University:
Massachusetts Institute of Technology
Course:
22.38 | Probability and Its Applications to Reliability, Quality Control, and Risk Assessment
Academic year:

2024
Views:

41

Pages:

5
Author:

Dorian White

3-61) b) marginal PMF of X: Px(x) = Σ Px,y(x,y) all Y Px(0) = .18045 Px(1) = .1789 Px(2) = .1819 Px(3) = .19697 Px(4) = .2677 c) P(A|B) = P(AB) P(B) => X = 3, Y(v) = ? Px(3) = Σ P(3,v) = .19697 P(v|x=3) = P(3,v) / .19697 P[v|x=3] = [P(0)=.354 P(1)=.352 P(2)=.252 P(3)=.118 P(4)=.071] d) P(X|y=4) = P(x,4) / P(y=4) → P(y=4) = .24812 P(X=4|Y=4) = .02752 / .24812 = .0303 e) cov(X,Y) = E(XY) - E(X)E(Y) E(X) = ΣxP(x,y) all Y = 2.186 E(Y) = ΣyP(x,y) all X = 2.13 E(XY) = ΣΣxyP(x,y) = 3.344 cov(X,Y) = 3.344 - 2.186(2.13) = -1.31991 X & Y are not statistically independent. P = cov(X,Y) / (σxσy) → σx = √(Σ(x-μx)^2P(x,y)) = 2.136 σy = √(Σ(y-μy)^2P(x,y)) = 2.144 P = -1.3199 / (2.136*2.144) = -.618 X & Y exhibits a fairly strong negative correlation, as expected. 3-62) fxy(x,y) = 2ye^(-y(2+x)) a) P(X≤100k, Y≤200k) = ? Fxy(x,y) = ∫[∫ fxy(u,v)du]dv = ∫[∫ 2ve^(-v(2+u))du]dv = ∫[2e^(-v(2+x)) - e^(-2v)]dv = -2e^(-v(2+x)) / (2+x) - e^(-2v) / 2 |_0^y Fxy(x,y) = 2/x(e^(-y(2+x)) - 1) - 1/2(e^(-2y) - 1) F(1,2) = .316671 b) PDFx: fx(x) = ∫[fxy(x,y)]dy = ∫[2ye^(-y(2+x))]dy fx(x) = 2e^(-y(2+x)) / -(2+x) - 2/(2+x)^2 * e^(-y(2+x)) |_0^∞ fx(x) = 2/(2+x)^2 c) PDFy: Fy(y) = ∫[fxy(x,y)]dx = 2e^(-2y) d) E(XY) = ∫∫[xy*fxy(x,y)]dxdy ≠ ∫[x*fx(x)]dx * ∫[y*fy(y)]dy ⇒ E(X)E(Y) They are not independent. e) P(Y>2|X=2) = 1 - P(Y≤2|X=2) = 1 - P(2,2) / Px(2) P(2,2) = F(2,2) = .4819 Px(2) = ∫[fx(x)]dx = 2/(2+x) |_2^∞ = 1 - 1/2 = .5 P(Y>2|X=2) = 1 - (.4819/.5) = .0363 R&H: 2.2) a) X50=100 => ∫[0 to ∞] λe^(-λt)dt = .5 => λ = 6.9x10^(-3) b) P(t< 500) = R(500) = e^(-λ(500)) = .032 c) P(+1000 | surviving at 500) = P(t=1500) = .032 Note: for an exponential distribution the failure rate is constant, and so the function is "memoryless" 2.8) a) P(T>200) = R(200) = exp(-∫[0 to 200] ktdt) = .961 b) MTTF = ∫[0 to ∞] R(t)dt = ∫[0 to ∞] e^(-kt^2/2)dt = 886.2 hours c) R(x|t) = P(T>x+t | T>t) = P(T>x) / P(T>t) where R(t) = e^(-kt^2/2) R(200|200) = R(400) / R(200) = .8868 d) R(t) = e^(-kt^2/2) = e^(-(√(k/2)*t)^2) It looks like a Weibull distribution where λ = √(k/2), α = 2 (specifically it looks like a Rayleigh distribution where MTTF = √(π/λ) * Γ(1.5) = √(π/√(k/2)) * Γ(1.5) = 886.2 hr which is what we expect from (b) 2.9 (R&H) a) R(t) = exp(-∫[0 to t] z(u)du) = exp(-∫[0 to t] (λt+αt²)dt) = exp(-λt²/2 - αt³/3) => e^(-λt²/2 - αt³/3) = R(t) b) MTTF = ∫[0 to ∞] R(t)dt = ∫[0 to ∞] e^(-λt²/2 - αt³/3)dt = √(π/λ) * Erf[(3λ/2α)^(1/3)] / 2α c) z(t) is linear, so this models fatigue of parts over time (the end of the bath tub curve). The distribution looks like a product of an exponential and Weibull distribution. 2.27) f(t) = b-a for a< t< b R(t) = ∫[t to b] f(u)du = (b-a) - (b-t) = b-t for a< t< b z(t) = f(t)/R(t) = b-a / b-t for a< t< b z(t)

22.38 - PS#5 Solutions

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