Answer:
Let X the random variable that represent the miles per gallon in cars of a population, and for this case we know the distribution for X is given by:
Where
and
We can calculate the coeffcient of variation for this cae like this:
![CV= \frac{\sigma}{\bar X}= \frac{2}{30}= 0.0667=6.7\%](https://tex.z-dn.net/?f=%20CV%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B2%7D%7B30%7D%3D%200.0667%3D6.7%5C%25)
Let Y the random variable that represent the miles per gallon in trucks of a population, and for this case we know the distribution for X is given by:
Where
and
We can calculate the coeffcient of variation for this cae like this:
![CV= \frac{\sigma}{\bar X}= \frac{3}{17}= 0.176=17.6\%](https://tex.z-dn.net/?f=%20CV%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B3%7D%7B17%7D%3D%200.176%3D17.6%5C%25)
So then we can conclude that the mpg for trucks have more variation since the coefficient of variation is larger than the value obtained for cars.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the miles per gallon in cars of a population, and for this case we know the distribution for X is given by:
Where
and
We can calculate the coeffcient of variation for this cae like this:
![CV= \frac{\sigma}{\bar X}= \frac{2}{30}= 0.0667=6.7\%](https://tex.z-dn.net/?f=%20CV%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B2%7D%7B30%7D%3D%200.0667%3D6.7%5C%25)
Let Y the random variable that represent the miles per gallon in trucks of a population, and for this case we know the distribution for X is given by:
Where
and
We can calculate the coeffcient of variation for this cae like this:
![CV= \frac{\sigma}{\bar X}= \frac{3}{17}= 0.176=17.6\%](https://tex.z-dn.net/?f=%20CV%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B3%7D%7B17%7D%3D%200.176%3D17.6%5C%25)
So then we can conclude that the mpg for trucks have more variation since the coefficient of variation is larger than the value obtained for cars.