Question

an insurance company crashed four cars in succession at 5 miles per hour. the cost of repair for each of the four crashes was $419, $454, $409, $238. compute the range, sample variance, and sample standard deviation cost of repair.

the range is $□.

Explanation:

Step1: Find range (max - min)

First, identify the maximum value (\$454) and minimum value (\$238) from the data set: \$419, \$454, \$409, \$238.
Expression: $454 - 238$

Step2: Calculate sample mean

Sum all values, divide by sample size $n=4$.
Expression: $\bar{x} = \frac{419 + 454 + 409 + 238}{4} = \frac{1520}{4} = 380$

Step3: Compute sample variance

Use sample variance formula $s^2 = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}$.
First calculate squared deviations:
$(419-380)^2 = 39^2 = 1521$
$(454-380)^2 = 74^2 = 5476$
$(409-380)^2 = 29^2 = 841$
$(238-380)^2 = (-142)^2 = 20164$
Sum of squared deviations: $1521 + 5476 + 841 + 20164 = 28002$
Variance: $\frac{28002}{4-1} = \frac{28002}{3}$

Step4: Compute sample standard deviation

Take square root of sample variance.
Expression: $s = \sqrt{\frac{28002}{3}}$

Answer:

Range: $\$216$
Sample Variance: $9334$
Sample Standard Deviation: $\approx \$96.61$