Question

the environmental protection agency (epa) requires that the exhaust of each model of motor vehicle be tested for the level of several pollutants. the level of oxides of nitrogen (nox) in the exhaust of one light truck model was found to vary among individual trucks according to an approximately normal distribution with mean $mu = 1.45$ grams per mile driven and standard deviation $sigma = 0.40$ gram per mile. which of the following best estimates the proportion of light trucks of this model with nox levels greater than 2 grams per mile? 0.0846 0.04256 0.9154 0.0228

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 2\), \(\mu=1.45\), and \(\sigma = 0.40\).

$$z=\frac{2 - 1.45}{0.40}=\frac{0.55}{0.40}=1.375$$

Step2: Find the proportion using the standard normal distribution

We want to find \(P(X>2)\), which is equivalent to \(P(Z > 1.375)\) (since \(X\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\), and \(Z=\frac{X-\mu}{\sigma}\) is standard normal).
Using the property \(P(Z>z)=1 - P(Z\leq z)\). Looking up \(P(Z\leq1.375)\) in the standard - normal table (or using a calculator with a normal - distribution function).
Using a calculator (for example, in R: 1 - pnorm(1.375) or in Excel: 1 - NORM.S.DIST(1.375,TRUE)), \(P(Z\leq1.375)\approx0.9154\)

$$P(Z>1.375)=1 - 0.9154 = 0.0846$$

Answer:

0.0846