Chapter 3

Probability Distributions

In this chapter, characteristics of discrete probability distributions such as binomial and poisson as well as common continuous probability distributions such as exponential, weibull, normal and lognormal are discussed [1-6].

These distributions are needed to formulate:

• The probability that a component will start (fail) on demand.
• The probability that a component will run for a period of time given a successful start.
• The impact of repair on these probabilities.
• The frequency of initiating events.

In general, it can be defined that:

• P[failure to start on demand] q unavailability
• P[successful start on demand] p availability
• Requirement: p + q = 1

3.1 Binomial

• Binomial probability distribution can be applied to an “experiment” that can have only two outcomes: “success” and “failure” or {0, 1} with probabilities p and q, respectively.
• Consider N “trials,” i.e., repetitions of this experiment with constant q. These are called Bernoulli trials.
• Define a new Discrete Random Variable (DRV): X = number of 1’s in N trials
• Sample space of X: {0,1,2,…,N}
• The probability that there will be k 1’s (failures) in N trials is

(3.1)

• This is the probability mass function (PMF) of the Binomial Distribution.
• It is the probability of exactly k failures in N demands.
• The binomial coefficient is:

(3.2)

• Mean number of failures:

(3.3)

• Variance:

  (3.4)

  Normalization:

  (3.5)

  P[at most m failures] =

  (3.6)

  Example: 2-out-of-3 system
• We found that the structure function is (using rain cut sets): XT = (XAXB+XBXC+XCXA) − 2XAXBXC
• The failure probability is P(failure) = P(XT=1) = 3q2 − 2q3
• Using the binomial distribution:

3.2 Poisson

• Used typically to model the occurrence of initiating events.
• DRV: number of events in (0, t)
• Rate is constant; the events are independent.
• The probability of exactly k events in (0, t) is (PMF):

(3.7)

k! 1*2*…*(k-1)*k 0! = 1

(3.8)

(3.9)

Example of the Poisson Distribution

• A component fails due to “shocks” that occur, on the aver age, once every 100 hours. What is the probability of exactly one replacement in 100 hours? Of no replacement?
• λt =10-2*100 = 1
• Pr[1 repl.] = e−λt = e−1 = 0.37 = Pr[no replacement]
• Expected number of replacements: 1

Pr[2repl] =

Pr[k≤2] = 0.37 + 0.37 + 0.185 = 0.925

Failure while running

• T: the time to failure of a component.
• F(t) = P[T < t]: failure distribution (unreliability)
• R(t) 1-F(t) = P[t < T]: reliability
• m: mean time to failure (MTTF)
• f(t): failure density, f(t)dt = P{failure occurs between t and t+dt} = P [t < T < t+dt]

3.3 Exponential

(3.10)

(3.11)

(3.12)

• h(t) = constant (no memory; the only pdf with this property) with useful life on bathtub curve

(3.13)

(3.14)

Example: 2-out-of-3 system

Each sensor has a MTTF equal to 2,000 hours. What is the unreliability of the system for a period of 720 hours?

• Step 1: System Logic.
  
• Step 2: Probabilistic Analysis.

For nominally identical components:

P(XT) = 3q2 − 2q3

But q(t) = 1 − e−λt = F(t) with λ = 5×10−4 hr−1

System Unreliability: FT(t) = 3(1− e−λt)2 − 2(1 − e−λt)3

If

For λ = 5×10−4 hr−1 and t = 720 hrs

q(t) = 1 − e−λt = 0.30 FT(720) = 3 × 0.302 − 2 × 0.303 = 0.22

Since λt = 0.36 > 0.1 the rare-event approximation does not apply.

