Suppose that vehicle speeds on a particular section of the i

Suppose that vehicle speeds on a particular section of the interstate are normally distributed with a mean of 67 MPH and a standard deviation of 6 MPH. What percentage of vehicles are traveling:

5. If there are 2000 vehicles on the roadway, how many are traveling above 80 MPH?

6. How fast would I have to drive in order to go faster than 88.88% of the vehicles?

7. How slow would I have to drive in order to go slower than 88.88% of the vehicles?

8. What range of speeds are within 1.96 standard deviations of the mean?

9. What are the population proportions P(z) associated with the z values +/- 1.96?

10. State patrol does extra speed traps on roadways when more than 20% of drivers are traveling over the speed limit. The speed limit for this section of interstate is 75 mph. Should state patrol use extra speed traps on this section of road?

Solution

Mean ( u ) =67
Standard Deviation ( sd )=6
Number ( n ) = 2000
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
P(X > 80) = (80-67)/6/ Sqrt ( 2000 )
= 13/0.134= 96.8963
= P ( Z >96.8963) From Standard Normal Table
= 0                  
b)
P ( Z < x ) = 0.8874
Value of z to the cumulative probability of 0.8874 from normal table is 1.213
P( x-u/s.d < x - 67/6 ) = 0.8874
That is, ( x - 67/6 ) = 1.21
--> x = 1.21 * 6 + 67 = 74.278                  
                  
c)
P ( Z > x ) = 0.8874
Value of z to the cumulative probability of 0.8874 from normal table is -1.21
P( x-u/ (s.d) > x - 67/6) = 0.8874
That is, ( x - 67/6) = -1.21
--> x = -1.21 * 6+67 = 59.722