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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.)

1 Answer

5 votes

Answer:

0.6915

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 = (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]</p><p><strong>What is the probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG?</strong></p><p>This is 1 subtracted by the pvalue of Z when X = 86. So</p><p>[tex]Z = (X - \mu)/(\sigma)


Z = (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

User Colm Prunty
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