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An insurance company crashed four cars in succession at 5 miles per hour. The cost of...

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.

Math   Prealgebra   
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Answer #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 ( 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 (n - 1).

Here, the difference of the data points from the mean are (423-379) = 44, ( 468-379) = 89, (403-379) = 24 and (222-379) = - 157. Hence the variance is [ 442+892+242+(-157)2]/3 = (1936+7921+576+24649)/3 = 35082/3 = 11694.

The standard deviation is the square root of the average squared deviation from the mean.

Here, the standard deviation is √ 11694 = 108.14 ( on rounding off to 2 decimal places).

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