The proportion of traffic fatalities for each U.S. state resulting from drivers with high alcohol blood levels in 1982 was approximately normally distributed, with mean 0.569 and standard deviation 0.068.

a. What proportion of states would you expect to have more than 65% of their traffic fatalities from drunk driving?

b. What proportion of deaths due to drunk driving would you expect to be at the 25th percentile of this distribution?

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".

Part a

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

[tex]X \sim N(0.569,0.068)[/tex]

Where [tex]\mu=0.569[/tex] and [tex]\sigma=0.068[/tex]

We are interested on this probability

[tex]P(X>0.65)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X>0.65)=P(\frac{X-\mu}{\sigma}>\frac{0.65-\mu}{\sigma})=P(Z>\frac{0.65-0.569}{0.068})=P(z>1.19)[/tex]

And we can find this probability using the complement rule:

[tex]P(z>1.19)=1-P(z<1.19)=1-0.883=0.117[/tex]

Part b

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.75[/tex] (a)

[tex]P(X<a)=0.25[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.674<\frac{a-0.569}{0.068}[/tex]

And if we solve for a we got

[tex]a=0.569 -0.674*0.068=0.523[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 0.523.