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:

1. below 65 MPH?

2. above 70 MPH?

3. between 60 and 70 MPH?

4. below 50 MPH or above 80 MPH?

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

1.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    65      
u = mean =    67      
          
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) / s =    -0.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.333333333   ) =    0.36944134 or 36.94% [ANSWER]

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2.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    70      
u = mean =    67      
          
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) / s =    0.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.5   ) =    0.308537539 or 30.85% [ANSWER]

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3.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    60      
x2 = upper bound =    70      
u = mean =    67      
          
s = standard deviation =    6      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.166666667      
z2 = upper z score = (x2 - u) / s =    0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.121672505      
P(z < z2) =    0.691462461      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.569789957 or 56.98% [ANSWER]

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4.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    50      
x2 = upper bound =    80      
u = mean =    67      
          
s = standard deviation =    6      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.833333333      
z2 = upper z score = (x2 - u) / s =    2.166666667      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.002303266      
P(z < z2) =    0.98486986      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.982566594      

Thus, those outside this interval is the complement =    0.017433406 OR 1.743% [ANSWER]  
      

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