Question

The mean gas mileage for a hybrid car is 56 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.2 miles per gallon.​

(a) What proportion of hybrids gets over 60 miles per​ gallon?


(b) What proportion of hybrids gets 52 miles per gallon or​ less?


(c )What proportion of hybrids gets between 57 and 62 miles per​ gallon?


(d) What is the probability that a randomly selected hybrid gets less than 45 miles per​ gallon?

Answer 1

a)
`P(X>60)=0.106`

b)
`P(X<52)=0.106`

c)
`P(57<X<62)=0.347`

d)
`P(X<45)=0.00029`

Explanation:

1) 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".

Let X the random variable that represent the gas mileage for a hybrid car of a population, and for this case we know the distribution for X is given by:

`X \sim N(56,3.2)`

Where
`\mu=56` and
`\sigma=3.2`

(a) What proportion of hybrids gets over 60 miles per​ gallon?

We are interested on this probability

`P(X>60)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

`P(X>60)=P(z>(60-56)/(3.2))`

`=P(z>(60-56)/(3.2))=P(z>1.25)`

And we can find this probability on this way:

`P(z>1.25)=1-P(z<1.25)=0.106`

(b) What proportion of hybrids gets 52 miles per gallon or​ less?

We are interested on this probability

`P(X<52)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

`P(X<52)=P(z<(52-56)/(3.2))`

`=P(z<(52-56)/(3.2))=P(z<-1.25)`

And we can find this probability on this way:

`P(z<-1.25)=0.106`

(c )What proportion of hybrids gets between 57 and 62 miles per​ gallon?

We are interested on this probability

`P(57<X<62)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

`P(57<X<62)=P((57-56)/(3.2)<z<(62-56)/(3.2))`

`=P(0.3125<z<1.875)`

And we can find this probability on this way:

`P(0.3125<z<1.875)=P(z<1.875)-P(z<0.3125)=0.347`

(d) What is the probability that a randomly selected hybrid gets less than 45 miles per​ gallon?

We are interested on this probability

`P(X<45)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

`P(X<45)=P(z<(45-56)/(3.2))`

`=P(z<(45-56)/(3.2))=P(z<-3.44)`

And we can find this probability on this way:

`P(z<-3.44)=1-P(z<-3.44)=0.00029`