The following table presents the time taken to review articles that were submitted for publication to a particular journal during a recent year. A few articles took longer than 6 months to review; these are omitted from the table. Consider the data to be a population.

To calculate the variance and standard deviation of the given data, we can utilize the method of grouped data.

Step 1: Determine Midpoints (x)

First, we need to calculate the midpoint of each time interval. The midpoint is calculated as the average of the lower and upper bounds of each interval.

| Time Interval | Midpoint (x) | Frequency (f) | |---------------|--------------|---------------| | 0.0 - 0.9 | \(0.45\) | \(44\) | | 1.0 - 1.9 | \(1.45\) | \(16\) | | 2.0 - 2.9 | \(2.45\) | \(17\) | | 3.0 - 3.9 | \(3.45\) | \(20\) | | 4.0 - 4.9 | \(4.45\) | \(12\) | | 5.0 - 5.9 | \(5.45\) | \(15\) |

Step 2: Calculate the Mean (μ)

The mean (\(\mu\)) of the population can be calculated using the formula:

\[ \mu = \frac{\sum f \cdot x}{\sum f} \]

Calculating \(\sum f \cdot x\) and \(\sum f\):

• \( \sum f \cdot x = (44 \times 0.45) + (16 \times 1.45) + (17 \times 2.45) + (20 \times 3.45) + (12 \times 4.45) + (15 \times 5.45) \)
• \( \sum f = 44 + 16 + 17 + 20 + 12 + 15 = 124 \)

Calculating \( \sum f \cdot x \):

\[ \sum f \cdot x = 19.8 + 23.2 + 41.65 + 69 + 53.4 + 81.75 = 288.9 \]

Now, calculate the mean:

\[ \mu = \frac{288.9}{124} \approx 2.33 \]

Step 3: Calculate the Variance (σ²)

Using the formula for variance:

\[ \sigma^2 = \frac{\sum f \cdot (x - \mu)^2}{\sum f} \]

First, compute \( (x - \mu)^2 \) for each midpoint:

• For \(0.45\): \( (0.45 - 2.33)^2 = 3.5049 \)
• For \(1.45\): \( (1.45 - 2.33)^2 = 0.7744 \)
• For \(2.45\): \( (2.45 - 2.33)^2 = 0.0144 \)
• For \(3.45\): \( (3.45 - 2.33)^2 = 1.2549 \)
• For \(4.45\): \( (4.45 - 2.33)^2 = 4.5296 \)
• For \(5.45\): \( (5.45 - 2.33)^2 = 9.1849 \)

Now we multiply by frequency (\(f\)) and sum this:

\[ \sum f \cdot (x - \mu)^2 = (44 \cdot 3.5049) + (16 \cdot 0.7744) + (17 \cdot 0.0144) + (20 \cdot 1.2549) + (12 \cdot 4.5296) + (15 \cdot 9.1849 \]

Calculating:

\[ = 154.2206 + 12.3904 + 0.2448 + 25.098 + 54.3552 + 137.7235 = 383.0225 \]

Now calculate the variance:

\[ \sigma^2 = \frac{383.0225}{124} \approx 3.08 \]

Step 4: Calculate the Standard Deviation (σ)

The standard deviation can be calculated as:

\[ \sigma = \sqrt{\sigma^2} \]

Calculating:

\[ \sigma \approx \sqrt{3.08} \approx 1.75 \]

Final Answers

(a) The variance is 3.08.

(b) The standard deviation is 1.75.

Let's revisit the calculations for the variance step-by-step to ensure we get the correct result.

Step 1: Calculate the Mean (μ)

From the previous response, we found:

• \(\sum f = 124\)
• \(\sum f \cdot x = 288.9\)

The mean (\(\mu\)) was calculated as:

\[ \mu = \frac{288.9}{124} \approx 2.33 \]

Step 2: Calculate \( (x - \mu)^2 \) values

The midpoints \(x\) and corresponding \( (x - \mu)^2 \) calculations are as follows:

1. For 0.45: \[ (0.45 - 2.33)^2 \approx (1.88)^2 = 3.5344 \]
2. For 1.45: \[ (1.45 - 2.33)^2 \approx (0.88)^2 = 0.7744 \]
3. For 2.45: \[ (2.45 - 2.33)^2 \approx (0.12)^2 = 0.0144 \]
4. For 3.45: \[ (3.45 - 2.33)^2 \approx (1.12)^2 = 1.2544 \]
5. For 4.45: \[ (4.45 - 2.33)^2 \approx (2.12)^2 = 4.4944 \]
6. For 5.45: \[ (5.45 - 2.33)^2 \approx (3.12)^2 = 9.7344 \]

Step 3: Compute \(f \cdot (x - \mu)^2\)

Now let's compute \(f \cdot (x - \mu)^2\):

1. For 0.45: \[ 44 \cdot 3.5344 \approx 155.8016 \]
2. For 1.45: \[ 16 \cdot 0.7744 \approx 12.3904 \]
3. For 2.45: \[ 17 \cdot 0.0144 \approx 0.2448 \]
4. For 3.45: \[ 20 \cdot 1.2544 \approx 25.088 \]
5. For 4.45: \[ 12 \cdot 4.4944 \approx 53.9328 \]
6. For 5.45: \[ 15 \cdot 9.7344 \approx 145.016 \]

Step 4: Calculate \(\sum f \cdot (x - \mu)^2\)

\[ \sum f \cdot (x - \mu)^2 \approx 155.8016 + 12.3904 + 0.2448 + 25.088 + 53.9328 + 145.016 \approx 392.474 \]

Step 5: Calculate the Variance (\(\sigma^2\))

Now, substituting back into the variance formula:

\[ \sigma^2 = \frac{\sum f \cdot (x - \mu)^2}{\sum f} = \frac{392.474}{124} \approx 3.17 \]

Step 6: Standard Deviation (\(\sigma\))

Now, calculating the standard deviation:

\[ \sigma = \sqrt{3.17} \approx 1.78 \]

Final Results

(a) The corrected variance is 3.17.

(b) The standard deviation is 1.78.

Thank you for your patience! Please let me know if you need further clarification or assistance!