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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 .
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...
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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.
- Spring '13
- Civil Engineering