Question

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x > 1. Determine the following (round all of your answers to 3 decimal places):
a) P(x < 5)=

b) P(x > 8)=

c) P(6 < x < 10)=

d) P(x < 6 or x > 10)=

3) Determine x such that P(X < x) = 0.75

Answer 1

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

Firstly we will calculate the general probability that of P(a < X < b)

       P(a < X < b) =  $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

                            = $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$    { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

                            = $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

                            = $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5)  {because x > 1 }

     Comparing with general probability we get,

     P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/$8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

                                = $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/$10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

               ⇒  P(1 < X < x) = 0.75

               ⇒  $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

               ⇒  $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

               ⇒  $x^{2}$ = $\frac{1}{0.25}$   ⇒ $x^{2}$ = 4 ⇒ x = $2$  

Therefore, value of x such that P(X < x) = 0.75 is 2.