Average person who drives car in United States drives 12 400

Average person who drives car in United States drives 12, 400 which is 50% more than an average driver in Europe. We assume that the number of yearly miles by U.S. drivers is approximately a normal random variable of standard deviation of 3800 miles. Calculate percent of drivers who traveled between 10,000 to 15,000 miles in a year.

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    10000      
x2 = upper bound =    15000      
u = mean =    12400      
          
s = standard deviation =    3800      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.631578947      
z2 = upper z score = (x2 - u) / s =    0.684210526      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.263831023      
P(z < z2) =    0.753078882      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.489247858 = 48.92% [ANSWER]