To compute the sample variance ([tex]\(s^2\)[/tex]) and sample standard deviation ([tex]\(s\)[/tex]) cost of repair, given the repair costs for the four crashes: \[tex]$413, \$[/tex]460, \[tex]$413, and \$[/tex]212, we follow these steps:
### 1. Compute the Mean (Average) Cost:
The mean cost is the sum of all costs divided by the number of data points (crashes).
[tex]\[
\bar{x} = \frac{413 + 460 + 413 + 212}{4} = \frac{1498}{4} = 374.5 \text{ dollars}
\][/tex]
### 2. Compute the Sample Variance:
The sample variance ([tex]\(s^2\)[/tex]) is calculated using the formula:
[tex]\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}
\][/tex]
Where [tex]\(x_i\)[/tex] is each individual cost, [tex]\(\bar{x}\)[/tex] is the mean cost, and [tex]\(n\)[/tex] is the number of data points.
First, compute each squared deviation from the mean:
[tex]\[
(413 - 374.5)^2 = (38.5)^2 = 1482.25
\][/tex]
[tex]\[
(460 - 374.5)^2 = (85.5)^2 = 7310.25
\][/tex]
[tex]\[
(413 - 374.5)^2 = (38.5)^2 = 1482.25
\][/tex]
[tex]\[
(212 - 374.5)^2 = (-162.5)^2 = 26406.25
\][/tex]
Now, sum these squared deviations:
[tex]\[
1482.25 + 7310.25 + 1482.25 + 26406.25 = 36681
\][/tex]
Then divide by [tex]\(n-1\)[/tex] (which is [tex]\(4-1 = 3\)[/tex]):
[tex]\[
s^2 = \frac{36681}{3} = 12227 \text{ dollars}^2
\][/tex]
### 3. Compute the Sample Standard Deviation:
The sample standard deviation ([tex]\(s\)[/tex]) is the square root of the sample variance.
[tex]\[
s = \sqrt{12227} \approx 110.57 \approx 111 \text{ dollars}
\][/tex]
Summarizing, we have the following values:
- Range: [tex]$248
- Sample Variance (\(s^2\)): \(12227 \text{ dollars}^2\)
- Sample Standard Deviation (\(s\)): \(111 \text{ dollars}\)
So, the sample variance is \( \$[/tex]12227 \text{ dollars}^2 \).
