# Weibull Distribution

Definition 1: The Weibull distribution has the probability density function (pdf)

for x ≥ 0. Here β > 0 is the shape parameter and α > 0 is the scale parameter.

The cumulative distribution function (cdf) is

The inverse cumulative distribution function is I(p) =

Observation: If x represents “time-to-failure”, the Weibull distribution is characterized by the fact that the failure rate is proportional to a power of time, namely β – 1. Thus β can be interpreted as follows:

• A value of β < 1 indicates that the failure rate decreases over time. This happens if there is significant “infant mortality”, or where defective items fail early with a failure rate decreasing over time as the defective items are weeded out of the population.
• A value of β = 1 indicates that the failure rate is constant over time. This might suggest random external events are causing mortality or failure.
• A value of β > 1 indicates that the failure rate increases with time. This happens if there is an “aging” process; e.g. if parts are more likely to wear out and/or fail as time goes on.

1/α can be viewed as the failure rate. The mean of the Weibull distribution is the mean time to failure (MTTF) or mean time between failures (MTBF) = $\alpha \Gamma \! \left( \! 1+\frac{1}{\beta} \! \right)$.

Key statistical properties of the Weibull distribution are:

• Mean = $\alpha \Gamma \! \left( \! 1+\frac{1}{\beta} \! \right)$
• Median = $\alpha ({ln 2})^{1/\beta}$
• Mode (when β > 1) = $\alpha \! \left( \! \frac{\beta-1}{\beta} \! \right)^{1/\beta}$
• Variance = $\alpha^2 \Gamma \! \left( \! 1+\frac{2}{\beta} \! \right)- \mu^2$

Excel Function: Excel provides the following function in support of the Weibull distribution.

WEIBULL(x, β, αcum) where α and β are the parameters in Definition 1 and cum = TRUE or FALSE

WEIBULL(x, β, α, FALSE) = the value of the Weibull pdf f(x) at x

WEIBULL(x, β, α, TRUE) = the value of  the Weibull cumulative distribution function F(x) at x

Excel 2010/2013/2016 also provide the additional function WEIBULL.DIST which is equivalent to WEIBULL.

Example 1: The time to failure of a very sensitive computer screen follows a Weibull distribution with α = 1,000 hours and β = .6. What is the probability that the screen will last more than 5,000 hours? What is the mean time to failure?

The probability that the screen will last no more than 5,000 hours

= WEIBULL(5000, .6, 1000, TRUE) = 0.92767.

Thus, the probability that the screen will last more than 5,000 hours = 1 – 0.92767 = 7.2%

MTTF = αΓ(1+1/β) = 1000Γ(1+1/.6) = 1000*EXP(GAMMALN(1 + 1/.6)) = 1,504.575 hours

Example 2: If the mean time to failure for a component which follows a Weibull distribution is 1,000 hours with a standard deviation of 400 hours, what is the probability that the component will last more than 2,000 hours?

Thus

We now solve these equations for α and β. First we simplify the second equation and then we take the natural log of both sides of both equations to get

Eliminating α

which is equivalent to

The above equation takes the form h(β) = 0, which we solve using Excel’s Goal Seek capability by selecting Data > What If Analysis|Goal Seek and filling in the dialog box that appears as shown in Figure 2.

Once we obtain the value for β, we can calculate α using the equation

and then we can calculate the probability that the component will last more than 2,000 hours using the WEIBULL.DIST function.

Figure 2 – Goal Seek initial guess

After clicking on the OK button, the result is shown in Figure 3.

Figure 3 – Goal Seek results

The values for α and β are shown in cells B5 and B3. The probability that the component will last more than 2,000 hours is 0.91% (cell B6).

Real Statistics Function: Since Excel doesn’t provide an inverse function, you can use the following function provided by the Real Statistics Resource Pack instead.

