The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution. What is the probability that the truck driver goes more than 650 miles in a day?
+128407 The mean is [ 300 + 700] / 2 = [1000 ] / 2 = 500 miles
One standard deviation is 34.13% above the mean = ( 1 + .3413) (500) - 500 = 170.65
So....the z score we need can be computed as
[ 650 - 500 ]
___________ = .2236
670.65
The z score table associated with this value is .58706
So....the probability that he drives more than 650 miles per day = 1 - .58706 ≈ 41.3%
