Solution HW4

IE 4521 fall 2011 5.1.13

223.8, 1.28μ σ= =

(a) 23 23.8( 23.0) ( ) 0.23981.28

P X −≤ = Φ =

(b) 24 23.8( 24.0) ( ) 0.42981.28

P X −≤ = Φ =

(c) 24.5 23.8 24.2 23.8(24.2 24.5) ( ) ( ) 0.09371.28 1.28

P X − −≤ ≤ = Φ −Φ =

(d) 23.8( ) 0.75 ( ) 0.751.28

xP X x This means that −≤ = Φ = By solving this we get

24.5631x =

(e) 23.8( ) 0.95 ( ) 0.951.28

xP X x This means that −≤ = Φ = By solving this we get

25.661x = 5.2.1

(a) 0 1.1~ (3.2 2.1,6.5 3.5) ( 0) 1 ( ) 0.636010

X Y N So P X Y −+ − + + ≥ = −Φ =

(b) 20 22.92 ~ ( 22.9,40) ( 2 20) ( ) 0.676740

X Y Z N So P X Y Z − ++ − − + − ≤ − = Φ =

(c) 1 0.93 5 ~ ( 0.9,146) (3 5 1) 1 ( ) 0.4375146

X Y N So P X Y ++ − + ≥ = −Φ =

(d) (4 3.2) (4 ( 2.1)) (2 12) 45.2The mean is × − × − + × =

2 2 2The variance is (4 6.5) (4 3.5) (2 7.5) 190× + × + × =

25 45.24 4 2 ~ (45.2,190), (4 4 2 25) ( ) 0.0714190

So X Y Z N P X Y Z −− + − + ≤ = Φ =

(e) 2The mean is 2.6, The variance is 140μ σ= =

1

2 2.6( 6 2) ( ) 0.3487140

P X Y Z − −+ + ≤ − = Φ =

2 2.6( 6 2) 1 ( ) 0.5202140

P X Y Z −+ + ≥ = −Φ =

( 6 2) 0.3487 0.5202 0.8689P X Y Z+ + ≥ = + =

(f) 2The mean is 2 3.2 2.1 6 2.5, The variance is 29.5μ σ= × + − = =

1 2.5(2 6 1) ( ) 0.391229.5

P X Y −− − ≤ = Φ =

1 2.5(2 6 1) ( ) 0.259729.5

P X Y − −− − ≤ − = Φ =

( 2 6 1) 0.3912 0.2597 0.1315P X Y− − ≤ − = − =

5.3.8 (a) By using a normal approximation

630 0.5 60 0.25( 30) 1 ( 30) 1 ( ) 8 1060 0.25 0.75

P X P X −− − ×≥ = − ≤ = −Φ = ×

× ×

(b) Given 0.35 0.5 0.25( ) 0.990.25 0.75

n nn

+ −Φ ≥

× ×

By solving this, we can get 92n ≥ 5.3.14

0.056, 2.5aλ = =

2.5(0.056 20)( 20) 0.265~ (500 0.265,500 0.265 (1 0.265))

124.5 132.5( 124.5) 1 ( ) 0.7997.3875

P X eY N

P X

− ×≥ = =× × × −

−≥ = −Φ =

5.4.15

(a) 10,50( 2.5) 0.016P F ≥ =

(b) 217( 12) 0.200P χ ≤ =

(c) 24( 3) 0.003P t ≥ =

2

(d) 14( 2) 0.967P t ≥ − =

4.2.12

0.0065λ = ~ (0.0065)X Exponential

When n=10, ( 150) 1 ( 150) 1 (150) 0.3772P X P X F≥ = − ≤ = − =

~ (10,0.3772)Y Bionomial

( 8) ( 8) ( 9) ( 10) 0.008151P Y P Y P Y P Y≥ = = + = + = =

17.1.5 By using both the equation of components in series and parallel, we can get that r=0.8198 17.2.3

1 2 3 41 1 1 1

125 60 150 100λ λ λ λ λ= + + + = + + +

1 24.19 minλ=

17.2.8 (a) The component has a Weibull failure time distribution, so the probablily that the

component is still operating at time 12 is

( 12) 1 (12) 0.103P T F≥ = − =

(b) The probability that the component fails before time 8 is

( 8) (8) 0.307P T F≤ = =

(c) The median failure time satisfies that ( ) 0.5F t = , which is (0.1 ) 4.51 0te− ×− = .5

We can get t=9.22 (d) The hazard rate is

4.5 3.5 3.5( )( ) 4.5 (0.1) 0.0001421 ( )

f th t t tF t

= = × × =−

(e) (12) 0.8518 4.13(8) 0.2061

hh

= = times likely to suddenly fail.

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