WEIBULL_INV(p, β, α) = x such that WEIBULL.DIST(x, β, α, TRUE) = p; i.e. the inverse of  WEIBULL.DIST(x, β, α, TRUE)

### 86 Responses to Weibull Distribution

1. Colin says:

Sir
Are mean time to failure (MTTF) and mean time before failure (MTBF) the same?
I just heard of mean time to repair (MTTR), I haven’t heard of MTTF.

• Charles says:

Colin,
MTTF = the average time to the first failure. This measure is particularly important for items that can’t repaired (e.g. a typical light bulb).
MTBF = the average time between failure. This measure is used with repairable items.
Charles

2. belachew says:

I need more application by using weibull distribution on health related area.

• Charles says:

I’ll try to add an application of Weibull to healthcare area in the future.
Charles

3. Hassan Elsafty says:

I think that, in example 1 the value of Alpha is being Substituted with Beta , and this is really confusing for me , Could you honored confirm my assertion.

Thanks alot
Hassan Elsafty

• Charles says:

Hassan,
I just checked and it seems that everything is correct in Example 1.
Charles

• Evan says:

Something is still off. “Beta = 1000 hours”. According to the tutorial above, “beta” should be unit-less. “alpha” should have units of time (1/alpha is mean failure rate). Your use of “beta” and “alpha” in the tutorial is opposite to the use of “beta” and “alpha” in the Excel function.

• Charles says:

Evan,
Thank you for identifying this problem. Excel interchanges the roles of alpha and beta from what is used in other sources. Beta is commonly used as the shape (or slope) parameter and alpha as the scale parameter, although other symbols are used by various authors. I have now corrected the referenced webpage to be consistent with this choice of symbols, and so the Excel Weibull function becomes WEIBULL(x, beta, alpha, cum).
Charles

4. Lisa says:

thank you sir,
please would you like to give me how can i calculate alpha and beta from standard deviation and average using excel (target value).
Lisa

• Charles says:

Lisa,
The referenced webpage shows how to calculate the mean and variance (and therefore standard deviation) from the alpha and beta parameters. Theoretically you should be able to solve for alpha and beta if you know the values of the mean and standard deviation (two equations in two unknowns). Given that the equations involve the gamma function this may not be so easy to do analytically. You should be able to come up with approximate solutions by using Solver, however.
I don’t know what you mean by “target value”, however.
Charles

5. Siva says:

Hi,
Thanks for the above information.
I have a problem , in which I need to use the Weibull distribution to generate a few data points to simulate a systems. This basically means, given the MTTF, alpha and beta, I need to come up with various time instances at which the system faults occur and use these time instances to simulate the response of the system.
Could you please suggest a way to get these time “fault time instances” such that it follows a Weibull distribution with given parameters.
Any help is appreciated. Thanks in advance. 🙂

• Charles says:

Siva,

To generate a random data element which follows a Weibull distribution with fixed alpha and beta values you can use the Real Statistics formula

=WEIBULL_INV(RAND(),alpha,beta)

Alternatively, you can use the Excel formula

=alpha*(-LN(1-RAND()))^(1/beta)

Finally note that MTTF = alpha*Gamma(1+1/beta), and so if you have any two of the values alpha, beta and MTTF the third is determined. Thus, if you are given all three you need to make sure that the three values are compatible.

Charles

6. Lara says:

Hi, Sir.

How can I compute the values of shape and scale parameters given that I have the values of mean and standard deviation?

• Charles says:

Lara,
I have just added Example 2 to the referenced webpage. This example shows how to calculate the shape and scale parameters from the mean and standard deviation.
Charles

7. Mark Bahner says:

Hi,

I’m trying to analyze data in Table 4 of this report on electronic waste (e-waste) in the U.S.:
U.S. EPA, 2011

I’m trying to determine what Weibull scaling factor and shape factor would best describe the data in Table 4. For example, for portable computers, we have:

20% are discarded in 4 years
35% are discarded in 5 years
55% are discarded in 6 years, and
100% are discarded in 7 years.

What scaling factor and shape factor would come closest to those values? Just by trial and error and eyeballing, I came up with a shape factor of 4.1 and a scale factor of 5.8. With those values, I get cumulative discards in years 4 through 7 of:

Year 4 = 20% (versus 20%)
Year 5 = 42% (versus 35%)
Year 6 = 68% (versus 55%), and
Year 7 = 88% (versus 100%).

