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An insurance company crashed four cars in succession at 5 miles per hour. The cost of repair for each of the four crashes was ​$419419​, ​$452452​, ​$404404​, ​$221221 . Compute the​ range, sample​ variance, and sample standard deviation cost of repair

Respuesta :

Note that Range = Max – Min.
In this case, Max = 452452, Min = 221221.
Hence, Range = 452452 – 221221 = 228230.

Sample Variance:
[tex]s^2= \frac{1}{n-1} (x-\bar{x})^2[/tex]

where n = 4,
[tex] \bar{x}= \frac{1}{4}( 419419 + 452452+ 404404 + 221221)=374374[/tex]
Hence
[tex] s^2= \frac{1}{4-1}[( 419419 - 374374)^2 + (452452-374374)^2+\\ (404404-374374)^2+ (221221-374374)^2)]\\ = 1.0828 \times 10^{10} [/tex]
Then standard deviation = [tex]s=\sqrt{1.0828 \times10^{10}}=104055.864[/tex]

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