Your Reliability Engineering Professional Development Site
by Fred Schenkelberg Leave a Comment
Define and use terms such as population, parameter, statistic, sample, the central limit theorem, etc., and compute their values.
This lesson takes a close look at a few more details and concepts.
The Law of Large Numbers and the Gambler’s Fallacy (article)
by Fred Schenkelberg Leave a Comment
by Fred Schenkelberg Leave a Comment
by Fred Schenkelberg Leave a Comment
Define and describe homogeneous and non-homogeneous Poisson process models (HPP and NHPP)
One more test you may want to run when considering repairable system data analysis.
by Fred Schenkelberg 2 Comments
Define and describe homogeneous and non-homogeneous Poisson process models (HPP and NHPP)
Do you have a trend or not?
Laplace’s Trend Test (article)
by Fred Schenkelberg Leave a Comment
Define and describe homogeneous and non-homogeneous Poisson process models (HPP and NHPP)
Lets check a few assumptions when using HPP or NHPP.
Mann Reverse Arrangement Test (article)
by Fred Schenkelberg Leave a Comment
Define and describe homogeneous and non-homogeneous Poisson process models (HPP and NHPP)
Now a bit more complex model for repairable system data.
Non-Homogeneous Poisson Process following a Power law (NIST Engineering Statistics Handbook)
Non-Homogeneous Poisson Process following an Exponential law (NIST Engineering Statistics Handbook)
by Fred Schenkelberg Leave a Comment
Define and describe homogeneous and non-homogeneous Poisson process models (HPP and NHPP)
This lesson discusses a few more terms, this time related to Poisson process models.
by Fred Schenkelberg 1 Comment
by Fred Schenkelberg Leave a Comment
Define and describe homogeneous and non-homogeneous Poisson process models (HPP and NHPP)
This lesson starts with the basic homogeneous Poisson process (HPP).
Homogeneous Poisson Process (NIST Engineering Statistics Handbook)
by Fred Schenkelberg Leave a Comment
by Fred Schenkelberg Leave a Comment
Compare and contrast various distributions (binomial, Poisson, exponential, Weibull, normal, log-normal, etc.) and their functions (e.g., cumulative distribution functions (CDFs), probability density functions (PDFs), hazard functions), and relate them to the bathtub curve.
This lesson takes a close look at the common pattern a product or system reliability travels through over its life time.
Failure modes and mechanisms (articles)
Why the Drain in the Bathtub Curve Matters (article)
by Fred Schenkelberg Leave a Comment
Compare and contrast various distributions (binomial, Poisson, exponential, Weibull, normal, log-normal, etc.) and their functions (e.g., cumulative distribution functions (CDFs), probability density functions (PDFs), hazard functions), and relate them to the bathtub curve.
This lesson takes a close look at the discrete distributions commonly used in reliability engineering.
The Poisson Distribution (article)
Poisson Distribution Calculation (article)
Binomial Cumulative Density Function (article)
Binomial Probability Density Function (article)
Hypergeometric Distribution (article)
OC Curve with Hypergeometric Method (article)
OC Curve with Binomial Method (article)
1-19. An airline has a fleet of four-engine aircraft. Maintenance records indicate that, on average, an engine fails twice in 10,000 operating hours with normal preventive maintenance. Calculate the Poisson distributed probability that two or more engines on an aircraft will fail during a typical flight time of 7 hours.
(A) 0.001398
(B) 0.001399
(C) 0.0000009786
(D) 0.0000009791
(D) 0.0000009791
Use the Poisson as directed in the question and calculate the probability of exactly zero or one failure, then subtract the combined values from one to find the chance of 2 or more failures.
The expected number of failures, mu, over 7 hours given a failure rate of 2 per 10000 hours is ( 2 /10,000 ) * 7 = 0.0014.
P(0,0.0014) = (exp[-0.0014 * 0.0014^0) / 0! = 0.99860098
P(1,0.0014) = (exp[-0.0014 * 0.0014^1) / 1! = 0.00139902
For two or more we subtract the chance of zero and one from one (compliment) 1 – 0.99860098 – 0.00139902 = 0.0000009791
1-33 A manufacturing plant operates 9 units, at least 7 of which must be operating for production volume requirements to be met. If there is a .23 probability that a malfunction will occur for any particular unit, what is the probability that 7 units can remain operating throughout the day?
(A) 0.1605
(B) 0.1628
(C) 0.3510
(D) 0.4960
(C) 0.3510
My first thought was to use the k-out-of-n redundancy formula. I think it’s possible to use that formula if we convert the probability of failure, 0.23, to a reliability. The formula reduces to the binomial though and adds an extra step. So, let’s just use the binomial directly.