But I’d like to do something that’s quicker than trial and error, and hopefully matches Table 4 a little better. (Though I realize Table 4 may contain data that simply don’t follow a realistic Weibull distribution.)

Thanks!
Mark

• Charles says:

You can calculate the mean and standard deviation of the data that you have and then use the approach described in Example 2 of the referenced webpage to estimate the values of the shape and scaling factors.
Charles

8. Ricardo says:

Very good explanation, it helped me a lot! Thanks!

9. Ron says:

Hello,
I would like to run a Weibull analysis to determine how many cycles to run a selected sample set to. For example, if I have a hydraulic cylinder that is designed for 2000 cycles, how many cycles do I have to run a sample set of ‘X’ to have a confidence level of ‘Y’ with a β of ‘Z’? How would the Weibull equation be set up to perform this function. Thanks in advance.

• Charles says:

Ron,

1. I have a hydraulic cylinder that is designed for 2000 cycles.
[CZ] I assume that you mean that the MTTF = 2000. Is this correct?

2. How many cycles do I have to run a sample set of ‘X’ to have a confidence level of ‘Y’ with a β of ‘Z’?
[CZ] I understand that Y takes a value such as 95%, meaning that you want a 95% confidence interval. I assume that Z takes a value such as .6. Is this correct? I don’t understand what sort of values X takes; can you explain this better?

Charles

10. Mohd Radzian Abdul Rahman says:

Hi Charles ,

You wrote an interesting subject.

I am trying to find the excel formula for variance of the Weibull distribution given you have the alpha, beta, and mean. Can you share with us ?

Thanks.

• Charles says:

The formula for the variance from the alpha, beta, and mean µ can be found on the referenced webpage.
Charles

11. Nugroho says:

Sir
In variance equation (example 2) why you use 12/beta, instead of 2/beta, as it is told at “key statistical properties of the Weibull distribution” section?
Thanks alot
Nugroho

• Charles says:

Nugroho,
Thanks for catching this error. The value in the example should be 2/beta. In fact, there are a couple of errors in this example, which hopefully I have now corrected.
Thanks for your help in making the website more accurate.
Charles

12. Zach L says:

Charles, I’m confused about how you calculated the excel values in Figures 2 and 3. For the F(2000) calculation using the excel WEIBULL.DIST function, which takes WEIBULL.DIST(x, Alpha, Beta, Cumulative), you seem to have selected your Beta value (B3) for Alpha, and your Alpha value (B5) for Beta. Is something wrong there, or did you intentionally swap the two values?

Any clarification would be very appreciated!

-Zach

• Charles says:

Zach,
Excel swapped the usual definitions of alpha and beta. I just flipped then back. In any case, either way, you should have WEIBULL.DIST(x, scale, shape, cum).
Charles

• Zach L says:

ahhh, I see what you mean. Good old Microsoft…

Cheers,

-Zach

13. Zach L says:

Charles,
One more question: In the final example 2 equation, after you substitute for alpha, how are you simplifying ln(1,160,000)-2*ln(1,000) into ln(1.6)? I get ln(1,160,000)-2*ln(1,000) = .14842, but ln(1.6) = .470004.
I apologize in advance if I’ve made some simple arithmetic error, or if I’m forgetting some natural log rule… 🙂

Thanks,

Zach

• Charles says:

Hi Zach,

You are correct. It looks like I made a typing error. ln(1,160,000)-2*ln(1,000) should be ln(1.16), calculated as follows.

In general, ln(x)-2ln(y) = ln(x)-ln(y^2) = ln(x/y^2). Thus, ln(1,160,000)-2*ln(1,000) = ln(1,160,000/1,000^2) = ln(1,160,000/1,000,000) = ln(1.16).

Thanks for catching this. I will correct the referenced webpage shortly.

Charles

• Zach L says:

Glad I could help! Thanks for the explanation and quick response.

