Question

The Environmental Protection agency requires that the exhaust of each model of motor vehicle be tested for the level of several pollutants. The level of oxides of nitrogen (NOX) in the exhaust of one light truck model was found to vary among individually trucks according to a Normal distribution with mean 1.45 grams per mile driven and standard deviation 0.40 grams per mile.

(a) What is the 99th percentile for NOX exhaust, rounded to four decimal places?

(b) Find the interquartile range for the distribution of NOX levels in the exhaust of trucks rounded to four decimal places.

Answer 1

Answer:

a) The 99th percentile for NOX exhaust is 2.3820 grams per mile driven.

b) The interquartile range is 0.54 grams per mile driven.

Step-by-step explanation:

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

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

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 percentile of this measure.

In this problem, we have that:

Normal distribution with mean 1.45 grams per mile driven and standard deviation 0.40 grams per mile, so $\mu = 1.45, \sigma = 0.40$.

(a) What is the 99th percentile for NOX exhaust, rounded to four decimal places?

This is the value of X when Z has a pvalue of 0.99.

$Z = 2.33$ has a pvalue of 0.99. So

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

$2.33 = \frac{X - 1.45}{0.40}$

$X - 1.45 = 2.33*0.40$

$X = 2.382$

The 99th percentile for NOX exhaust is 2.3820 grams per mile driven.

(b) Find the interquartile range for the distribution of NOX levels in the exhaust of trucks rounded to four decimal places.

The interquartile range is the subtraction of X when Z has a pvalue of 0.75, that is, the third quartile, by X when Z has a pvalue of 0.25, that is, the first quartile.

Third quartile

Z has a pvalue of 0.75 between 0.67 and 0.68, so we use $Z = 0.675$.

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

$0.675 = \frac{X - 1.45}{0.40}$

$X - 1.45 = 0.675*0.40$

$X = 1.72$

The third quartile is 1.72 grams per mile driven.

First quartile

Z has a pvalue of 0.25 between -0.67 and -0.68, so we use $Z = -0.675$.

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

$-0.675 = \frac{X - 1.45}{0.40}$

$X - 1.45 = -0.675*0.40$

$X = 1.18$

The first quartile is 1.18 grams per mile driven.

The interquartile range is 1.72-1.18 = 0.54 grams per mile driven.

Answer 2

The 99th percentile for NOX exhaust is 2.382 grams per mile while the interquartile range is 1.45 grams per mile

Z score

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (x - μ)/σ

where x is raw score, σ is standard deviation and μ is mean a

μ = 1.45, σ = 0.4

a) The 99th percentile corresponds with a z score of 2.33, hence:

2.33 = (x - 1.45)/0.4

x = 2.382 grams per mile

b) The interquartile range (50%) corresponds with a z score of 0, hence:

0 = (x - 1.45)/0.4

x = 1.45 grams per mile

The 99th percentile for NOX exhaust is 2.382 grams per mile while the interquartile range is 1.45 grams per mile

Find out more on Z score at: https://brainly.com/question/25638875