session7

# Measured resistance of wires produced by a company

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ȃ( t ) = 1 ( a + b)( a 2 + b 2 ) 4 € 1 ∞ (b) 1 = ∞ ∞ ∫ ce− x dx = c ∫ e− x dx = c [−e− x ] = c ⇒ c = 1 0 F ( x) = P( X ≤ x) = ∫ ∞ M (t) = 0 0 ∫e x −t 0 e dt = 1 − e− x ∞ ∞ tx − x e dx = 0 ∫e ( t −1)x 0 1 (1 − t ) 2 M ʹȃʹȃʹȃ(0) = 6 M ʹȃ( t ) = M ʹȃʹȃ( t ) = ⎡ྎ 1 ( t −1)x ⎤ྏ 1 dx = −⎢ྎ e ⎥ྏ = 1 − t ⎣ྏ t − 1 ⎦ྏ0 −2 (1 − t ) 3 M ʹȃʹȃʹȃ( t ) = 6 (1 − t ) 4 € 2. Measured resistance of wires produced by a company have normal probability distribution, with a mean of 0.13 Ohms and a standard deviation of 0.005 ohms. Find the third moment of the resistance (the third moment means E(X3)). M ( t ) = e µt +σ 22 t /2 M ' ( t ) = (µ + σ 2 t )e µt +σ M ' ' ( t ) = σ 2e µt +σ 22 t /2 22 t /2 + e µt +σ M (''' ) ( t ) = σ 2 (µ + σ 2 t )e µt +σ 22 t /2 22 t /2 (µ + tσ 2 ) 2 + (µ + σ 2 t )e µt +σ 22 t /2 (µ + tσ 2 ) 2 22 + 2σ 2 (µ + tσ 2 )e µt +σ t / 2 Evaluated at t = 0, this last expression becomes M't''= 0 = 2σ 2µ + µ 3 + µσ 2 = µ 3 + 3µσ 2 Substituting for the values of µ and σ , we get E(X 3 ) = M't''= 0 = (0.13) 3 + 3(0.13)(0.005) 2 = 0.0022 € 3. Let X1, . . . , X20 be independent Poisson random variables with mean 1. Find the probability distribution of the sum of these random variables. Prove your result. The moment generating function of sum of random variable is the product of the moment generating functions of each random variab...
View Full Document

## This note was uploaded on 03/04/2014 for the course STAT 100A taught by Professor Wu during the Winter '10 term at UCLA.

Ask a homework question - tutors are online