• Charles says:

Zach,
Your help is very appreciated. I have just corrected the referenced webpage. I hope I got it correct this time.
Charles

14. Gami Nasir says:

Hello Charles:
How can I solve the following problems in Excel using the “WEIBULL.DIST” function?

1) Given a Weibully distributed population with a shape parameter of 3.93 and a scale parameter of 151. 34.2% is the probability of being greater than what value?

2)What is the probability that a randomly selected item from a population having a Weibull distribution with a shape parameter of 3 and a scale parameter of 109.3 has a value between 85.6 and 97.5?

Thank you

Gami Nasir

• Charles says:

Gami,

1) WEIBULL(x, β, α, TRUE) = the probability that the distribution has a values less than or equal to x, where alpha is the scale parameter and beta is the shape parameter.

2) The probability that the distribution has a value between x1 and x2 is WEIBULL(x2, β, α, TRUE) – WEIBULL(x1, β, α, TRUE).

Charles

• Gami Nasir says:

Thank you very much Sir!

• Gami Nasir says:

Hello Dr. Charles,

For this problem:

Given a Weibully distributed population with a shape parameter of 4.66 and a scale parameter of 52. 12.6% is the probability of being greater than what value?

Using WEIBULL(12.6, 4.66, 52, TRUE) I get 0.00135

I am not sure how to get the correct amswer

Thanks

Gami

15. Gami Nasir says:

Hello Dr. Charles,

Happy Easter

For this problem:

Given a Weibully distributed population with a shape parameter of 4.66 and a scale parameter of 52. 12.6% is the probability of being greater than what value?

Using WEIBULL(12.6, 4.66, 52, TRUE) I get 0.00135

I am not sure how to get the correct amswer

Thanks

Gami

• Charles says:

Gami,
I think you are asking me, what is the value of x such that WEIBULL(x, 4.66, 52, TRUE) = 1 – .126.
You can try to find the value of 60.79592 by trial and error, guessing at the correct value of x.
You can also use the Real Statistics formula =WEIBULL_INV(1-.126,52,4.66).
Charles

• Gami Nasir says:

Perfect!!
The WEIBULL_INV worked like a charm

Thanks again

Gami

16. James Zouris says:

Is it possible to create a formula of the shape (a) and scale (B) parameters for a Weibull distribution to have a fixed mean but allow the user to change the variance?

For example Weibull(a,B) where EX=Fixed Number and VarX=d where d is all real numbers

I understand that the shape and scale parameters will change every time.

• Charles says:

James,
I don’t understand your question. Once you specify the shape and scale parameters the mean and variance are determine (as described on the referenced webpage). If you specify only the mean and variance you can calculate the shape and scale parameters as shown in Example 2; in this case the function would Weibull1(mean, var).
Charles

• James Zouris says:

Thank You. I finally figured it out. Your example was very helpful!

17. George says:

A widget has Weibull distribution with shape 2.0 and scale 700. There are three identical widgets in parallel in my system such that a minimum of one is required. Given that all widgets have survived 200 hours so far, need the probability that one of the three widgets fails in the next 3 hours, another of the remaining two widgets fails within 15 minutes of the preceding widget failure, and the third widget survives the entire 3 hours and 15 minutes? Thanks

• Charles says:

Sorry George,
This sounds like a homework assignment, and I have a rule that I won’t answer such questions.
Charles

• George says:

Not a homework assignment. Its a real problem and a proposed solution is Prob(one failure in next 3 hours given survived 200 hours) * Prob(one failure in next 3 hours and 15 minutes given survived 200 hours) * Prob(no failure in next 3 hours and 15 minutes given survived 200 hours) * a multinomial coefficient of 6. The result seems unattractive compared to no failures at all which is modeled as (Prob(no failure in next 3 hours and 15 minutes given survived 200 hours))^3. I believe that the two failures should be calculated as the probability of two failures within 15 minutes of each other given both survived 200 hours, but do not know how to model such, or how to then incorporate the non failure.

