Atrucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 30 thousand miles and a standard deviation of 10 thousand miles. complete parts (a) through (c) below. a. nbsp what proportion of trucks can be expected to travel between 18 and 30 thousand miles in a year?

To solve this problem, we have to make use of the z statistic. The formula for z score is:

z = (x – u) / s

where x is the sample value = 18,000 to 30,000, u is the sample mean = 30,000, and s is the standard deviation = 10,000

at x = 18,000

z = (18,000 – 30,000) / 10,000 = - 1.2

at x = 30,000

z = (30,000 – 30,000) / 10,000 = 0

So find for the P at values of z = -1,2 and z = 0.

P (z = - 1.2) = 0.1151

P (z = 0) = 0.5000

The proportion in between is the difference:

P (18,000 < x < 30,000) = 0.5000 – 0.1151 = 0.3849

So about 38.49% can be expected to travel 18,000 to 30,000 miles in a year

here is your mark brainliest