The Environmental Protection Agency requires that the exhaust of each model of motor vehicle be tested for the level of several pollutants. The level of oxides of nitrogen (NOX) in the exhaust of one light truck model was found to vary among individual trucks according to a Normal distribution with mean 1.45 grams per mile driven and standard deviation 0.40 grams per mile.
What is the 45th percentile for NOX exhaust, rounded to four decimal places?

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".

Solution to the problem

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

[tex]X \sim N(1.45,0.4)[/tex]

Where [tex]\mu=1.45[/tex] and [tex]\sigma=0.4[/tex]

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

[tex]P(X>a)=0.55[/tex] (a)

[tex]P(X<a)=0.45[/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 0.45 of the area on the left and 0.55 of the area on the right it's z=-0.126. On this case P(Z<-0.126)=0.45 and P(z>-0.126)=0.55

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

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.45[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.45[/tex]

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

[tex]z=-0.126<\frac{a-1.45}{0.4}[/tex]

And if we solve for a we got

[tex]a=1.45 -0.126*0.4=1.3996[/tex]

So the value of height that separates the bottom 45% of data from the top 55% is 1.3996.