In: Math
A state highway goes through a small town where the posted speed limit drops down to 40MPH, but which out of town drivers don’t observe very carefully. Based on historical data, it is known that passenger car speeds going through the city are normally distributed with a mean of 47 mph and a standard deviation of 4MPH. Truck speeds are found to be normally distributed with a mean of 45MPH and a standard deviation of 6MPH. The town installed a speed camera and wants to set a threshold for triggering the camera to issue citations. If the camera is triggered, the driver is mailed a flat $50 ticket for cars and a flat $75 for trucks. On average 100 cars and 25 trucks go through the city in a day.
Let X be the speed of any given passenger car going through the city. X is normally distributed with mean and standard deviation
Let Y be the speed of any given truck going through the city. Y is normally distributed with mean and standard deviation
The probability that any given passenger car going through the city triggers the camera by going over 50 MPH is
Number of cars, out of 100/day that trigger the camera = 0.2266*100=22.66
revenue per day from ticketing cars = 22.66*50=1133
The probability that any given truck going through the city triggers the camera by going over 50 MPH is
Number of truck, out of 25/day that trigger the camera = 0.2033*25=5.0825
revenue per day from ticketing trucks = 5.0825*75=381.1875
Total revenue per day = 1133+381.1875=1514.1875
Revenue per month is 1514.1875*30= $45,425.63
ans: If the town sets the camera triggering speed at 50MPH, total revenue it will make in a month is $45,425.63
The town wants to set the triggering speed at a value such that the fastest 10% of truck drivers get ticketed. At what value should they set the trigger?
Let Y=y be the value such that, the probability that any given truck passing through get ticketed is 0.10
That is
We will find the z score for which the standard normal probability of P(Z<z)=0.90.
Using the standard normal table we can get for z=1.28, we get P(Z<1.28)=0.8997, close enough to 0.90
Now we will equate z=1.28 to the z score of y and solve for y
ans: The town needs to set the trigger at 52.68 MPH, such that the fastest 10% of truck drivers get ticketed.
At this trigger value, what percentage of cars are ticketed and what is the monthly revenue for the city?
The probability that any given car passing through the city triggers the camera by going over 52.68 MPH is
ans: The percentage of cars that are ticketed is 7.78%
Revenue per day from cars = 0.0778*100*50=389
Revenue per day from trucks = 0.10*25*75=187.5
Total revenue per day is 389+187.5=576.5
Total revenue per month is $17,295.00
ans: the monthly revenue for the city is $17,295.00