To approximate the variance and standard deviation from the given frequency distribution, we will first calculate the midpoints for each interval, multiply these midpoints by their corresponding frequencies, and then use this information to find the mean and variance.
Step 1: Calculate the midpoints for each interval.
- \(0 - 0.9\) → Midpoint = \(0.45\)
- \(1.0 - 1.9\) → Midpoint = \(1.45\)
- \(2.0 - 2.9\) → Midpoint = \(2.45\)
- \(3.0 - 3.9\) → Midpoint = \(3.45\)
- \(4.0 - 4.9\) → Midpoint = \(4.45\)
- \(5.0 - 5.9\) → Midpoint = \(5.45\)
Step 2: Create a table with midpoints and frequencies.
| Time (months) | Midpoint | Frequency (f) | |---------------|----------|---------------| | 0 - 0.9 | 0.45 | 41 | | 1.0 - 1.9 | 1.45 | 18 | | 2.0 - 2.9 | 2.45 | 14 | | 3.0 - 3.9 | 3.45 | 17 | | 4.0 - 4.9 | 4.45 | 12 | | 5.0 - 5.9 | 5.45 | 11 |
Step 3: Calculate the total number of articles.
\[ N = 41 + 18 + 14 + 17 + 12 + 11 = 113 \]
Step 4: Calculate the mean (\( \mu \)).
The mean can be calculated using the formula:
\[ \mu = \frac{\sum (f \cdot x)}{N} \]
Where \(f\) is the frequency and \(x\) is the midpoint.
\[ \begin{align*} \sum (f \cdot x) &= (41 \cdot 0.45) + (18 \cdot 1.45) + (14 \cdot 2.45) + (17 \cdot 3.45) + (12 \cdot 4.45) + (11 \cdot 5.45) \ &= 18.45 + 26.1 + 34.3 + 58.65 + 53.4 + 59.95 \ &= 250.85 \end{align*} \]
Now calculate the mean:
\[ \mu = \frac{250.85}{113} \approx 2.22 \]
Step 5: Calculate the variance (\( \sigma^2 \)).
The variance can be calculated using:
\[ \sigma^2 = \frac{\sum f(x - \mu)^2}{N} \]
We first need \( (x - \mu)^2 \) for each midpoint:
- For \(0.45\): \( (0.45 - 2.22)^2 = 2.95^2 = 8.7025 \)
- For \(1.45\): \( (1.45 - 2.22)^2 = 0.77^2 = 0.5929 \)
- For \(2.45\): \( (2.45 - 2.22)^2 = 0.23^2 = 0.0529 \)
- For \(3.45\): \( (3.45 - 2.22)^2 = 1.23^2 = 1.5129 \)
- For \(4.45\): \( (4.45 - 2.22)^2 = 2.23^2 = 4.9729 \)
- For \(5.45\): \( (5.45 - 2.22)^2 = 3.23^2 = 10.4289 \)
Now, calculate \( f \cdot (x - \mu)^2 \):
\[ \begin{align*} (41)(8.7025) &= 356.7625 \ (18)(0.5929) &= 10.6502 \ (14)(0.0529) &= 0.7406 \ (17)(1.5129) &= 25.6923 \ (12)(4.9729) &= 59.6758 \ (11)(10.4289) &= 115.7189 \ \end{align*} \]
Now sum these values:
\[ \sum f(x - \mu)^2 = 356.7625 + 10.6502 + 0.7406 + 25.6923 + 59.6758 + 115.7189 = 569.5393 \]
Now calculate the variance:
\[ \sigma^2 = \frac{569.5393}{113} \approx 5.04 \]
Final Results
(a) The approximate variance of the times is 5.04.
(b) The standard deviation \( \sigma \) is the square root of the variance:
\[ \sigma = \sqrt{5.04} \approx 2.24 \]
Summary
The variance is 5.04, and the standard deviation is 2.24.