = 2\)[/tex]
Total Frequency ([tex]\(N\)[/tex]):
[tex]\[ N = 2 + 2 + 2 = 6 \][/tex]
Mean ([tex]\(\bar{x}\)[/tex]):
[tex]\[ \bar{x} = \frac{(4.0 \cdot 2) + (7.0 \cdot 2) + (10.0 \cdot 2)}{6} = \frac{8 + 14 + 20}{6} = \frac{42}{6} = 7.0 \][/tex]
### Step 3: Calculate the variance
Variance ([tex]\(\sigma^2\)[/tex]) is calculated by summing the squared differences between each midpoint and the mean, then dividing by the total frequency.
[tex]\[ \sigma^2 = \frac{\sum_{i=1}^{n} f_i \cdot (x_i - \bar{x})^2}{N} \][/tex]
Where [tex]\(f_i\)[/tex] and [tex]\(x_i\)[/tex] represent the frequency and midpoint of the [tex]\(i\)[/tex]-th interval.
Calculations for each interval:
- For [tex]\(3-5\)[/tex]:
[tex]\[ (4.0 - 7.0)^2 = (-3.0)^2 = 9.0 \][/tex]
Contribution to variance from this interval:
[tex]\[ 2 \cdot 9.0 = 18.0 \][/tex]
- For [tex]\(6-8\)[/tex]:
[tex]\[ (7.0 - 7.0)^2 = 0^2 = 0.0 \][/tex]
Contribution to variance from this interval:
[tex]\[ 2 \cdot 0.0 = 0.0 \][/tex]
- For [tex]\(9-11\)[/tex]:
[tex]\[ (10.0 - 7.0)^2 = 3.0^2 = 9.0 \][/tex]
Contribution to variance from this interval:
[tex]\[ 2 \cdot 9.0 = 18.0 \][/tex]
Summing these contributions:
[tex]\[ \sigma^2 = \frac{18.0 + 0.0 + 18.0}{6} = \frac{36.0}{6} = 6.0 \][/tex]
### Step 4: Calculate the standard deviation
Standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance.
[tex]\[ \sigma = \sqrt{6.0} \approx 2.45 \][/tex]
Thus, the standard deviation of the distribution is approximately 2.45.
### Step 1: Determine the class midpoints
First, find the midpoint of each class interval. The midpoint (or class mark) is calculated as the average of the lower and upper boundaries of each interval.
For the intervals and calculations:
- For [tex]\(3-5\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{3 + 5}{2} = 4.0 \][/tex]
- For [tex]\(6-8\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{6 + 8}{2} = 7.0 \][/tex]
- For [tex]\(9-11\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{9 + 11}{2} = 10.0 \][/tex]
Thus, the midpoints are [tex]\(4.0\)[/tex], [tex]\(7.0\)[/tex], and [tex]\(10.0\)[/tex].
### Step 2: Calculate the mean of the distribution
The mean ([tex]\(\bar{x}\)[/tex]) is calculated by summing the product of each midpoint and its corresponding frequency, then dividing by the total frequency.
Given frequencies:
- [tex]\(\text{Frequency of } 3-5 = 2\)[/tex]
- [tex]\(\text{Frequency of } 6-8 = 2\)[/tex]
- [tex]\(\text{Frequency of } 9-11 = 2\)[/tex]
Total Frequency ([tex]\(N\)[/tex]):
[tex]\[ N = 2 + 2 + 2 = 6 \][/tex]
Mean ([tex]\(\bar{x}\)[/tex]):
[tex]\[ \bar{x} = \frac{(4.0 \cdot 2) + (7.0 \cdot 2) + (10.0 \cdot 2)}{6} = \frac{8 + 14 + 20}{6} = \frac{42}{6} = 7.0 \][/tex]
### Step 3: Calculate the variance
Variance ([tex]\(\sigma^2\)[/tex]) is calculated by summing the squared differences between each midpoint and the mean, then dividing by the total frequency.
[tex]\[ \sigma^2 = \frac{\sum_{i=1}^{n} f_i \cdot (x_i - \bar{x})^2}{N} \][/tex]
Where [tex]\(f_i\)[/tex] and [tex]\(x_i\)[/tex] represent the frequency and midpoint of the [tex]\(i\)[/tex]-th interval.
Calculations for each interval:
- For [tex]\(3-5\)[/tex]:
[tex]\[ (4.0 - 7.0)^2 = (-3.0)^2 = 9.0 \][/tex]
Contribution to variance from this interval:
[tex]\[ 2 \cdot 9.0 = 18.0 \][/tex]
- For [tex]\(6-8\)[/tex]:
[tex]\[ (7.0 - 7.0)^2 = 0^2 = 0.0 \][/tex]
Contribution to variance from this interval:
[tex]\[ 2 \cdot 0.0 = 0.0 \][/tex]
- For [tex]\(9-11\)[/tex]:
[tex]\[ (10.0 - 7.0)^2 = 3.0^2 = 9.0 \][/tex]
Contribution to variance from this interval:
[tex]\[ 2 \cdot 9.0 = 18.0 \][/tex]
Summing these contributions:
[tex]\[ \sigma^2 = \frac{18.0 + 0.0 + 18.0}{6} = \frac{36.0}{6} = 6.0 \][/tex]
### Step 4: Calculate the standard deviation
Standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance.
[tex]\[ \sigma = \sqrt{6.0} \approx 2.45 \][/tex]
Thus, the standard deviation of the distribution is approximately 2.45.
