Let's calculate the range, sample variance, and sample standard deviation for the given costs of repair.

The costs of repair for the four crashes are:

• $428
• $462
• $402
• $240

1. Range

The range is the difference between the maximum and minimum values in a data set.

• Maximum value: $462
• Minimum value: $240

$\text{Range} = \text{Maximum} - \text{Minimum} = 462 - 240 = 222$

Thus, the range is $222.

2. Sample Variance

The sample variance is calculated using the formula:

$s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2$ where:

• $n$ is the number of data points (4 in this case).
• $x_i$ is each data point.
• $\bar{x}$ is the sample mean.

Step 1: Find the mean ($\bar{x}$) of the data.

$\bar{x} = \frac{428 + 462 + 402 + 240}{4} = \frac{1532}{4} = 383$

Step 2: Find the squared differences between each data point and the mean.

• For $428: (428 - 383)^2 = 45^2 = 2025$
• For $462: (462 - 383)^2 = 79^2 = 6241$
• For $402: (402 - 383)^2 = 19^2 = 361$
• For $240: (240 - 383)^2 = 143^2 = 20449$

Step 3: Sum these squared differences.

$\text{Sum of squared differences} = 2025 + 6241 + 361 + 20449 = 28976$

Step 4: Calculate the sample variance.

$s^2 = \frac{28976}{4 - 1} = \frac{28976}{3} = 9660.67$

Thus, the sample variance is $9660.67.

3. Sample Standard Deviation

The sample standard deviation is the square root of the sample variance:

$s = \sqrt{9660.67} \approx 98.31$

Thus, the sample standard deviation is $98.31.

Final Answers:

• Range: $222
• Sample Variance: $9660.67
• Sample Standard Deviation: $98.31

Would you like further details or explanations on any of these calculations?

Here are some related questions for you:

1. How do you compute the population variance versus the sample variance?
2. What are some methods to interpret the meaning of the standard deviation in real-life situations?
3. How does the range compare to other measures of dispersion such as the variance and standard deviation?
4. How would you calculate the cost of repair for an additional crash in this scenario?
5. How can you interpret the spread of the costs using the variance and standard deviation?

Tip: Remember that the standard deviation is more sensitive to extreme values in the dataset, which is why it's often preferred for understanding overall data spread!