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The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life ( 150,000 miles of driving) of cars of a particular model varies Normally with mean 85 mg/mi and standard deviation 6 mg/mi. A company has 16 cars of this model in its fleet. Using Table A, find the level such that the probability that the average NOX + NMOG level ¯ for the fleet greater than is only 0.01 ?

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Answer:

Mg/mi. A Company Has 16 Cars Of This Model In Its Fleet. ... The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in d ... organic gas (NMOG) in the exhaust over the useful life (150,000 miles of driving) of cars of a particular model varies Normally with mean 85 mg/mi and standard deviation 6 mg/mi.

Step-by-step explanation:

Using the normal distribution and the central limit theorem, it is found that the level is 88.49 mg/mi.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, for the sampling distribution of sample means of size n, the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem:

  • Mean of 85 mg/mi, thus [tex]\mu = 85[/tex]
  • Standard deviation of 6 mg/mi, thus [tex]\sigma = 6[/tex]
  • Sample of 16 cars, thus [tex]n = 16, s = \frac{6}{\sqrt{16}} = 1.5[/tex]

This level is the 1 - 0.01 = 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327. Then:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]2.327 = \frac{X - 85}{1.5}[/tex]

[tex]X - 85 = 2.327(1.5)[/tex]

[tex]X = 88.49[/tex]

The level is 88.49 mg/mi.

A similar problem is given at https://brainly.com/question/12403209

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