Weibull Distribution: Overview, Questions, Preparation

The Weibull Distribution is a part of probability distributions that help analyze the item’s constancy and life data analysis as the item’s failure. Parameters are used for measuring Weibull’s distribution reliability. 

There are two types of Weibull probability density function, and they are:

• Two parameter 
• Three parameter 

The Weibull Distribution is a continuous function because the time failure for an item is always positive. The general of Weibull Distribution for three-parameter pdf is:
f(x)=γ/α((x−μ)/α)^γ−1 exp(−((x−μ)/α)^γ)   α,γ>0; x≥μ 

• γ = shape parameter(also known as threshold parameter or Weibull Slope)
• α = scale parameter(also known as characteristic life parameter).
• μ = location parameter(also known shift or waiting time parameter)

The formula gets reduced for deriving the standard Weibull distribution when α becomes 1, and μ becomes 0. And the formula derived is: f(x) = γ(x)^γ−1 exp^(−(x)^γ), x >= 0 and y > 0. The formula for cumulative distribution function is: F(x) = 1 - exp^(−(x/α)^γ).

Properties of Weibull Distribution

Some Weibull distribution’s properties are:

• Probability density function
• Moments and Moment generating function
• Shannon entropy
• Cumulative distribution function

Weibull Distribution is a part of Probability in class XII, but majorly the topic is crucial in higher studies like engineering and many others. Probability and its applications are also present in various chapters of mathematics. In the class XII mathematics exam, the chapter has a weightage of nearly 7 to 8 marks. Apart from Weibull Distribution, the other topics that are majorly important and are covered in the chapter on probability are:

• Conditional Probability
• Multiplication Theorem on Probability 
• Independent Events
• Bayes’ Theorem 
• Random Variables and its Probability Distributions 
• Bernoulli Trials 
• Binomial Distribution

3. What will be the probability of a magnetic disk failing before 500 hours if the disk is exposed to corrosive gas. The value of γ and α is 300 and 0.5.

Solution. Applying the values in the formula of the distribution function of x i.e.

F(x) = 1 - exp^(−(x/α)^γ),

Now, the probability for the failure of the disk before 500 hours will be

P(x0.5) 

= 1 - exp^(-(1.6667)^0.5) 

= 0.725.

2. What will be the probability of the magnetic disk to last 600 hours or more in the above question?
Solution. F(600) = 1 - exp^(-(600/300)^0.5),

P(x>=600) = 1 - P(x1 - exp^(-(600/300)^0.5)

= 0.2431.

3. If a bearing has a Weibull distribution for its time failure (in hours) with the parameters γ = 0.5, α = 5000 and μ = 0. Then determine the probability of a bearing last for 6000 hours.
Solution. We know the formula for distributive function of x i.e., F(x) = 1 - exp^(-(x/α)^γ) 
= 1 - exp^((6000/5000)^0.5) 
= 0.666.

Q: Why do we use Weibull Distribution?

Q: Mention one example of Weibull Distribution?

Q: What is the formula of Inverse Weibull Distribution?

Q: Mention the Two-parameter Weibull Distribution formula.

Q: What is the importance of γ in the Two-parameter Weibull Distribution formula?