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 66th 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

Solution:-

Given that,

mean = = 1.45 grams

standard deviation = = 0.40 grams

a) Using standard normal table,

P(Z < z) = 66%

= P(Z < z) = 0.66

= P(Z < 0.4125 ) = 0.66  

z = 0.4125

Using z-score formula,

x = z * +

x = 0.4125 * 0.40 + 1.45

x = 1.6150 grams.

b) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 0.40 + 1.45

x = 1.1802

First quartile =Q1 = 1.1802 grams

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.6745 ) = 0.75

z = 0.6745

Using z-score formula,

x = z * +

x = 0.6745 * 0.40 + 1.45

x = 1.7198

Third quartile =Q3 = 1.7198 grams

IQR = Q3 - Q1

IQR = 1.7198 - 1.1802

IQR = 0.5396 grams.