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 92 milligrams per mile (mg/mi) of nitrogen oxides (NOX) and non methane 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 88 mg/mi and standard deviation 4 mg/mi. (a) What is the probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG? (Enter your answer rounded to four decimal places.)

Answer 1

Answer:

0.6915

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 88, \sigma = 4[/ex]What is the probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG?This is 1 subtracted by the pvalue of Z when X = 86. So[tex]Z = \frac{X - \mu}{\sigma}$

$Z = \frac{86 - 88}{4}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085

1 - 0.3085 = 0.6915

The answer is 0.6915