NAME: GTID: Spring 2012, MATH 3770D, Final Exam, 5/3/12

1. The automatic opening device of a military cargo parachute has been designed to open when the parachute is 200 mabove the ground. Suppose opening altitude actually has a normal distribution with mean value 200 m and standarddeviation 30 m. Equipment damage will occur if the parachute opens at an altitude of less than 100 m. What is theprobability that there is equipment damage to the payload of at least one of five independently dropped parachutes?(8 points)

Solution. Let Xi be thte altitude where the ith parachute is open. Then we know Xi ∼ i.i.d. N(200, 302). Therefore

P (the ith device is damaged) = P (Xi < 100) = P (200 + 30Z < 100)= P (Z < −10/3) = 1− Φ(10/3)

where Z ∼ N(0, 1) is standard normal random variable and Φ is the cdf of Z. Furthermore, we know

P (the ith device is not damaged) = P (Xi ≥ 100) = 1− P (Xi < 100) = Φ(10/3).

Based on these, we have

P (at least one device is damaged)=1− P (none of the devices is damaged)=1− P (X1 ≥ 100, . . . , X5 ≥ 100)=1− P (X1 ≥ 100)× · · · × P (X5 ≥ 100) because Xi’s are independent

=1− (Φ(10/3))5

2. Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y :

f(x, y) =

{xe−x(1+y) x ≥ 0 and y ≥ 00 otherwise

(a) What is the probability that the lifetime X of the first component exceeds 3?

(b) What are the marginal pdf’s of X and Y ? Are the two lifetimes independent? Explain.

(c) What is the probability that the lifetime of at least one component exceeds 3?

(8 points)

Solution. (a) The marginal pdf of X is

fX(x) =∫ ∞

0

f(x, y)dy =∫ ∞

0

xe−x1+y)dy = e−x, (x ≥ 0).

So there isP (X ≥ 3) =

∫ ∞

3

fX(x)dx =∫ ∞

0

e−xdx = e−3

(b) The marginal pdf of X is fX(x) = e−3 as given above. The marginal pdf of Y is

fY (y) =∫ ∞

0

f(x, y)dx =∫ ∞

0

xe−x(1+y)dx =1

(1 + y)2, (y ≥ 0).

The two lifetimes X and Y are not independent because

f(x, y) = xe−x(1+y) 6= e−x

(1 + y)2= fX(x)fY (y).

(c) The probability that at least one of X and Y exceeds 3 is

P (X ≥ 0 or Y ≥ 3) =1− P (X < 3, Y < 3) = 1−∫ 3

0

∫ 3

0

f(x, y)dydx

=1−∫ 3

0

∫ 3

0

xe−x1+y)dydx =14(1− e−12) + e−3.

3. Annie and Alvie have agreed to meet for lunch between noon (12:00PM) and 1:00PM. Denote Annie’s arrival timeby X, Alvie’s by Y , and suppose X and Y are independent with pdf’s

fX(x) =

{3x2 0 ≤ x ≤ 10 otherwise

and fY (y) =

{2y 0 ≤ y ≤ 10 otherwise

What is the expected amount of time that the one who arrived first must wait for the other person? [Hint: h(X, Y ) =|X − Y |.] (6 points)

Solution. Since X and Y are independent, we know that the joint pdf is

f(x, y) = fX(x)fY (y) = 6x2y, (0 ≤ x, y ≤ 1).

Let h(X, Y ) = |X − Y |, the time that one waits for the other, then

E(h(X, Y )) =∫ 1

0

∫ 1

0

|x− y|6x2ydydx

=∫ 1

0

∫ x

0

(6x3y − 6x2y2)dydx +∫ 1

0

∫ 1

x

(6x2y2 − 6x3y)dydx

=14.

4. Consider a random sample X1, X2, . . . , Xn from the shifted exponential pdf

f(x;λ, θ) =

{λe−λ(x−θ) x ≥ θ

0 otherwise

Taking θ = 0 gives the pdf of the exponential distribution considered previously (with positive density to the rightof zero). Obtain the maximum likelihood estimators of θ and λ. (6 points)

Solution. The likelihood function of θ and λ is

L(θ, λ) =∏

i

f(xi; θ, λ) =∏

i

λe−λ(xi−θ) = λne−λ(P

i xi−nθ).

