Weibull Distribution (statistics)
Michelle
Problem 1. Suppose that the lifetime of a certain kind of an emergency backup battery (in hours) is a random variable X having a Weibull distribution with alpha=0.1 and beta=0.5. Find (1) The probability that such a battery will last more than 300 hours. (2) The mean lifetime of these batteries.
Correct answers
(1) 0.177
(2) 200 hours
Problem 2: Suppose that the service life (in hours) of a semiconductor is a random variable having the Weibull distribution with alpha=0.025 and beta=0.5. What is the probability that such a semiconductor will still be in operating condition after 4000 hours?
Correct answer: 0.2057
I used wblpdf and wblcdf but my answers are way off. Please help. Thanks.
Tom Lane
> textbook (Probability and Statitics fo Engineers, Miller and Freund's)
> I used wblpdf and wblcdf but my answers are way off. Please help.
> Thanks.
Different authors use equivalent but different definitions for the Weibull
distribution. For example, the cdf might be defined as
1 - exp(-A*x^B) or 1 - exp(-(x/A)^B)
You just have to compare the book's definition with the one MATLAB uses.
Here's a hint:
>> 1-wblcdf(300,(1/.1)^(1/.5),.5)
ans =
0.1769
-- Tom
Nasser M. Abbasi
> I am getting werid answers from Matlab for two problems that are in the
> textbook (Probability and Statitics fo Engineers, Miller and Freund's)
>
> Problem 1. Suppose that the lifetime of a certain kind of an emergency
> backup battery (in hours) is a random variable X having a Weibull distribution
> with alpha=0.1 and beta=0.5. Find (1) The probability that such a battery
> will last more than 300 hours.
fyi, This is what Mathematica shows, assuming I did not mess something up:
--------------------
alpha = 0.1;
beta = 0.5;
Probability[x>300,x ~ WeibullDistribution[alpha,beta]]
----------------------
Out[99]= 0.1501833070963304
(2) The mean lifetime of these batteries.
>
> Correct answers
> (1) 0.177
> (2) 200 hours
>
> Problem 2: Suppose that the service life (in hours) of a semiconductor is a
>random variable having the Weibull distribution with alpha=0.025 and beta=0.5.
> What is the probability that such a semiconductor will still be in operating
> condition after 4000 hours?
>
> Correct answer: 0.2057
>
> I used wblpdf and wblcdf but my answers are way off. Please help. Thanks.
-----------------------------
alpha = 0.025;
beta = 0.5;
Probability[x>4000, x ~ WeibullDistribution[alpha,beta]]
------------------------------
Out[130]= 0.28595468135239815
I guess there is sligthly different definitions for this pdf. Mathematica
defines it as
"The probability density for value x in a Weibull distribution is
proportional to x^(Alpha-1) Exp[ -(x/Beta)^Alpha ] for x>0, and is zero for x<0"
hth,
--Nasser
Michelle
Michelle
Why are we using an expression (1/.1)^(1/.5) inside wblcdf. According to the documentation, P = wblcdf(X,A,B). A is alpha and B is beta. Can you please clarify? I have never used this function before. Thanks.
Tom Lane
> Why are we using an expression (1/.1)^(1/.5) inside wblcdf. According to
> the documentation, P = wblcdf(X,A,B). A is alpha and B is beta. Can you
> please clarify? I have never used this function before. Thanks.
Okay, it's just that the wblcdf parameters, which are also called A and B,
are not the same A and B as in your book apparently. Using less ambiguous
notation:
1 - exp(-A*x^B) or 1 - exp(-(x/C)^B)
just substitute C = (1/A)^(1/B) into the expression on the right.
-- Tom