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 $216.
$s^{2} = square$ dollars$^{2}$
(round to the nearest whole number as needed.)

Explanation:

Step1: Calculate sample mean

First, find the average of the repair costs.
$\bar{x} = \frac{419 + 454 + 409 + 238}{4} = \frac{1520}{4} = 380$

Step2: Compute squared deviations

Find each cost's squared difference from the mean.
$(419-380)^2 = 39^2 = 1521$
$(454-380)^2 = 74^2 = 5476$
$(409-380)^2 = 29^2 = 841$
$(238-380)^2 = (-142)^2 = 20164$

Step3: Sum squared deviations

Add all the squared deviations together.
$1521 + 5476 + 841 + 20164 = 28002$

Step4: Calculate sample variance

Divide the sum by $n-1$ (where $n=4$).
$s^2 = \frac{28002}{4-1} = \frac{28002}{3} = 9334$

Answer:

The range is $\$216$
$s^2 = 9334$ dollars$^2$