Solution Found!
An article in IEEE Transactions on Dielectrics and
Chapter 4, Problem 169E(choose chapter or problem)
Problem 169E
An article in IEEE Transactions on Dielectrics and Electrical Insulation [“Statistical Analysis of the AC Breakdown Voltages of Ester Based Transformer Oils” (2008, Vol. 15(4))] used Weibull distributions to model the breakdown voltage of insulators. The breakdown voltage is the minimum voltage at which the insulator conducts. For 1 mm of natural ester, the 1% probability of breakdown voltage is approximately 26 kV, and the 7% probability is approximately 31.6 kV. Determine the parameters δ and β◊of the Weibull distribution.
Questions & Answers
QUESTION:
Problem 169E
An article in IEEE Transactions on Dielectrics and Electrical Insulation [“Statistical Analysis of the AC Breakdown Voltages of Ester Based Transformer Oils” (2008, Vol. 15(4))] used Weibull distributions to model the breakdown voltage of insulators. The breakdown voltage is the minimum voltage at which the insulator conducts. For 1 mm of natural ester, the 1% probability of breakdown voltage is approximately 26 kV, and the 7% probability is approximately 31.6 kV. Determine the parameters δ and β◊of the Weibull distribution.
ANSWER:
Solution :
Step 1 of 1:
Let X denotes break voltage with weibull distribution.
Given the 0.01 probability of breakdown voltage is approximately 26 kV, and the 0.07 probability is approximately 31.6 kV.
Our goal is:
We need to determine the parameters δ and β of the Weibull distribution.
Now we need to determine the parameters δ and β of the Weibull distribution.
P(X26) = 0.01 and P(X31.6) = 0.07.
P(Xx) = 1-
First we are finding P(X26) = 0.01.
P(X26) = 1-= 0.01
P(X26) = = 1-0.01
P(X26) =