Indeed, FT ≅ 3(λt)2 − 2(λt)3 FT(720) ≅ 3×0.362 − 2×0.363 = 0.295 > 0.2

3.4 Weibull

The Weibull instantaneous failure rate or hazard function is given by:

(3.15)

where:

t0 determines the position of the origin

(R(t0) = 1 at t = t0

η is the scale parameter

β is the shape parameter

(3.16)

If we assume τ = (t − t0)/η,
where τ is normalized time, then:

3.5 Normal

Probability density function (PDF) of normal distribution can be defined as below:

(3.17)

Where:

Mean: μ Standard Deviation: σ
Standard Normal Variable:

(3.18)

Standard Normal Distribution:

(3.19)

Example of the normal distribution:

• μ = 10,000 hr (MTTF) = 1,000 hr
• Pr[X > 11,000 hr] = Pr[Z > 1] = 0.50 − 0.34 = 0.16

An Example

A capacitor is placed across a power source. Assume that surge voltages occur on the line at a rate of one per month and they are normally distributed with a mean value of 100 volts and a standard deviation of 15 volts. The breakdown voltage of the capacitor is 130 volts.

i. Find the mean time to failure (MTTF) for this capacitor.

λSV = 1 per month

Pd/sv = conditional probability of damage given a surge voltage = P (surge voltage > 130 volts/surge voltage) =

= P(Z > 130-100/15) = P(Z > 2) =

= 1 − P(Z < 2) = 1 − 0.9772 = 0.0228

Therefore, the rate of damaging surge voltages is

λd = λSV x Pd/sv= 2.28 × 10−2 per month

Equivalently, the capacitor’s failure time follows an exponential distribution with rate

MTTF = 43.86 months

ii. Find the capacitor’s reliability for a time period of three months. R(3 mos) = exp(− λd X 3) = exp(−2.28 × 10−2 x 3) = 0.934

3.6 Lognormal

(3.20)

(3.21)

(3.22)

(3.23)

(3.24)

Error Factor:

(3.26)

Relationship with the Normal Distribution

If λ is a lognormal variable with parameters μ and σ,

then:

is a normal variable with parameters μ (mean) and σ (standard deviation).

The 95th percentile

Since Y is a normal variable, its 95th percentile is

An example of the Lognormal Distribution:

Suppose that μ = −6.91 and σ = 1.40

• Median: λ50 = exp(−6.91) ≅ 10−3
• Mean: m = exp(−6.91 + 1.402/2) = 2.65×10−3
• 95th percentile: = λ95 exp(−6.91 + 1.645×1.40) ≅ 10−2
• 5th percentile: λ05 = exp(−6.91 − 1.645×1.40) ≅ 10−4
• Error Factor: EF = 10

3.7 Mean Time To Failure (MTTF)

Reliability (R) is the probability that the product continues to meet the specification, over a given time period, subject to given environmental conditions.

Unreliability (F) is the probability that the product fails to meet the specification.

R(t) + F(t) = 1

MTTF for non-repairable and repairable items are presented [7].

Non-repairable items

Assume that all N items fail during a test interval T and the ith failure occurs at time Ti

Ti is the survival time or up time for the ith failure.

Mean time to failure (MTTF):

(3.26)

The mean failure rate :

(3.27)

For the continuous function R(t):

(3.28)

Repairable items

The down time TDj associated with the jth failure is the total time that elapses between the occurrence of the failure and the repaired item being put back into normal operation.

Mean down time (MDT):

(3.29)

(3.30)

The mean up time or the mean time between failures (MTBF) is therefore given by:

(3.31)

The mean failure rate is given by:

(3.32)

References

1. Lewis, E. E., Introduction to Reliability Engineering, Second Edition, John Wiley & Sons, Ltd., 1994.

2. O’Connor, P.D.T. and Kleyner, A., Practical Reliability Engineering, Fifth Edition, John Wiley & Sons, Ltd., 2012.

3. Walpole, R.E., Myers, R.H., and Myers, S.L., Probability and Statistics for Engineers and Scientists, Sixth edition, Prentice Hall, 1998.

4. Montgomery, D.C. and Runger, G.C., Applied Statistics and Probability for Engineers, Second edition, John Wiley & Sons, Inc., 1999.

5. Ross, S.M., Introduction to Probability Models, Sixth edition, Academic Press, 1999.

6. Montgomery, D.C., Runger, G.C., and Hubele, N.F., Engineering Statistics, John Wiley & Sons, Inc., 1998.

7. Bentley, J., Introduction to Reliability...