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Material Type: Exam; Class: Introduction to Probability Theory I; Subject: Statistics; University: California State University-East Bay; Term: Unknown 1989;
Typology: Exams
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3.1. Suppose that an object obeys an EFL with rate
λ^ = 2 per month.
(a) Sketch the density function, the cumulative distribution function, the reliability function andthe hazard function of its length of life
lam <- 2t <- seq(.01,
3,^ by=.01) dens <- dexp(t,
rate=lam) cdf^ <- pexp(t,
rate=lam) reli <- 1
-^ cdf hazd <- dens/relipar(mfrow=c(2,
plot(t,^
dens,^ type="l") plot(t,^
cdf,^ type="l") plot(t,^
reli,^ type="l") plot(t,^
hazd,^ ylim=c(0,
4),^ type="l")
par(mfrow=c(1,
EFL with^
λλλλ^ = 2
A Component With Quadratic Hazard Rate
5.1. A communications satellite can be regarded as a parallel array of two independent EFLcomponents with mean lifetime 3 years.(a) Plot (on the same axes) both the reliability function for this system and the reliability functionfor one component. F^ ( t ) = P( SS
≤^ t ) = P( T^1
≤^ t ,^ T ≤^ t ) = P( 2
T ≤^ t ) P( T^1
≤^ t ) = [ F ( 2
(^2) t )]= [1 – exp(–
(^2) λ t )]
= 1 – 2 exp(–
λ t ) + exp(–
λ t ) R^ ( t ) = 2 exp(– S
λ t ) – exp(–
λ t ) f^ ( t ) = 2λ^ exp(– S
λt) – 2λ^ exp(–
λt) lam <- 1/3t <- seq(.01,
10,^ by=.01) reli.1 <-
1 -^ pexp(t,
rate=lam)
hazd.1 <-
dexp(t,
rate=lam)/reli. reli.2 <-
2exp(-lamt)
-^ exp(-2lamt)
hazd.2 <-
2lam(exp(-lam*t)
-^ exp(-2lamt))/reli.
par(mfrow=c(1,
plot(t,^
reli.1,
ylim=c(0,1),
type="l",
ylab="Reli (Syst = red)")
lines(t,
reli.2,
ylim=c(0,1),
col="red")
plot(t,^
hazd.1,
ylim=c(0,.4),
type="l",
ylab="Hazd (Syst = red)")
lines(t,
hazd.2,
ylim=c(0,.4),
col="red")
par(mfrow=c(1,
6.1. Three components obey WFLs with
λ^ = 1. Their values of
α^ are 0.5, 1, and 2, respectively.
(c) Sketch the three density curves on the same axes. ...Note that R (along with most other statistical packages) uses the true scale parameter
–1/α β = λ.
When^ λ^ = 1,
β^ = 1 also, so the distinction between Poisson rate
λ^ and Weibull scale
β^ is moot here.
But the program below is written to use Weibull shape and scale parameters. t <- seq(.01,
3,^ by=.01) alp <- 0.5;
lam^ <-^
1;^ bet^
=^ lam^(-1/alp)
plot(t, dweibull(t,
alp,^ scale=bet),
type="l",
ylab="Density",
ylim=c(0,2), main = "Weibull Densities with Shapes .5 (black), 1 (red), 2 (blue), 4 (green)")alp <- 1;
lam^ <-
1;^ bet^
=^ lam^(-1/alp)
lines(t, dweibull(t, alp,
scale=bet)
,^ col="red")
alp <- 2;
lam^ <-
1;^ bet^
=^ lam^(-1/alp)
lines(t, dweibull(t, alp,
scale=bet)
,^ col="blue")
alp <- 4;
lam^ <-
1;^ bet^
=^ lam^(-1/alp)
lines(t, dweibull(t, alp,
scale=bet)
,^ col="darkgreen")
Goodness
of^ Fit^
Test
Distribution
AD^
P
Normal^
0.^
Exponential
<0. Weibull^
0.251^ >0. Gamma^
0.^
ML^ Estimates
of^ Distribution
Parameters
Distribution
Location
Shape
Scale
Threshold
Normal*^
4.^
Exponential
Weibull^
2.^
Gamma^
3.^
*^ Scale:
Adjusted
ML^ estimate
C 1 P er cent
(^105) (^0) - 99.9^999050101 0.
C 1 P er cent
100.00010.0001. 0.1000. 99.9^9050101 0.10. C 1 P er cent
10.01. 99.9^9050101 0.10.
C 1 P er cent
10.01. 99.9^999050101 0.10.
G oodness of F it TestN ormalA D = 0.457P -V alue = 0.260E xponentialA D = 9.962P -V alue < 0.003WeibullA D = 0.251P -V alue > 0.250G ammaA D = 0.732P -V alue = 0.
Probability Plot for C N ormal - 95% C I
E xponential - 95% C I
Weibull - 95% C I
G amma - 95% C I