Question

A company maintains a fleet of vehicles that consist of both cars and trucks. The miles per gallon realized in the cars follows a normal distribution with mean of 30 and a standard deviation of 2 while the miles per gallon realized in the trucks follows a normal distribution with mean of 17 with a standard deviation of 3.Which distribution is more variable?
a. they have the same variability
b. the mpg for the trucksc. the mpg for the carsd. there is insufficient information to answer the questi

Answer 1

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:

$X \sim N(30,2)$  

Where $\mu=30$ and $\sigma=2$

We can calculate the coeffcient of variation for this cae like this:

$CV= \frac{\sigma}{\bar X}= \frac{2}{30}= 0.0667=6.7\%$

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:

$X \sim N(17,3)$  

Where $\mu=17$ and $\sigma=3$

We can calculate the coeffcient of variation for this cae like this:

$CV= \frac{\sigma}{\bar X}= \frac{3}{17}= 0.176=17.6\%$

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:

$X \sim N(30,2)$  

Where $\mu=30$ and $\sigma=2$

We can calculate the coeffcient of variation for this cae like this:

$CV= \frac{\sigma}{\bar X}= \frac{2}{30}= 0.0667=6.7\%$

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:

$X \sim N(17,3)$  

Where $\mu=17$ and $\sigma=3$

We can calculate the coeffcient of variation for this cae like this:

$CV= \frac{\sigma}{\bar X}= \frac{3}{17}= 0.176=17.6\%$

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.