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Plotting Probability Distributions
A machine cuts copper bars into lengths of 6 m. Using random
sampling, a quality control engineer measures 20 of these cut
lengths (in metres) as
5.97
5.99
6.00
6.01
5.97
5.99
6.00
6.01
5.98
5.99
6.00
6.01
5.98
5.99
6.00
6.02
5.98
5.99
6.00
6.02
Length (m), x:
Frequency:
Probability, f(x):
5.97
2
0.10
5.98
3
0.15
5.99
5
0.25
6.00
5
0.25
6.01
3
0.15
6.02
2 = 20
0.10 = 1.0
Cum. Prob, F(x):
0.10
0.25
0.50
0.75
0.90
1.00
Data Processing:
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Probability Distributions
Probability Distribution Function
Cum. Probability, F(x)
Probability, f(x)
Probability Density Function
0.30
0.25
0.20
0.15
0.10
0.05
0.00
5.97 5.98 5.99 6.00 6.01 6.02
Length (m), x
1.0
0.8
0.6
0.4
0.2
0.0
5.97 5.98 5.99 6.00 6.01 6.02
Length (m), x
2
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Example using Discrete Distribution
Total number of samples = N0
Number of failures at time t = Nf(t)
Number of survivors at time t = Ns(t) = N0 - Nf(t)
Number of failures in interval ∆t = ∆Nf (t)
Failure Density Function, f(t) = ∆Nf (t) / N0
Failure Distribution Function, Q(t) = Nf (t) / N0 (Probability of Failure)
Survivor Function R(t) = Ns (t) / N0
(Probability of Success)
Hazard Rate, λ(t) = f(t) / R(t)
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Example: Exponential Distribution
Find the mean time to failure of a component which has a failure rate of
2 failures per year. Calculate its reliability for different mission times,
e.g. 10, 1000, 10000 hours.
MTTF = 1/λ
λ = ½ = 0.5 yrs = 0.5 x 8760 = 4380 hrs
R(t) = e
−λt
R(10)=0.997719, R(1000)=0.795877, R(10000)=0.101967
1.0
0.8
Rt)
0.6
0.4
0.2f/yr
0.2
1f/yr
2f/yr
0.0
0
2
4
6
8
10
5
time in 1000 hrs
Example:
A simple electronic circuit consists of 6 transitors each having a failure rate of
10-6 f/hr, 4 diodes each having a failure rate of 0.5 x 10-6 f/hr, 3 capacitors each
having a failure rate of 0.2 x 10-6 f/hr, 10 resistors each having a failure rate of
5 x 10-6 f/hr and 2 switches each having a failure rate of 2 x 10-6 f/hr. Assuming
connectors and wiring are 100% reliable, evaluate the equivalent failure rate
of the system and the probability of the system surviving 1000 and 10000 hours
if all components must operate for system success.
n
Equivalent failure rate of the system, λe =
=
6(10-6 )
+ 4(0.5 x
10-6 )
+ 3(0.2 x
−λ t
− 6.26 x 10
Rs(1000 hr) = e e = e
10-6 )
−6
−6
i =1
+ 10(5 x 10-6 ) + 2(2 x 10-6 ) = 6.26 x 10-5 f/hr
x1000
−λ t
− 6.26 x10
Rs(10,000 hr) = e e = e
∑ λi
= 0.9393
x10000
= 0.5347
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Example:
Consider a system comprising of 4 identical units each having a failure rate of
0.1 f/yr. Evaluate the probability of the system surviving 0.5 years and 5 years
if at least two units must operate successfully.
Using Binomial Expansion,
[R(t) + Q(t)] 4 = R4(t) + 4 R3(t)Q(t) + 6 R2(t)Q2(t) + 4 R(t)Q3(t) + Q4(t)
where, R(t) =
e − λt
and
Q(t) = 1 - e − λt
For 2 out of 4 system,
Rs(t) = R4(t) + 4 R3(t)Q(t) + 6 R2(t)Q2(t)
− 4 λt
− 3 λt
− λt
− 2 λt
− λt 2
=e
+4e
(1 - e ) + 6 e
(1 - e
)
For t = 0.5 years, λt = 0.1 x 0.5 = 0.05
For t = 5 years,
Rs(0.5) = 0.9996
λt = 0.1 x 5 = 0.5
Rs(5) = 0.8282
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Normal Distribution: Example
What is the probability of electric lamps failing in the first 700 burning
hours, if the average life is 1000 burning hours with a standard deviation of
200 hours? Assume that the failure density function is a normal distribution.
µ = 1000 hr & σ = 200 hr
z = (x – µ) / σ
For x = 700,
z = (700 – 1000) / 200 = -1.5
From the Table:
F(z) = 0.4332 for z = 1.5
Area up to 700 hours = 0.5 – 0.4332 = 0.0668
which is the probability of failing in the first 700 burning hours.
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Example
The useful life failure rate of a component is 10 f/106 hours and the mean
wear our life is 1000 hours with a standard deviation of 100 hours. What is
the component reliability for a mission of 750 hours starting from 300 hours
from now?
λ = 10-5 f/hr
Μ = 1000 hr
σ = 200 hr
T = 300 hr
t = 750 hr
R(t) = Ru(t) x Rw(t)
e −λt = e −750 x 10 = 0.9925
-5
Ru(t) =
R (T + t)
R (1050)
Rw(t) = Rw (T) = Rw (300)
w
w
z = (t – Μ) / σ
From the Table:
z = 0.5
F(z) = 0.1915
= 7.0
= 0.5
R(1050) = 0.5 – 0.1915 = 0.3085
R(300) = 0.5 + 0.5 = 1.0
Rw(t) = 0.3085 / 1.0 = 0.3085
R(t) = 0.9925 x 0.3085 = 0.306
Poisson Distribution Example
If the average number of cable faults per year per 10 km of cable is 0.05,
evaluate the probabilities of 0, 1, 2, .. faults occurring in
(a) 20 year period
(λt ) x e − λt
(b) 40 year period
Px(t) =
x!
Failure Rate, λ = 0.05 f/yr
(a) For a 20 year period, t = 20 yr
(1) x e −1
Px(t) =
x!
Expected # of failures, E(x) = λt = 0.05 x 20 = 1.0
# of failures
Probability
0
0.36788
1
0.36788
2
0.18394
3
0.06131
Failure Distribution Function
Failure Density Function
1
probability
0.4
probability
4
0.01533
0.3
0.2
0.1
0.8
0.6
0.4
0.2
0
0
0
1
2
# of failures
3
4
0
1
2
3
410
# of failures
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Poisson Distribution Example
(b) For a 40 year period, t = 40 yr
Expected # of failures, E(x) = λt = 0.05 x 40 = 2.0
# of failures
Probability
0
0.13534
1
0.27067
2
0.27067
3
0.18045
1
0.4
20-year
20-year
0.8
probability
0.3
probability
4
0.09022
40-year
40-year
0.2
( 2) x e − 2
x!
Px(t) =
0.6
0.4
0.1
0.2
0
0
0
1
2
3
4
# of failures
Failure Density Functions
0
1
2
# of failures
3
4
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Failure Distribution Functions
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