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, samplevariance, and sample standard deviation cost of repair.

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​$423​, ​$468​, ​$403, ​$222 . Compute the​ range, sample​variance, and sample standard deviation cost of repair.

Bradley Sherman · Accepted Answer

Step 1The 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.Step 2The 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).

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, samplevariance, and sample standard deviation cost of repair.

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