Suppose a normal model describes the fuel efficiency of cars currently registered in your state. the mean is 24 mpg, with a standard deviation of 6 mpg. a. sketch the normal model. b. what percent of all cars get less than 15 mpg? c. what percent of all cars get between 20 and 30 mpg? d. what percent of cars get more than 40 mpg? e. describe the fuel efficiency of the worst 20% of all cars. f. what gas mileage represents the third quartile? g. describe the gas mileage of the most efficient 5% of all cars. h. what gas mileage would you consider unusual? why?

A: (It´s an attachment)

B:  P(x≤15)=0.0668⇒6.68%

C: P(20≤x≤30)=0.5899⇒58.99%

D: P(x≥40)=0.0039⇒0.39%

E: The fuel efficiency of those cars are less than 18.9 mpg

F: The third quartile is represented with 28.08 mpg

G: The fuel efficiency of that group of cars are greater than 34.5 mpg

H: Any fuel efficiency less than 6 and greater than 42

Step-by-step explanation:

It´s useful to know beforehand how to standardize a Normal distribution, that means, to transform the data to use the Standard Normal distribution formulas in the given problems, because there´s already tools that are pretty useful in that distribution, it´ll be easy to give an solution by transforming a Normal distribution into a Standard Normal distribution.

A: To sketch the model in a graph, you center the "gaussian bell" at 24 in the X-axis and stretch it by the sides according to the Standard deviation given (the height it´s always less than 1)

B: At first, we don´t aim for the percent that was asked because it´ll be easy to find with the probability that someone´s car fuel efficiency is less than 25 mpg

Let´s begin with the transformation that i referred above:

We are looking for P(x≤15)  with x being the possible fuel efficiency of a car, but that value isn´t meant to be found directly. The only Normal distribution which values´ probabilities can be found directly is the Standard Normal distribution and we do that transformation like this:

P(a≤x≤b)=P( (a-μ)/σ ≤ z ≤ (b-μ)/σ )

where μ is the mean value and σ is the Standard deviation, this problem it´s approached with that formula:

P(x≤15)=P( z ≤ (15-24)/6 )=P(z≤-1,5)

We changed the goal of our problem and now we can look for that value, because that P(z≤-1.5) belongs to a Standard Normal distribution and there are methods to look for it´s values. My favorite is a chart which has the predetermined results for so many common points around 0, there it is:

(It´s an attachment)

There we look for the value -1.5, in the first column of each section are the whole part and the first decimal, and in every single row appears the second decimal. the value of those coordinates (-1.5,0) it´s the value of P(z≤-1.5)

P(x≤15)=P(z≤-1.5)=0.0668

and finally we find the percent by multiplying it by 100

0.0668*100=6.68%

C: The mechanic it´s pretty the same but with a little change at the beginning.

P(20≤x≤30)=P( (20-24)/6 ≤ z ≤ (30-24)/6 )=P(-0.66... ≤z≤1)

For some properties of probability, P(a≤x≤b)=P(x≤b)-P(x≤a) and we need this because the chart only work with one value at the time

P(-0.66... ≤z≤1)=P(z≤1)-P(z≤-0.66...)

And using the chart, we have

P(-0.66... ≤z≤1)=0.8413-0.2514

P(20≤x≤30)=0.5899

and the answer is 0.5899*100=58.99%

D: We proceed as before but, knowing that the chart only work with P(x ≤ value) we need to exchange the direction of the inequality as follows:

By a property of probabilities:

P(x≥40)=1-P(x≤40)

Now, by chart:

P(x≥40)=1-0.9961=0.0039

And the answer is 0.0039*100=0.39%

E: The dynamic that we had was to "Standardize" the probability that we looked for in order to take it from the chart, but know, with this question we proceed backwards (We look the value given in the chart, we express it like P(z≤y) and we Standardize backwards, P(z≤y)=P( x ≤ (σ*y + μ)) )

We want the worst 20%, that means we need some fuel efficiency value that its probability it´s the closest to 0.2 (20/100) from below. Looking in the chart we know that:

P(z ≤ -0.85)=0.1997

P(z ≤ -0.85)=P(x ≤ (-0.85*6 + 24) )=P(x≤18.9)

and our final answer is that the worst 20% of the fuel efficiency is less than 18.9 mpg

F. It´s similar to E problem, but we look for the first value that its probability is greater than 0.75. looking in the chart we find that:

P(z≤0.68)=0.7517

P(z≤0.68)=P(x≤28.80)

and our final answer is that the third quartile is represented by 28.08

G. This problem it´s meant to use the trick of D and E at the same time.

P(z≥y)=0.05

P(z≥y)=1-P(z≤y) (D, the chart won´t work with "≥" so we change it)

P(z≤y)=1-0.05=0.95

And by looking for it in the chart:

P(z≤1.65)=0.9505 (E, to look directly for the value 0.05 in the chart)

P(z≥1.65)=1-P(z≤1.65)=1-0.9505=0.05

We conclude that those 5% of the most efficient fuels are greater than 1.65 mpg

H. The standard deviation is the measure of the dispersion of the value with respect to the mean, so it´s common that some value it´s closer to the mean, closer than the standard deviation.

It´s weird that values are farther than 3*6 from the mean (looking in the A graph)

So i consider that x≤6 and x≥42 are pretty unusual.