Question

2. An important factor in solid missile fuel is the particle size distribution. Significant pro- blems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by (5x-6, X >1, elsewhere. Find F(x). 3. Use the same probability distribution as in Question 2. What is the probability that a random particle from the manufactured fuel exceeds 3 micrometers? (You could use either the PDF or CDF.) 4. Use the same probability distribution as in Question 2. (a) Find the expected particle size. (b) Find E (X2).

Answer Q4 only.

Answer 1

Given

$\small f(x)=5x^{-6}\; ,\; \; \; x>1$

= 0 elsewhere

(a) Expected particle size = E(X)

$\small \\E(X)=\int xf(x)=\int_{1}^{\infty }x\times \left (5x^{-6} \right )\;\\\\=\int_{1}^{\infty }5x^{-5}=5\left [ \frac{x^{-4}}{-4} \right ]_{1}^{\infty }=5\left [ 0-\left ( -\frac{1}{4} \right )\right ]=\frac{5}{4}=1.25$

Expected particle size = 1.25

(b) E(X²)

$\small \\E(X^{2})=\int x^{2}f(x)=\int_{1}^{\infty }x^{2}\times \left (5x^{-6} \right )\;\\\\=\int_{1}^{\infty }5x^{-4}=5\left [ \frac{x^{-3}}{-3} \right ]_{1}^{\infty }=5\left [ 0-\left ( -\frac{1}{3} \right )\right ]=\frac{5}{3}=1.6667$

E(X²) = 5/3 = 1.6667