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

Answer:

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

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

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

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

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X>60)=P(z>\frac{60-56}{3.2})$

$=P(z>\frac{60-56}{3.2})=P(z>1.25)$

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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X<52)=P(z<\frac{52-56}{3.2})$

$=P(z<\frac{52-56}{3.2})=P(z<-1.25)$

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:

$z=\frac{x-\mu}{\sigma}$

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

$P(57<X<62)=P(\frac{57-56}{3.2}<z<\frac{62-56}{3.2})$

$=P(0.3125<z<1.875)$

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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X<45)=P(z<\frac{45-56}{3.2})$

$=P(z<\frac{45-56}{3.2})=P(z<-3.44)$

And we can find this probability on this way:

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