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.

asigurato7 2022-07-31 Answered
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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Answers (1)

Bradley Sherman
Answered 2022-08-01 Author has 17 answers
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 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 2
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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