• Charles says:

George,

Let’s try to tackle the first of these problems, namely that “Given that all widgets have survived 200 hours so far, need the probability that one of the three widgets fails in the next 3 hours”.

We begin by tackling the simpler problem that “Given widget 1 has survived 200 hours so far, what is the probability that it fails in the next 3 hours?” Let A = widget 1 has survived 200 hours so far and B = widget 1 fails in the next 3 hours. You want P(B|A), which by Bayes’ theorem is equal to P(A and B) / P(A). Now P(A) = 1-WEIBULL(200,2.700,TRUE) = 1-.07839 = .92161. You can use WEIBULL.DIST instead of WEIBULL for versions of Excel after 2007. P(A and B) = WEIBULL(203,2.700,TRUE)-WEIBULL(200,2.700,TRUE) = .08066 – .07839 = .00227. Thus, P(B|A) = .00227/.92161 = .002464. This also means that given widget 1 has survived 200 hours so far, the probability that it does not fail in the next 3 hours is 1 – .002464 = .997536.

Assuming that the survivability of all the widgets are independent of each other, then the same calculation holds for widgets 2 and 3 as well. Since we want exactly one widget to fail and the other two to not fail, there are C(3,1) = 3 ways to choose the failing widget and so given that all widgets have survived 200 hours so far, the probability that exactly one of the three widgets fails in the next 3 hours = 3 * .002464 * .997536^2 = .007356.

Assuming I haven’t made a mistake in logic or arithmetic, this should take care of the first problem. I’ll let you handle the others.

Charles

• George says:

Thanks

18. daniel says:

hello…

how we can calculate the parameter for weibull distributions. can someone help me

• Charles says:

Daniel,
You can use the approach shown in Example 2 of the referenced webpage.
Charles

19. daniel says:

tq charles for reply my email..

during my reading, i have got some idea that to find the parameter of weibull distribution, we can find it by several methods such as maximum likelihood estimation MLE, modified maximum likelihood estimation (MMLE), energy patterm method (EPF) and etc…

my question is, how we can use that several methods in excel to find the parameter

• Charles says:

Daniel,
Sorry, but I have not implemented these techniques for the Weibull distribution. With as little work you could use Excel’s Solver to implement this. I have shown how to do this sort of thing a number of times on the website.
Charles

• daniel says:

tq mr charles…do you have any suggestion how to use the method..

• Charles says:

Daniel,
Is this a follow up on my reply from June 2016? If so, which method are you referring to? MLE?
Charles

20. bhuvan says:

I have monthly demand data for a certain product, it is following Weibull distribution (2 parameter,I calculated the 2 parameters). Now I want to know the 5 month distribution of demand based on the above information.

Does 5 month demand distribution also follows weibull, how to know that ?

21. Reinoud says:

Very nice and elegant. I compared your result of the Weibull parameter estimate from Mean and stdDev to the Weibull estimate function of .Net lib MathNet.Numerics.Distributions (also called Math.Net Numerics). Yours give a better estimate. I can’t paste pictures here, but if you drop me an email, I’ll email them to you.
Rgds, Reinoud

• Charles says:

Reinoud,
Charles

22. Deepika says:

Hi Charles, the article was very informative. I’m, working on a project to revise the maintenance schedule for a machine. I believe the time between failures follows a Weibull distribution as the chance of failure increases with increase in time after each maintenance. The average time of failure with the current maintenance schedule (every 700 hours) is 500 hours. So, am suggesting a revision of the current schedule to do a maintenance at every 500 hours. I calculated the parameters for the Weibull distribution. What I’m stuck now is at how I can quantify my results financially. In other words, for the new maintenance schedule, I have to show that the average time of failure if higher than what it is for the current schedule. Any insights on how I should go about this?

