Determine the probability of the column solution i let

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: load, R S T . Then the failure is represented as R 0 . follows a normal distribution, therefore, we get, R S T 1.5 70 70 37.5 kN . Now, 2 2 R S T 0.15 1.5 70 2 4.85 2 16.4798 Hence the probability of failure is, from the CDF table for standard Normal Distribution given below: 0 X 37.5 P X 0 16.4798 R 2.2755 1 2.2755 1 0.98855 0.01145 3. Maximum of Two or More Variables Let X 1 , X 2 ,...., X n are n variables and defined as Y maxX 1 , X 2 ,...., X n . Then CDF of Y is given by, PY y PX 1 y , X 2 y ,..., X n y FX 1 ,X 2 ,...,X n y , y ,..., y and X 1 , X 2 ,...., X n are independent of each other, then n n i 1 i 1 FY y PX i y FX i y If we consider that, these variables have the same probability distribution, FX 1 y FX 2 y ... FX n y FX y The previous equation can be rewritten as, n FY y FX i y FX y n i 1 Now, by differentiating we obtain the pdf as, fY y d FY y dy nFX y n 1 fX y 4. Minimum of Two or More Variables Similar to the expression for maximum of two random variables, we may find the minimum of two or more random variables. Let, Z minX 1 , X 2 ,...., X n and if X 1 , X 2 ,...., X n are independent of each other, then n n i 1 i 1 FZ z 1 PX i z 1 1 FX i z Now, assuming that the variables have the same probability distribution, FX 1 y FX 2 y ... FX n y FX y...
View Full Document

This note was uploaded on 03/18/2014 for the course CE 5730 taught by Professor Dr.rajibmaity during the Spring '13 term at Indian Institute of Technology, Kharagpur.

Ask a homework question - tutors are online