Final answer:
To determine the mileage that at least 90% of the trucks will travel, the z-score method is used on a normal distribution with a mean of 40,000 miles and standard deviation of 12,000 miles. The z-score for the 90th percentile is 1.28, which gives a mileage of 55,360 miles or less for at least 90% of the trucks.
Step-by-step explanation:
The student is asking how to determine the mileage that at least 90% of the trucks will travel, given a normal distribution of miles traveled by trucks for a trucking company. To solve this, we need to identify the z-score that corresponds to the bottom 90% of a normal distribution. Looking at a standard normal distribution table, we find that the z-score for 90% (or the 10th percentile from the upper end) is approximately 1.28. We then use the formula:
x = μ + zσ
Where μ is the mean, z is the z-score corresponding to the desired percentile, and σ is the standard deviation. Substituting the values for the trucking company, μ = 40,000 miles and σ = 12,000 miles, we get:
x = 40,000 + (1.28)(12,000)
x = 40,000 + 15,360
x = 55,360 miles
Therefore, we can conclude that at least 90% of the trucks will travel 55,360 miles or less per year.