• Charles says:

Deepika,
This sounds like a financial break-even analysis (and not a statistical analysis). You need to calculate the cost for each scheduled maintenance and then calculate the total cost over a sufficiently long period of time (this could be one year) based on whatever MTTF values you want to study. You then need to estimate the cost of one failure (in terms of repair costs, lost productivity, impact of the rest of the operation, etc.) and then based on the average number of failures per year (for each MTTF estimate) the total costs for failure.
Charles

23. Pedro says:

WEIBULL.DIST(x, β, α, TRUE) is only working with x=integer. Does anyone know a way to include numbers with fractional component? Regards, Pedro

• Charles says:

Pedro,
That is strange. When I use WEIBULL on my computer it handles non-integer values of x.
Charles

• Pedro says:

Hi Charles, thanks for your response.
More precisely, I am building a weibull distribution (no cumulative) with the WEIBULL.DIST(x; 3; 9.46; FALSE). First I let x goes from 1 to 25 (only integers) and so my distribution looks good when I sum each one of the outputs of the above function (sum = approx 100%). The problem is now that I need to plug franctions into the x variables. For instance, now I need to build the same distribution with an unit variation of 0,5 so that inserting 1, then 1.5, then 2, then 2.5, then 3, then 3.5…until 25,5. The sum of this distribution is now a number close to 200% and so an incorrect result. Is there anyway to overcome this situation? I tried to avoid the excel function applying the direct formula but the result remains incorrect. For instance; (beta/(alpha/0,89))*((x/(alpha/0,89))^(beta-1))*(EXP(-((x/(alpha/0,89))^beta))). I really appreciate your valuable thought on that.
Best regards,
Pedro (congratulation for this website I have gone through a very fruitful readings)

• Charles says:

Pedro,
If you have a discrete distribution, then you can sum pdf values as you have described. You can’t do this with a continuous distribution such as the Weibull distribution. Why do you want to do this sort of summing anyway?
Charles

• Pedro says:

Hi Charles,
Understood. The weibull pdf is for the wind distribution and I was trying to insert x with 0.5 unit because that’s the way that the turbine supplier is giving to me the power coefficient curve (so weibull distribution times 8760 hours in a year times the power curve will result in the annual energy production). I can do the job with the intergers only but the result should not be the same. While I am writing I am considering running a monte carlo like =(α*(-LN(1-Rand()))^(1/β))+min. I have to figure out the best way to do format that but this is a possibility. Do you have any advice?
Best regards,
Pedro

• Charles says:

Pedro,
Since I don’t understand the situation well enough, I am not able to comment further.
Charles

24. Shane Devenshire says:

Pedro,

My response is probably displays my ignorance, but… If I understand how Excel is handling the pdf calculation, it needs to give a value to the width at the points in question, since otherwise the pdf evaluates to 0 at any point, so Excel assigns a value of 1. You are breaking the integers into 1/2 but the area at each point continues to be assumed to have a width of 1. So, if you divide by 2 your answers will work out. If you breakup the range into 10ths then divide the results by 10 and so on.

25. Tuna Bulunmaz says:

Hi,

I have some measurement data depends on shoot.I want to identify an upper limit for data and estimate to shoot for upper limit.
Data don’t increase/decrease depends on time.

Have you any suggestions?

• Charles says:

Sorry, but I don’t understand your question.
Charles

26. TJ says:

This is only working for me if I swap alpha and beta between WEIBULL.FIT (or FITM or FITR) and WEIBULL.DIST. It appears that WEIBULL.FIT calls beta the shape parameter, but WEIBULL.DIST considers alpha the shape parameter.

• Charles says:

TJ,
Yes, this is true. This is why I reversed the roles of alpha and beta in the description of WEIBULL and WEIBULL.DIST on the referenced webpage.
Charles

27. Roger says:

The life of an X-ray machine in hours follows the Weibull distribution, with β = 3 and η=1000 hours.

Determine:

a) The mean and standard deviation of the life of the machine

b) The probability that the X-ray will fail before 250 hours.

• Charles says:

Roger,
How to solve these problems is described on the referenced webpage.
Charles

• Aoife Skelly says:

Roger, did you find a solution to your question? I would like to have a solution for it as well.

28. Miguel Hidalgo says:

Hi Charles, How are you?

