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 vari

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.)

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Julia 17 hours 2021-09-10T10:02:13+00:00 1 Answer 0

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    2021-09-10T10:03:54+00:00

    Answer:

    The level is L = 0.084

    Step-by-step 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 = \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.

    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 = \frac{\sigma}{\sqrt{n}}

    In this problem, we have that:

    \mu = 0.08, \sigma = 0.01, n = 36, s = \frac{0.01}{\sqrt{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 = \frac{X - \mu}{\sigma}

    By the Central limit theorem

    Z = \frac{X - \mu}{s}

    2.325 = \frac{X - 0.08}{0.0017}

    X - 0.08 = 2.325*0.0017

    X = 0.084

    The level is L = 0.084

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