Question

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?

Answer 1

Answer:

The % of drivers that will exceed this limit is 10.56 %

Step-by-step explanation:

Let's start defining the random variable :

$X$ : '' The speed on that road ''

We know that $X$ can be modeled with a Gaussian distribution ⇒

$X$ ~ $N$ ( μ , σ )

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 :  

$X$ ~ $N(67,4)$

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

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

We are going to substract the mean to $X$ and then divide by its standard deviation :

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

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

Z can be modeled as

$Z$ ~ $N(0,1)$

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

$P(Z>\frac{72-67}{4})$ = $P(Z>1.25)=1-P(Z\leq 1.25)$

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 $P(Z\leq 1.25)=0.8944$ ⇒

$P(X>72)=P(Z>1.25)=1-P(Z\leq 1.25)=1-0.8944=0.1056$ = 10.56 %

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