The level of nitrogen oxides (NOX) in the exhaust after 50,000 miles or fewer of driving of cars of a particular model varies Normally with mean 0.06 g/mi and standard deviation 0.01 g/mi. A company has 25 cars of this model in its fleet. What is the level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01? (Hint: This requires a backward Normal calculation. Round your answer to three decimal places.)

Answer :

Answer:

The level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01 is L = 0.065.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem:

Normal probability distribution:

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

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

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

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.

Central Limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, which is also called standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 0.06, \sigma = 0.01, n = 25, s = \frac{0.01}{\sqrt{25}} = 0.002[/tex]

What is the level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01?

This is X when Z has a pvalue of 1-0.01 = 0.99. So it is X when Z = 2.325.

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

By the Central Limit Theorem

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

[tex]2.325 = \frac{X - 0.06}{0.002}[/tex]

[tex]X - 0.06 = 2.325*0.002[/tex]

[tex]X = 0.065[/tex]

The level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01 is L = 0.065.

Other Questions