We need at least 7 units to operate of the 9. We can calculate the probability that none fail, x = 0, and the probability that exactly 1 unit will fail. Then add those together, which is the probability that 7 or more will operate for the day of production.
$$ P\left( 0,9,0.23 \right)=\left( \begin{array}{*{35}{l}}9 \\ 0 \\ \end{array} \right){{0.23}^{0}}{{\left( 1-0.23 \right)}^{9-0}}=0.0952$$
and for exactly one unit failing
$$ P\left( 1,9,0.23 \right)=\left( \begin{array}{*{35}{l}}9 \\1 \\\end{array} \right){{0.23}^{1}}{{\left( 1-0.23 \right)}^{9-1}}=0.2558$$
then add the two results to find the probability of at least 7 units operating for a day. 0.0952 + 0.2558 = 0.3510
by Fred Schenkelberg Leave a Comment
Compare and contrast various distributions (binomial, Poisson, exponential, Weibull, normal, log-normal, etc.) and their functions (e.g., cumulative distribution functions (CDFs), probability density functions (PDFs), hazard functions), and relate them to the bathtub curve.
This lesson takes a close look at the continuous distributions commonly used in reliability engineering.
Interpolation within Distribution Tables (article)
Reading a Standard Normal Table (article)
The Normal Distribution (article)
Lognormal Distribution (article)
Calculating Lognormal Distribution Parameters (article)
The Exponential Distribution (article)
Using The Exponential Distribution Reliability Function (article)
Weibull Distribution (article)
Calculate Weibull Mean and Variance (article)
1-18. Which distribution is used to describe the time between failures that occur independently at a constant rate?
(A) exponential
(B) gamma
(C) lognormal
(D) Weibull
(A) exponential
The give away is the term “constant”. While other distributions can model time between failures at a constant failure rate, the most common and this is the primary characteristic of the exponential distribution.
1-22. What is the approximate reliability at the mean time to failure for the exponential model?
(A) 34%
(B) 37%
(B) 50%
(C) 67%
(B) 37%
The reliability function of the exponential distribution is
$$ R\left( t \right)={{e}^{-\frac{t}{\theta }}}$$
Setting t = θ we have
$$ R\left( \theta \right)={{e}^{-\frac{t}{\theta }}}={{e}^{-1}}=0.3679$$
1-27. Consider a Weibull distribution. What is the scale parameter, as a characteristic of time to failure, as a percentile of the distribution?
(A) 31.6
(B) 36.7
(C) 63.2
(D) 63.3
(C) 63.2
“β is the shape parameter and η is the scale parameter, or characteristic life —it is the life at which 63.2% of the population will have failed.” (Practical Reliability Engineering, 5th ed., p. 38.)
1-28. A test shows four failures in 40 hours of operation. If the failure rate is constant, how many failures will the test show in 800 hours of operation?
(A) 4
(B) 8
(C) 80
(D) 160
(C) 80
The failure rate is constant, and assumed constant over the entire time of operation. Thus the chance to fail in any given hour of operation is 4 / 40 = 0.10. Given the assumption of a constant failure rate, then over 800 hours we would expect 10% of the hours to have a failure, thus 80 failures. This is not a question about the probability of surviving over 800 hours, instead how many failures will occur over 800 hours. We treat each hour as a separate chance of failing, as if we have 800 units running for one hour each.
1-29. A trans-African safari is to be made using a special custom-made four-wheeled vehicle equipped with five tires. The probability of failure for each tire on the safari follows a binomial distribution and is estimated to be 0.4. Calculate the probability that the safari can be completed successfully with the five available tires?
(A) 0.1296
(B) 0.2592
(C) 0.3370
(D) 0.4752
(C) 0.3370
This one may trip you up if you assume only four tires at at risk at a time assuming the spare is installed after the first failure. The question is worded such that there is a 0.4 chance of failure over the duration of the trip for all 5 tires on the specially equipped vehicle. In short the spare has the same chance of failure whether or not it is in use.
Since there are only four working tires on the four-wheeled vehicle if there are two failures we don’t make it (stranded). We can use the Binomial distribution PDF function to solve this by calculating the probability of exactly 0 failure and 1 failure, then sum those probability to get the chance of a successful trip (not stranded). The binomial pdf is
$$ P\left( x,n,p \right)=\left( \begin{array}{l}n\\x\end{array} \right){{p}^{x}}{{\left( 1-p \right)}^{n-x}}$$
where x is exact number of failures of the number, n, of tires, here n=5, and p is the probability of failure, p = 0.4.
