Question

According to the paper "Commuters' Exposure to Particulate Matter and Carbon Monoxide in Hanoi, Vietnam" (Transportation Research [2008]: 206-211), the carbon monoxide exposure of someone riding a motorbike for \(5 \mathrm{~km}\) on a highway in Hanoi is approximately normally distributed with a mean of 18.6 ppm. Suppose that the standard deviation of carbon monoxide exposure is 5.7 ppm. Approximately what proportion of those who ride a motorbike for \(5 \mathrm{~km}\) on a Hanoi highway will experience a carbon monoxide exposure of more than 20 ppm? More than \(25 \mathrm{ppm} ?\)

Short answer

Approximately 40.13% of riders experience carbon monoxide levels exceeding 20 ppm, and approximately 13.14% experience levels exceeding 25 ppm.

Step-by-step solution

Understand the Normal Distribution

The normal distribution is defined by two parameters: the mean (μ) and the standard deviation (σ). Readings are said to follow a normal distribution here, with μ=18.6 and σ=5.7. The goal is to calculate proportion of data falling over certain values - which would entail calculating the complementary cumulative distribution function (also known as 1-CDF). This involves converting the values to z-scores.

Calculate Z-scores

The z-score for a data point is given by the formula \(Z = (X - μ) / σ\), where X is the point, μ the mean and σ the standard deviation. Substituting X = 20 ppm, then \(Z1 = (20 - 18.6) / 5.7 ≈ 0.25\). Similarly, for X = 25 ppm, \(Z2 = (25 - 18.6) / 5.7 ≈ 1.12\).

Find Proportions by Calculating 1-CDF

Given the z-scores, the required proportions can be found as 1-CDF(Z). Consulting a standard normal (Z) table or appropriate software yields 1-CDF(0.25)=0.4013 and 1-CDF(1.12)=0.1314. Thus, approximately 40.13% of riders experience carbon monoxide levels exceeding 20 ppm, and approximately 13.14% experience levels exceeding 25 ppm.