Question

In 2017, the entire fleet of light‑duty vehicles sold in the United States by each manufacturer must emit an average of no more than 86 milligrams per mile (mg/mi) of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) over the useful life ( 150,000 miles of driving) of the vehicle. NOX + NMOG emissions over the useful life for one car model vary Normally with mean 80 mg/mi and standard deviation 5 mg/mi. A company has 25 cars of this model in its fleet. What is the probability that the average NOX + NMOG level ⎯⎯⎯ of these cars is above 86 mg/mi? Give your answer to four decimal places.

Answer 1

The probability that the average NOX + NMOG level of a fleet of 25 cars is above 86 mg/mi is almost negligible, as the Z-score for this scenario is 6, which is well into the tail of the standard normal distribution. The probability (to four decimal places) is approximately 0.0000.

Step-by-step explanation:

A student asked about the probability that the average NOX + NMOG level of a fleet of 25 cars exceeds 86 mg/mi, given that the emissions for one car model follow a normal distribution with a mean of 80 mg/mi and a standard deviation of 5 mg/mi. To answer this, we would calculate the Z-score of 86 mg/mi for the distribution of sample means and use the standard normal distribution to find the probability.

The Z-score is given by:

Z = (X - μ) / (σ / √ n)

Where X is the sample mean we are testing (86 mg/mi), μ is the population mean (80 mg/mi), σ is the standard deviation of the population (5 mg/mi), and n is the sample size (25).

Substituting the values, we get:

Z = (86 - 80) / (5 / √ 25) = (6) / (1) = 6

With a Z-score of 6, the probability of the sample mean being above 86 mg/mi becomes almost negligible since a Z-score of 6 lies far in the tail of the standard normal distribution. The probability (to four decimal places) is approximately 0.0000.