First let’s calculate the probability of having none of the five tires fail, x = 0.
$$ P\left( 0,5,0.4 \right)=\left( \begin{array}{l}5\\0\end{array} \right){{0.4}^{0}}{{\left( 1-0.4 \right)}^{5-0}}=0.0778$$
next when there is one of the five tires failing, x = 1
$$ P\left( 1,5,0.4 \right)=\left( \begin{array}{l}5\\1\end{array} \right){{0.4}^{1}}{{\left( 1-0.4 \right)}^{5-1}}=0.2592$$
The sum of the probability of zero or one failed tires is the probability of successfully completing the trip, 0.0778 + 0.2592 = 0.3370
1-30. An earthquake prediction network has been determined to have a mean time to failure of a constant 130 hours. Calculate its reliability at t = 135 hours?
(A) 0.354
(B) 0.368
(C) 0.632
(D) 0.646
(A) 0.354
The key word here is “constant”, thus we should use the exponential distribution. The exponential distribution reliability function is
$$ R\left( t \right)={{e}^{-\frac{t}{\theta }}}$$
and setting θ = 130 and t = 135, we find the reliability at 135 hours as
$$ R\left( 135 \right)={{e}^{-\frac{135}{130}}}=0.354$$
1-36. Which of the following probability distributions is continuous?
(A) binomial
(B) hypergeometric
(C) Poisson
(D) Weibull
(D) Weibull
This a classification or terminology problem. Binomial, Hypergeometric and Poisson are useful with count data, thus consided discrete distributions. Weibull is useful with time, length, cycles or other continuous datasets thus classified continuous.
1-43. If Z is a continuous random variable with a density distribution of 1 ≤ Z ≤ 5, what is the probability that Z = 4.0?
(A) 0.00
(B) 0.20
(C) 0.30
(D) 0.40
(A) 0.00
This sort of a trick question. For a continuous distribution the chance of any one specific (exact) value existing approaches zero. We use small intervals or less then/greater than statements to say something meaningful about a continuous distribution.
For example, we may be interested in the percentage of males taller than 2 meters based on a sample of 100 people. The chance that someone is exactly, precisely, 2 meters tall is very, very small given there is an infinite set of values between 1.99 meters and 2.01 meters tall on a continuous scale. Of course, we are not able to measure to the absolute precision this implies, thus within our measurement capability, which is a small range (say 1 mm wide) we do have a finite probability of someone being 2 meters tall, within 1 mm.
by Fred Schenkelberg Leave a Comment
Compare and contrast various distributions (binomial, Poisson, exponential, Weibull, normal, log-normal, etc.) and their functions (e.g., cumulative distribution functions (CDFs), probability density functions (PDFs), hazard functions), and relate them to the bathtub curve.
This lesson takes a close look at the four common functions used with statistical distributions.
The Four Functions (article)
Reliability Function (article)
Reliability from Hazard step function (article)
PDF to CDF with Brief Calculus Refresher (article)
1-12. Identify which of the following are normally accepted reliability engineering tools.
I. probability density function
II. hazard rate function
III. conditional reliability function
IV. mean life function
V. cumulative distribution function
(A) I and II only
(B) I, II, and III only
(C) II, III, IV, and V only
(D) I, II, III, IV, and V
(D) I, II, III, IV, and V
All five terms represent either ways to describe distributions (PDF, hazard rate, CDF, and conditional reliability) or the mean of a set of life data.
1-20. The hazard rate function h(t) as a function of time t in hours for a device is
h(t) = 0, for t ≤ 0 and 0.66t, for t > 0
Determine the reliability of this device at t = 3 hours?
(A) 0.0.00263
(B) 0.0.0513
(C) 0.138
(D) 0.372
(B) 0.0.0513
Given a distributions PDF you can derive the reliability function with
$$ R(t)={{e}^{-\int_{-\infty }^{x}{h\left( \tau \right)d\tau }}}$$
Then substitute the given hazard rate function and simplify
$$ R(t)={{e}^{-\int_{0}^{t}{0.66\left( \tau \right)d\tau }}}={{e}^{-0.66\int_{0}^{t}{\left( \tau \right)d\tau }}}$$
We can move the constant out of the integral due to the Constant Multiple Property, the solve the integral
$$ R(t)={{e}^{-0.66\int_{0}^{t}{\left( \tau \right)d\tau }}}={{e}^{-0.66\left( \frac{{{t}^{2}}}{2} \right)}}={{e}^{-0.33{{t}^{2}}}}$$
Finally, solve when t = 3
$$ R(3)={{e}^{-0.33{{\left( 3 \right)}^{2}}}}=0.0513$$
The distractors are answers when the integral is done incorrectly a few different ways.
Ask a question or send along a comment.
Please login to view and use the contact form.