Mathematics, 14.03.2020 01:29 hendo617

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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Answer from: nathanbrockdac

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

Answer from: Quest

answer

a,b, and d

explanation

a reflection angle is more than

$\pi \: radians$

but less than;

$2 \pi \: radians$

an angle that measures

$\frac{4 \pi}{3} \: radians$

is a reflex angle.(true)

an angle that measures less than

$\frac{\pi}{2}$

is acute.(true)

provided we are considering angles greater or equal to zero.

an acute angle lies in the range.

$0 \leqslant \theta \: < \frac{\pi}{2}$

an obtuse angle is less than 180° but more than 90°

an angle that measures 300° is obtuse (false)

an angle that measures 65° is an acute angle.(true)

Answer from: Quest

Answers: the graph is translated down 1 unit and the domain and range of g(x) is the same as the domain and range of f(x).

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