Question

A recent study indicates that the annual cost of maintaining and repairing a car in a town in Ontario averages 200 with a variance of 260. A tax of 20% is introduced on all items associated with the maintenance and repair of cars (i.e., everything is made 20% more expensive). Calculate the variance of the annual cost of maintaining and repairing a car after the tax is introduced.

Answer 1

Answer:

374.4

Step-by-step explanation:

All items related to the maintenance are 20% more expensive, it means that each datum is 20% bigger including the average.

The variance its a dispersion measurof the data and its calculated of this way:

$\sigma^{2} =\frac{1}{n} \sum\limits^n_{i=1} (x_{i}-\var{x})^2$

Here n is the number of data, $\var{x}$ is the average and $x_{i}$ represent each datum. The increment in 20% in each parameter can be represented multiplying for 1.2, of this way

$\sigma_{20\%}^{2} =\frac{1}{n} \sum\limits^n_{i=1} (1.2x_{i}-1.2\var{x})^2$

Factorizing the 1.2 we have:

$\sigma_{20\%}^{2} =\frac{1}{n} \sum\limits^n_{i=1} (1.2(x_{i}-\var{x}))^2$

$\sigma_{20\%}^{2} =\frac{1}{n} \sum\limits^n_{i=1}1.2^{2} (x_{i}-\var{x})^2$

$\sigma_{20\%}^{2} =\frac{1.2^{2}}{n} \sum\limits^n_{i=1} (x_{i}-\var{x})^2$

That is:

$1.2^{2}\sigma^{2}=\sigma_{20\%}^{2}$

The new variance is $1.2^{2} \sigma^{2} =1.44*260=374.4$

Answer 2

Using statistical concepts, it is found that the variance of the annual cost of maintaining and repairing a car after the tax is introduced will be of 374.4.

What happens to the variance when the distribution is multiplied by a constant?

The variance is multiplied by the square of the constant.

In this problem, the variance is of 260, and with the tax, the distribution will be multiplied by 1.2, hence:

V = 1.2^2 x 260 = 374.4

The variance of the annual cost of maintaining and repairing a car after the tax is introduced will be of 374.4.

More can be learned about the variance of a distribution at https://brainly.com/question/25639778