Question

In 2017, the entire fleet of light-duty vehicles sold in the United States by each manufacturer must emit an average of no more than 86 milligrams per mile (mg/mi) of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) over the useful life (150,000 miles of driving) of the vehicle. NOX +NMOG emissions over the useful life for one car model vary Normally with mean 82 mg/mi and standard deviation 5 mg/mi

(a) What is the probability that a single car of this model emits more than 86 mg/mi of NOX +NMOG? (Enter your answer rounded to four decimal places.)

probability:

(b) A company has 25 cars of this model in its fleet. What is the probability that the average NOX NMOG level cars is above 86 mg/mi? (Enter your answer rounded to four decimal places.)

probability:

Answer 1

Answer:

a) 0.2119 = 21.19% probability that a single car of this model emits more than 86 mg/mi of NOX +NMOG

b) 0% probability that the average NOX NMOG level cars is above 86 mg/mi

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 are 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:

$\mu = 82, \sigma = 5$

(a) What is the probability that a single car of this model emits more than 86 mg/mi of NOX +NMOG?

This is 1 subtracted by the pvalue of Z when X = 86. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{86 - 82}{5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881

1 - 0.7881 = 0.2119

0.2119 = 21.19% probability that a single car of this model emits more than 86 mg/mi of NOX +NMOG

(b) A company has 25 cars of this model in its fleet. What is the probability that the average NOX NMOG level cars is above 86 mg/mi?

Now we have that $n = 25, s = \frac{5}{\sqrt{25}} = 1$

This is 1 subtracted by the pvalue of Z when X = 86. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

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

$Z = \frac{86 - 82}{1}$

$Z = 4$

$Z = 4$ has a pvalue of 0.99998.

1-99998 = 0.00002

0% probability that the average NOX NMOG level cars is above 86 mg/mi