1. You work with traffic engineers for DOT (the department of transportation) and is in charge of performing measurements and analyzing speeding on a busy road. After measuring the speeds and collected a very large data sample, the speed on that road was found to have a Gaussian distribution with an average of 67 mph and standard deviation of 4 mph. If the highway patrol plans to ticket anyone driving faster than 72 mph, what is the % of drivers that will exceed this limit?

Where ''μ'' is the mean and ''σ'' is the standard deviation. Given that the average speed and the standard deviation of the problem are known we write :

[tex]X[/tex] ~ [tex]N(67,4)[/tex]

We are asked about [tex]P(X>72)[/tex] (which is the % of drivers that will exceed this limit).

To find this probability we are going to make a standardization of the variable [tex]X[/tex] (also called a change of variables).

We are going to substract the mean to [tex]X[/tex] and then divide by its standard deviation :

[tex]P(X>72)=[/tex] P [(X-μ) / σ > (72 - μ) / σ] (I)

The new variable [(X - μ) / σ] is called Z.

Z can be modeled as

[tex]Z[/tex] ~ [tex]N(0,1)[/tex]

⇒ Replacing in (I) the values of the mean and the standard deviation :

[tex]P(Z>\frac{72-67}{4})[/tex] = [tex]P(Z>1.25)=1-P(Z\leq 1.25)[/tex]

The convenience of this is that we can find the probabilities of Z (which is a N(0,1) ) in any table on internet ⇒

Looking at any table we will find that [tex]P(Z\leq 1.25)=0.8944[/tex] ⇒

[tex]P(X>72)=P(Z>1.25)=1-P(Z\leq 1.25)=1-0.8944=0.1056[/tex] = 10.56 %

We find that the % of drivers that will exceed this limits is 10.56 %