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 ​$419419​, ​$452452​, ​$404404​, ​$221221 . Compute the​ range, sample​ variance, and sample standard deviation cost of repair

Answer 1

To find the range, subtract the lowest value from the highest value. To find the sample variance, find the average of the squared differences between each value and the mean. To find the sample standard deviation, take the square root of the sample variance.

Step-by-step explanation:

To compute the range, we need to find the difference between the highest and lowest values in the sample. The highest value is $452,452 and the lowest value is $221,221, so the range is $452,452 - $221,221 = $231,231.

To compute the sample variance, we need to find the average of the squared differences between each value and the mean. The mean of the sample is ($419,419 + $452,452 + $404,404 + $221,221) / 4 = $374,624.75. The squared differences are ($419,419 - $374,624.75)^2, ($452,452 - $374,624.75)^2, ($404,404 - $374,624.75)^2, and ($221,221 - $374,624.75)^2. Summing up these squared differences and dividing by (n-1), where n is the sample size, we get the sample variance.

To compute the sample standard deviation, we take the square root of the sample variance.

Answer 2

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

Sample Variance:

`s^2= (1)/(n-1) (x-\bar{x})^2`

where n = 4,

`\bar{x}= (1)/(4)( 419419 + 452452+ 404404 + 221221)=374374`
Hence

`s^2= (1)/(4-1)[( 419419 - 374374)^2 + (452452-374374)^2+\\ (404404-374374)^2+ (221221-374374)^2)]\\ = 1.0828 * 10^(10)`
Then standard deviation =
`s=\sqrt{1.0828 *10^(10)}=104055.864`