Question

Particles are a major component of air pollution in many areas. It is of interest to study the sizes of contaminating particles. Let X represent the diameter, in micrometers, of a randomly chosen particle. Assume that in a certain area, the probability density function of X is inversely proportional to the volume of the particle ; that is, assume that fX(x) = c x 3 , x > 1, where c is a constant. (a) Find the value of c so that fX is a probability density function. (b) The term PM10 refers to particles 10 µ m or less in diameter. What proportion of contaminating particles are PM10 ? (c) The term PM2.5 refers to particles 2.5 µ m or less in diameter. What proportion of contaminating particles are PM2.5 ? (d) What proportion of the PM10 particles are PM2.5 ?

Answer 1

Answer:

a) c = 2

b) 0.99

c) 0.84

d) 0.8485

Step-by-step explanation:

We are given the following in the question:

$f(x) = \dfrac{c}{x^3}, x > 1$

a) Value of c

Property of probability density function

$\displaystyle\int^{\infty}_{-\infty} f(x) = 1$

Putting values, we get,

$\displaystyle\int^{\infty}_{1} \frac{c}{x^3} = 1\\\\\Rightarrow -\frac{c}{2}\bigg[\frac{1}{x^2}\bigg]^{\infty}_{1} = 1\\\\\Rightarrow \frac{c}{2} = 1\\\\\Rightarrow c = 2$

Thus, the value of c is 2.

$f(x) = \dfrac{2}{x^3}, x > 1$

b) proportion of contaminating particles are PM10

We have to evaluate

$P( x \leq 10) =\displaystyle\int ^{10}_{1}\frac{2}{x^3}dx\\\\=\bigg(\frac{2}{-2x^2}\bigg)^{10}_{1}\\\\=-(\frac{1}{100}-1)\\\\=0.99$

c) proportion of contaminating particles are PM2.5

$P( x \leq 2.5) =\displaystyle\int ^{2.5}_{1}\frac{2}{x^3}dx\\\\=\bigg(\frac{2}{-2x^2}\bigg)^{2.5}_{1}\\\\=-(\frac{1}{6.25}-1)\\\\=0.84$

d)  proportion of the PM10 particles are PM2.5

$P(PM ~2.5|PM~10) = \dfrac{0.84}{0.99} = 0.8485$