Charles, I utilized 3 different methods to model the Weibull (two from your site) and they give me relatively similar information MTTF of 20072, 21062, 23400 hours(MLE, MOM, MedRR with R2=0.92) with different parameters. All Betas >1
My question, I want to determine which one has the smallest MSE or RMSE with respect to my data. How do you get the predicted Ys so I can compare with the data (or in this case X=hours)?

• Charles says:

Hi Miguel,
To get the “predicted” y values just use the WEIBULL.DIST function with the estimated alpha and beta values.
Charles

• Miguel says:

Charles, thanks for the response. I used the Weibull with the data and estimated parameters to generate the Cum Prob, then I used the Weibull_Inv and generated the predictive Y’s.

Charles, I have a couple of questions related to what to do after modeling the Weibull for equipment failure.

From a population of 352 machines, 62 of them failed this year (~18%) and by modeling the Weibull distribution on the 64 failed machines, I estimated the MTTF around 16850 hours. If I understood correctly, after viewing a histogram, this tells me that 50% of the machines failed by the time the machines reached 16850 hours of operations. Some of these machines failed around 10600, 14700, 15835, 16083,…,, 18907 hours. Of the 62 that failed, 31 reached or passed the MTTF.

Of the 352 machine, 80 reached or passed the estimated MTTF (~23%), 31 failed (~39%) and 49 others did not failed after passing the MTTF.

If I want to estimate budget for next year maintenance/repair, I would estimate the hours of operations of all machines to determine which ones will reach or pass the MTTF. Assuming 11 will reach the MTTF + the 49 that already reached the MTTF, therefore 60 machines will reach the MTTF.

So, for what % of machines I budget?
– For the 18% of the population (proportion that failed last year)?
– For the 39% of the 60 that will reach the MTTF+ 9% of the total population (50%of the machines that failed before MTTF last year)?
– Any suggestions?

Thanks,

• Charles says:

Miguel,
1. MTTF means mean time to failure. If MTTF = 16850, this doesn’t mean that half the machines will fail in less than this time and half will fail after. That would be the median time to failure.
2. Since there are 8760 hours in a year, after one year, you should expect that not all 352 machine would fail in the first year. If the shape parameter < 1, then you would expect that the failure rate would decrease over time and so you would have fewer failures next year than this year. The reverse is true if the shape parameter > 1.
3. Regarding how many repairs you should budget for next year, first you need to decide whether you budget for the average number of failures in year 2 or allow for some cushion in case this number is higher than the average. You can also decide that you will budget for the median number of failures in year 2.
4. You also need to decide what happens with machines that fail in the first year. Do they get taken out of service or get repaired. If they get repaired does their MTTF change?
5. Let’s give an example. Suppose alpha = 1,000 hours and beta = .6 (as in Example 1 on the referenced webpage). Then MTTF = 1504.575. Now suppose you have 1000 screens assume that when a screen fails it is simply taken out of service. Now let’s estimate the number of failures in month 2, assuming each month has 30 days (or 720 hours). There are The probability that a screen will last through month 1 is =1-WEIBULL(720,.6,1000,TRUE), which is .44. Thus of the 1000 screens we expect that 440 will still be running after one month. The probability that a screen will last through month 2 is =1-WEIBULL(1440,.6,1000,TRUE), which is .288. Thus, we expect that 288 of the 440 screens that were running after one month are still running after two months, and so on average 440-288 = 152 fail during the second month.
Charles

• Miguel says:

Charles, I have another question :

I have data for 18 compressors with 8 cylinders per compressor. A total of 144 Cylinders/Pistons. During the last year they were overhauled and in accordance to the manufacturer their average life is 10000 hrs.

18 of the cylinders failed with hours between 2200 to 13250. I model the Weibull (With 3 methods as you instructed us in your site/ MRR/MLE/MOM). All of them provide similar parameters. Using MLE, B=2.32, Alpha=7492, MTTF=6638.