So the log-likelihood function isl(θ, λ) = ln(L(θ, λ)) = n lnλ− λn(x̄− θ),

where x̄ = (∑

x xi)/n is the sample mean.

For any λ, we know x̄− θ ≥ x̄−mini{xi} because all xi’s are no smaller than θ. Therefore, to maximize l(θ, λ), weneed λ = mini{xi}. Futhermore, we have

∂l(θ, λ)∂λ

=n

λ− n(x̄− θ) = 0,

which yields

λ =1

x̄− θ=

1x̄−mini{xi}

that maximizes l(θ, λ). Therefore, the maximum likelihood estimators are

θ = mini{Xi} and λ =

1X̄ −mini{Xi}

.

5. A sample of 25 pieces of laminate used in the manufacture of circuit boards was selected and the amount of warpage(in.) under particular conditions was determined for each piece, resulting in a sample mean warpage of 0.06 and asample variance of 0.052. For simplicity the measurements are assumed to be normally distributed with pdf N(µ, σ2).Refer to the tables in the end of this test for critical values. (6 points)

(a) Calculate the 95% confidence interval of µ.

(b) Calculate the 95% confidence intervals of σ2 and σ.

[Hint: refer to the critical values tables in the last two pages.]

Solution. (a) We know that (X̄ − µ)/(S/√

25) follows student’s t24 distribution with degrees of freedom 24. Giventhat X̄ = 0.06 and S = 0.05. Therefore,

P

(0.06− t0.025,24

0.05√25

< µ < 0.06 + t0.025,240.05√

25

)= 0.95. (1)

Find out that t0.025,24 = 2.064 in the critical values table of student’s t distribution. So the 95% CI of µ is

(0.06− 0.02064, 0.06 + 0.02064, )

(b) We know that (n− 1)S2/σ2 ∼ χ224, therefore

P

(χ2

0.975,24 <(n− 1)S2

σ2< χ2

0.025,24

)= 0.95. (2)

Note that n = 25, S2 = 0.052, and find out χ20.975,24 = 12.401 and χ2

0.025,24 = 39.364 from the critical values tableof the χ2 distribution. Then

P

(24× 0.052

39.364< σ2 <

24× 0.052

12.401

)= 0.95.

Hence the 95% CI of σ2 is (24× 0.052

39.364,24× 0.052

12.401

)and the 95% CI of σ is (√

24× 0.052

39.364,

√24× 0.052

12.401

).

6. The article “Reliability-Based Service-Life Assessment of Aging Concrete Structures” (J. Structural Engr., 1993:1600–1621) suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Supposethe mean time between occurrences of loads is .5 year. [Hint: Recall that a Poisson process is parameterized byµ > 0, the number of occurances during a unit time period. Then during any time period t > 0, the number ofoccurances, Xt, follows the Poisson(µt) distribution.] (6 points)

(a) What is the value of µ? [Hint: Let T be the time between occurrences, then E(T ) is known to be .5. On theother hand, the cdf of T is FT (t) = P (T ≤ t) = P (Xt ≥ 1) (Why?). Use this to derive the pdf of T and hencethe expression of E(T ) in terms of µ.]

(b) How many loads can be expected to occur during a 2-year period? Use the result of (a) to obtain this value.

(c) How long must a time period be so that the probability of no loads occurring during that period is at most .1?

Solution. (a) We let T be the time between occurrences. Then the cdf of T is

FT (t) = P (T ≤ t) = P (Xt ≥ 1) = 1− e−µt

where Xt ∼ Poisson(µt). Therefore the pdf of T is

fT (t) = F ′T (t) = µe−µt, (t > 0)

So T ∼ exp(µ) and E(T ) = 1/µ = .5. Therefore µ = 2.

(b) Let Y be the number of occurrences during a 2-year period, then µt = 4 and Y = X2 ∼ Poisson(4). So E(Y ) = 4.

(c) We have thatP (Xt = 0) = e−2t ≤ 0.1

Therefore, t ≥ −12 ln 0.1 = ln 10

2 .