Question

Ask Your Teacher 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 0.08 g/mi and standard deviation 0.01 g/mi. A company has 36 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 1

The level is L = 0.084

Explanation:

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

Normal probability distribution:

Problems of normally distributed samples can be 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 0.08, \sigma = 0.01, n = 36, s = (0.01)/(√(36)) = 0.0017`

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.

`Z = (X - \mu)/(\sigma)`

By the Central limit theorem

`Z = (X - \mu)/(s)`

`2.325 = (X - 0.08)/(0.0017)`

`X - 0.08 = 2.325*0.0017`

`X = 0.084`

The level is L = 0.084