Question

Arsenic occurs naturally in very low concentrations. In healthy human adults arsenic blood concentrations are approximately Normally distributed with mean 3.9 μg/dL (micrograms per decilitre) and standard deviation 1.4 μg/dL. For the purposes of this question, assume that the distribution of arsenic blood concentrations is exactly as just described.

(a) What proportion of healthy adults have arsenic blood concentrations between 2 and 4.5 μg/dL?

b) Choosing a healthy adult at random, what is the chance that their arsenic blood concentration exceeds 6.8 μg/dL?

(c) What are the lower and upper limits of the middle 80% of arsenic blood concentrations in healthy human adults?

Answer 1

Let X be the concentration of the arsenic in the blood of healthy human adults. Given X follows a normal distribution with mean 3.9 μg/dL and a standard deviation of 1.4 μg/dL

a)

where Z follows a standard normal distribution

From Z table P(Z < 0.43) - P(Z < -1.36) = 0.6664 - 0.0869 = 0.5795

Therefore the proportion of healthy adults who have arsenic blood concentrations between 2 and 4.5 μg/dL is 0.5795

b)

From Z table P(Z > 2.07) = 0.0192

Therefore the chance that a healthy human adult has an arsenic blood concentration exceeding 6.8 μg/dL is 1.92%

c) Let μ-a, μ+a be the lower and upper limits of the middle 80% of arsenic blood concentrations in healthy human adults

From Z table P(Z < -1.28) = 0.10

μ-a = 2.108, μ-a = 5.692

Therefore 2.108 μg/dL, 5.692 μg/dL are the lower and upper limits of the middle 80% of arsenic blood concentrations in healthy human adults