Question

Suppose the number of gallons of gasoline per day used by a car is normally distributed with a mean of 2.2 gallons and a standard deviation of 1.2 gallons. What is the difference in gallons per day used by a car with a z-score of 3 and another car that has a z-score of 0? 1.2 2.6 3.6 4.6

Answer 1

The difference between the two z-scores is 3 - 0 = 3 standard deviations. One standard deviation represents 1.2 gallons per day, so the difference is 3(1.2) = 3.6 gallons per day.

Answer 2

C. 3.6 gallons.

Explanation:

Let
`X_1` be the raw score corresponding to z-score 0 and
`X_2` be the raw score corresponding to z-score 3.

We have been given that the number of gallons of gasoline per day used by a car is normally distributed with a mean of 2.2 gallons and a standard deviation of 1.2 gallons.

We will use z-score formula to solve or given problem.

`z=(x-\mu)/(\sigma)`, where,

`z=\text{z-score}`,

`x=\text{Raw-score}`,

`\mu=\text{Mean}`,

`\sigma=\text{Standard deviation}`.

Now we will find raw scores corresponding to given z-scores as:

`0=(X_1-2.2)/(1.2)` and
`3=(X_2-2.2)/(1.2)`

Now we will multiply both sides of our equations by 1.2.

`0*1.2=(X_1-2.2)/(1.2)*1.2` and
`3*1.2=(X_2-2.2)/(1.2)*1.2`

`0=X_1-2.2` and
`3.6=X_2-2.2`

Now we will add 2.2 to both sides of our equations.

`0+2.2=X_1-2.2+2.2` and
`3.6+2.2=X_2-2.2+2.2`

`2.2=X_1` and
`5.8=X_2`

Now let us find the difference of our both raw scores to find the difference in gallons per day used by both cars.

`\text{Difference in gallons of gasoline used per day}=X_2-X_1`

`\text{Difference in gallons of gasoline used per day}=5.8-2.2`

`\text{Difference in gallons of gasoline used per day}=3.6`

Therefore, the difference in gallons of gasoline used per day by both cars is 3.6 gallons and option C is the correct choice.