an insurance company crashed four cars in succession at 5 miles per hour. the cost of repair for each of the four crashes was $417, $462, $402, $229. compute the range, sample variance, and sample standard deviation cost of repair

Let's go step by step to compute the range, sample variance, and sample standard deviation for the given costs of repair.

### 1. **Range**
The range is the difference between the largest and smallest value in the dataset. The costs of repair are:

\[ 417, 462, 402, 229 \]

- The largest value is $ 462 $.
- The smallest value is $ 229 $.

So, the range is:

\[
\text{Range} = 462 - 229 = 233
\]

### 2. **Sample Variance**
The formula for the sample variance is:

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

Where:
- $ n $ is the number of observations (4 in this case),
- $ x_i $ is each individual data point,
- $ \bar{x} $ is the sample mean.

#### Step 1: Compute the mean ($\bar{x}$):

\[
\bar{x} = \frac{417 + 462 + 402 + 229}{4} = \frac{1510}{4} = 377.5
\]

#### Step 2: Compute the squared differences from the mean:

\[
(417 - 377.5)^2 = (39.5)^2 = 1560.25
\]
\[
(462 - 377.5)^2 = (84.5)^2 = 7140.25
\]
\[
(402 - 377.5)^2 = (24.5)^2 = 600.25
\]
\[
(229 - 377.5)^2 = (-148.5)^2 = 22052.25
\]

#### Step 3: Sum the squared differences:

\[
1560.25 + 7140.25 + 600.25 + 22052.25 = 31353
\]

#### Step 4: Divide by $ n - 1 = 3 $:

\[
s^2 = \frac{31353}{3} = 10451
\]

So, the sample variance is:

\[
s^2 = 10451
\]

### 3. **Sample Standard Deviation**
The sample standard deviation is the square root of the variance:

\[
s = \sqrt{10451} \approx 102.22
\]

### Summary of results:
- **Range**: 233
- **Sample Variance**: 10451
- **Sample Standard Deviation**: 102.22

Would you like any details or clarifications?

Here are five related questions:
1. How does the range help in understanding the spread of data?
2. What is the significance of using sample variance instead of population variance?
3. Can you explain how the standard deviation interprets data spread?
4. How would results change if we had more cars and crashes involved?
5. How do outliers like the cost of $229 affect the sample variance?

**Tip:** The standard deviation gives a sense of how much data varies from the mean, making it a key measure of spread.

An insurance company crashed four cars in succession at 5 miles per hour. The cost of repair for each of the four crashes was $417, $462, $402, $229. Compute the range, sample variance, and sample standard deviation cost of repair.

Learn how to calculate the range, sample variance, and sample standard deviation for car crash repair costs of $417, $462, $402, and $229. This detailed guide covers key statistics concepts and formulas, ideal for students in Grades 9-12.

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