# An insurance company crashed four cars in succession at 5 miles per hour. The cost of repair for each of the four crashes was​$423​, ​$468​, ​$403, ​$222 . Compute the​ range, sample​variance, and sample standard deviation cost of repair.

An insurance company crashed four cars in succession at 5 miles per hour. The cost of repair for each of the four crashes was​$423​, ​$468​, ​$403, ​$222 . Compute the​ range, sample​variance, and sample standard deviation cost of repair.
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Step 1
The range of a data set is the highest data value minus the lowest data value. Here, the range of the cost of repair is $468-222=246$.
The mean of the given data set for the cost of repair is $\frac{423+468+403+222}{4}=1516/4=379$.
If there are n data points, then the Variance = sum of the squares of (Data points - mean) divided by $\left(n-1\right)$.
Here, the difference of the data points from the mean are . Hence the variance is $\left[442+892+242+\left(-157\right)2\right]/3=\left(1936+7921+576+24649\right)/3=35082/3=11694$.
Step 2
The standard deviation is the square root of the average squared deviation from the mean.
Here, the standard deviation is $\sqrt{11694}=108.14$ ( on rounding off to 2 decimal places).