An insurance company crashed four cars in succession at 5 miles per hour. The cost of repair for each of the four crashes was ​$411​, ​$466, ​$40​, ​$223 . Compute the​ range, sample​ variance, and sample standard deviation cost of repair.

To solve this problem, we need to compute the **range**, **sample variance**, and **sample standard deviation** for the cost of repair for the four crashes. The costs are:

- $411
- $466
- $40
- $223

### Step 1: Calculate the Range
The range is the difference between the maximum and minimum values.

\[
\text{Range} = \text{Max} - \text{Min} = 466 - 40 = 426
\]

### Step 2: Calculate the Sample Variance

The sample variance formula is:

\[
s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2
\]

Where:
- $ n = 4 $ (the number of data points),
- $ x_i $ are the individual data points,
- $ \bar{x} $ is the sample mean.

#### Step 2.1: Compute the Sample Mean ($\bar{x}$)

\[
\bar{x} = \frac{411 + 466 + 40 + 223}{4} = \frac{1140}{4} = 285
\]

#### Step 2.2: Compute the Squared Differences from the Mean

Now, calculate the squared differences of each data point from the mean:

- $ (411 - 285)^2 = (126)^2 = 15876 $
- $ (466 - 285)^2 = (181)^2 = 32761 $
- $ (40 - 285)^2 = (-245)^2 = 60025 $
- $ (223 - 285)^2 = (-62)^2 = 3844 $

#### Step 2.3: Sum of Squared Differences

\[
\sum (x_i - \bar{x})^2 = 15876 + 32761 + 60025 + 3844 = 112506
\]

#### Step 2.4: Calculate the Sample Variance

\[
s^2 = \frac{112506}{4-1} = \frac{112506}{3} = 37502
\]

### Step 3: Calculate the Sample Standard Deviation

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

\[
s = \sqrt{37502} \approx 193.65
\]

### Summary of Results:

- **Range**: 426
- **Sample Variance**: 37502
- **Sample Standard Deviation**: 193.65

Would you like more details or have any questions?

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Here are 5 related questions you might consider:

1. How is sample variance different from population variance?
2. What is the effect of outliers on the range and standard deviation?
3. How would the calculations change if there were more data points?
4. Can the standard deviation ever be negative?
5. Why do we divide by $n-1$ instead of $n$ for sample variance?

**Tip:** The standard deviation provides insight into how much the data varies from the mean; a smaller standard deviation indicates that data points are closer to the mean.

Learn how to compute the range, sample variance, and standard deviation for the cost of car repairs in a sequence of crashes. The costs analyzed are $411, $466, $40, and $223. Suitable for students in Grades 10-12, this problem covers key statistical concepts and provides a detailed solution.

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