Now, since B>1, it means the failure rate is increasing (Wear-out phase) even though they were overhauled during last year, do I suppose to use the Beta/Alpha from the Weibull estimated or I suppose to ignore that and use B=1 or less and assume they are at the increasing or constant failure rate? I asked because when comparing other colleagues using Minitab, their Reliability curve and MTTF are much larger.

Thanks again for sharing your knowledge!

• Charles says:

Miguel,
Please explain what the justification is for ignoring the beta values of 2.32 and using a beta values less than or equal to one.
Charles

29. Kuldeep Jain says:

Hello Charles,
i have plotted a weibull for the failure received and want to see the predicted trend. i have calculated the weibull parameter as suggested at https://www.qualitydigest.com/magazine/1999/jan/article/using-microsoft-excel-weibull-analysis.html#
and now i have some question which i am not able to find anywhere. i can sample excel if this helps.
1) As we have processed the data to be represented by linear equation by taking natural logs and got the parameter for this linear equation. Does the slope need to calculated for its confidence interval.
2) We calculated weibull plot CDF and 1-CDF is plotted on chart as reliability. How to plot the confidence interval and prediction interval.
3) i also tried plotting confidence interval and prediction interval for linear equation using t critical value but these are very narrow and just follow the trend-line. i have doubt that i am doing it correct?

• Charles says:

Jain,
I believe that the article is describing the procedure that I have described on the following webpage:
http://www.real-statistics.com/distribution-fitting/fitting-weibull-regression/
1. You don’t need to calculate a confidence interval for the slope, but you can do so. This is described on the referenced article and is the usual confidence interval for linear regression, but with a reverse transformation, undoing the ln(x) by taking exp(x).
2. To plot the confidence or prediction intervals for linear regression involves creating a two additional scatter plots, one for the upper limits and one for the lower limits. Thus you have three scatter plots on the same chart: one for the data, one for the upper limits and one for the lower limits. You can create a table with four rows: (1) x data, (2) y data, (3) upper limit for predicted y, (4) lower limit for predicted y; the plots are x data vs. each of the three types of y values. I don’t know whether this is the same thing as the CDF and 1-CDF charts that you are referring to.
3. The scatter plots described in item #2 should look something like that shown in Figure 1 on the following webpage
http://www.real-statistics.com/regression/confidence-and-prediction-intervals/
Charles

• Kuldeep Jain says:

Thanks for the response.
For linear regression i could plot the confidence bounds. Thanks however as asked in earlier question ” We have calculated weibull plot CDF and 1-CDF is plotted on chart as reliability. How to plot the confidence interval and prediction interval.” i do not find a clear article on this point as not the data is processed to plot the weibull. is there any sample workbook on internet using excel?

2) Another doubt is that the article i have used is using Benard’s approximation as =(X-0.3)/(N+.4) while in your article it is =(X-0.5)/N. Which one is more accurate and why ?

• karthi says:

I need your help to do Weibull analysis in our product

• Charles says:

Karthi,
What sort of help do you need?
Charles

• Charles says:

Kuldeep,

Based on all the questions I have been receiving lately about the Weibull distribution I plan to add more details to the website and Real Statistics software in the next release. For now, let me address your questions:

1. You can create a range with three rows: one for the predicted values, one for the lower end of the confidence interval and one for the upper end of the confidence interval. You then highlight this range and use Excel’s Line chart capability
2. I don’t see any reference to (X-.5)/N on this webpage. Where are you looking at? Bernard’s approximation is commonly used and I will address this issue in the next release.

Charles

Charles

• Kuldeep Jain says:

Thanks,
You have used (X-.5)/N as “=(A4-0.5)/A\$15 in cell E4” on http://www.real-statistics.com/distribution-fitting/fitting-weibull-regression/

i will try the plot the line as suggested and will revert if needed. thanks

• Charles says:

Kuldeep,
Thanks for the clarification. I am in the process of using the Benard’s approximation and also implementing Weibull distribution fitting even when there is censored data. I will try to also provide an example of how to graph the prediction intervals as well. These capabilities will be added when I issue the next software release.
Charles