Questions: An insurance company crashed four cars in succession at 5 miles per hour. The cost of repair for each of the four crashes was 420, 453, 402, 229. Compute the range, sample variance, and sample standard deviation cost of repair. The range is s^2= dollars (Round to the nearest whole number as needed.) s= (Round to two decimal places as needed.)

Solution Steps

The range of the repair costs is calculated as follows:

\[ \text{Range} = \max(x_i) - \min(x_i) = 453 - 229 = 224 \]

Thus, the range is \( \$224 \).

The mean \( \mu \) of the repair costs is given by:

\[ \mu = \frac{\sum x_i}{n} = \frac{420 + 453 + 402 + 229}{4} = \frac{1504}{4} = 376.0 \]

The sample variance \( s^2 \) is calculated using the formula:

\[ s^2 = \frac{\sum (x_i - \mu)^2}{n-1} \]

Calculating each term:

\[ \begin{align_} (420 - 376)^2 & = 1936 \\ (453 - 376)^2 & = 5929 \\ (402 - 376)^2 & = 676 \\ (229 - 376)^2 & = 21609 \\ \end{align_} \]

Summing these values:

\[ \sum (x_i - \mu)^2 = 1936 + 5929 + 676 + 21609 = 10050 \]

Thus, the sample variance is:

\[ s^2 = \frac{10050}{4 - 1} = \frac{10050}{3} = 3350.0 \]

The sample standard deviation \( s \) is the square root of the sample variance:

\[ s = \sqrt{s^2} = \sqrt{10050.0} \approx 100.25 \]

Final Answer