Value of exponential term in Weibull too large

[Helpless]

Dear All,

I have this problem here. Hope some kind soul can help me solve this.

Problem: Weibull distribution with gamma=0
Shape parameter=1.4
Scale parameter=3000 hours and 9000 hours

I wish to determine the probability that the system will fail after 72 hours. However, it seems that if I will to substitute the value using calculator. I cannot get an answer. Is there a way to normalize the value of the exponential term as it gets too large for a calculator to churn?

[RS Support]

The probability of failure for the Weibull distribution is given by:

P = 1 - exp(-(t/eta)^beta)

Therefore, you just need to plug your values into this equation. Additional information on Weibull distribution statistical properties can also be found at http://www.weibull.com/LifeDataWeb/weibull_statistical_properties.htm. Hope this helps.

[Dennis Craggs]

The Weibull life equation is

F(t) = 1 – e ^[- (t/theta)^beta] where t is the hours, theta is the characteristic life, and beta is the shape factor.

Use a series approximation of the exponential term in the Weibull equation:

e^x = 1 + x/1! + x^2/2! + x^3/3! + ... where x can be any value.

This approximation is valid when x is small. Most of the time only the first order term is necessary. This means a good approximation is

F(t) = (t/theta)^beta

For t=72, theta = 3000, beta = 1.4, then for the approximation F(70)=0.005399 vs. the exact answer of 0.005384.

For t=72, theta = 9000, beta =1.4, then for the approximation F(70)=0.001160 vs. the exact answer of 0.001159.

Note: This was done both ways on an HP15C calculator.

[Pantelis]

Dennis,
Just out of curiosity – why the approximation?
Using the exact equation requires just a couple more button clicks on a calculator.

[Dennis Craggs]

Helpless claimed he could not perform the calculation on his calculator (hard to believe). I performed the calculation on my HP15C and a Palm RPN calculator. I wanted to show there are multiple paths to performing the calculation and provide an approximation if his calculator was experiencing rounding errors. The approximation result was very close to the exact answer.

I have to confess that it was a mathematical exercise. Helpless should recognize the calculation is simple and any calculator with x^y expontiation function and an e^x function should be able to solve the problem. The rounding error should not be significant.