### II. Probability and Statistics for Reliability

A. Basic concepts

## 3. Discrete and continuous probability distributions (Analyze)

*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.

## Additional References

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)

## Quick Quiz

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, 5^{th} 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.

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