NAME:
MAE 108 – Probability and Statistical Methods for Engineers - Winter 2011
Final exam - March 16, 2011
3 hours, open book, open notes, calculator allowed, no cell phones.
Write your answers directly on the exam.
100 points total. Each of the 10 problems is worth 10 points.
(1) A satellite system consists of 6 independent components and can function adequately if at least
4 of the 6 components are in working condition. If each component is, independently, in working
condition with probability 60%, what is the probability that the system functions adequately?
The number of working components X is a binomially distributed random variable with n = 6
and p = 0.6 hence we have
P (X6 ≥ 4) = P (X6 = 4 ∪ X6 = 5 ∪ X6 = 6)
= P (X6 = 4) + P (X6 = 5) + P (X6 = 6)
6 4
6 5
=
p (1 − p)2 +
p (1 − p) + p6
4
5
= 15(0.6)4 (0.4)2 + 6(0.6)5 (0.4) + (0.6)6 = 0.5443
(2) In a given year, the winter in Chicago can be cold (C) and wet (W). On average, 50% of the
winters in Chicago are cold, and 30% of the winters are wet. Moreover, 40% of the cold winters
are also wet. An unpleasant winter is one when the weather is either cold or wet or both.
Given P (C) = 0.5, P (W ) = 0.3, P (W |C) = 0.4.
(a) Are the events C and W statistically independent? Why?
No because P (W |C) 6= P (W ).
(b) What is the probability of an unpleasant winter in a given year?
P (C ∪ W ) = P (C) + P (W ) − P (CW )
= P (C) + P (W ) − P (W |C)P (C)
= 0.5 + 0.3 − 0.4 × 0.5 = 3/5 = 0.6
(c) What is the probability that the winter in any given year will be cold but not wet?
P (C W̄ ) = P (W̄ |C)P (C)
= (1 − P (W |C))P (C)
= 0.6 × 0.5 = 3/10 = 0.3
(d) If the winter in a given year is unpleasant, what is the probability that it will be both cold and
wet?
P (CW )
P (CW ∩ (C ∪ W ))
=
P (C ∪ W )
P (C ∪ W )
P (C)P (W |C))
0.5 × 0.4
=
=
= 1/3 = 0.3333
P (C ∪ W )
0.6
P (CW |C ∪ W ) =
(3) A postmix Pepsi machine is adjusted to release a certain amount of Pepsi syrup into a chamber
where it is mixed with carbonated water. A random sample of 25 beverages was found to have
an estimated mean syrup content of x̄ = 1.1 fluid ounces, and an estimated standard deviation
s = 0.015 fluid ounces.
Given n = 25
(a) Derive the two-sided 99% confidence interval for the syrup volume in a beverage.
With α = 0.01 we obtain
hµi0.99
s
= x̄ ∓ √ t0.995,24
n
0.015
= (1.1 ∓
2.7969) = (1.0916, 1.1083)
5
(b) If we actually know that the true standard deviation for the syrup volume is given by the value
of 0.015, compute the new confidence interval and compare with the one in (a).
With σ = 0.015 we obtain
σ
x̄ ∓ √ k0.995
n
0.015
= (1.1 ∓
2.576) = (1.09227, 1.10773)
5
hµi0.99 =
This interval is smaller by a factor k0.995 /t0.995,24 = 0.921
(4) An assembly consists of three mechanical components. Suppose that the probabilities that the
first, second, and third components meet specifications are 0.5, 0.9 and 0.99, respectively. Assume
that the three components are independent.
The probabilities of the components meeting specification are P (C1 ) = 0.5, P (C2 ) = 0.9 and
P (C3 ) = 0.99.
(a) Determine the probability mass function (PMF) for the random variable X measuring the
number of components in the assembly that meet specifications.
pX (0) = P (C1 )P (C2 )P (C3 ) = 0.5 × 0.1 × 0.01 = 0.0005
pX (1) = P (C1 )P (C2 )P (C3 ) + P (C1 )P (C2 )P (C3 ) + P (C1 )P (C2 )P (C3 )
= 0.5 × 0.1 × 0.01 + 0.5 × 0.9 × 0.01 + 0.5 × 0.1 × 0.99 = 0.0545
pX (2) = P (C1 )P (C2 )P (C3 ) + P (C1 )P (C2 )P (C3 ) + P (C1 )P (C2 )P (C3 )
= 0.5 × 0.9 × 0.01 + 0.5 × 0.1 × 0.99 + 0.5 × 0.9 × 0.99 = 0.4995
pX (3) = P (C1 )P (C2 )P (C2 ) = 0.5 × 0.9 × 0.99 = 0.4455
Check
P3
x=0 pX (x)
= 0.0005 + 0.0545 + 0.4995 + 0.4455 = 1
(b) What is the mean number of components in the assembly that meet specifications?
3
X
x=0
xpX (x) = pX (1) + 2pX (2) + 3pX (3) = 0.0545 + 2 × 0.4995 + 3 × 0.4455 = 2.39
(5) In Ohio, 62% of the population has no college degree, while 38% are college graduates. In the
2004 presidential election, exit polls in Ohio show that half of the people with no college degree voted
for Bush, while the other half voted for Kerry. In contrast, 53% of college graduates voted for Bush
while 47% voted for Kerry. What is the probability a randomly selected respondent voted for Bush?
Given P (C) = 0.38 and P (C̄) = 0.62 and P (B|C̄) = 0.5, and P (B|C) = 0.53.
P (B) = P (B|C)P (C) + P (B|C̄)P (C̄) = 0.53 × 0.38 + 0.5 × 0.62 = 0.5114
(6) An engineering company E is submitting bids for two projects A and B; the probabilities of
winning are estimated to be, respectively, 0.5 and 0.3. Also, if the company wins one bid, its chance
of winning the other bid is reduced to one-half of the original probability.
Given P (A) = 0.5, P (B) = 0.3, P (A|B) = 0.25 and P (B|A) = 0.15.
(a) What is the probability of Company E winning at least one of its bids?
P (A ∪ B) = P (A) + P (B) − P (A|B)P (B) = 0.5 + 0.3 − 0.25 × 0.3 = 29/40 = 0.725
(b) If Company E wins at least one of its bids, what is the probability that it will be for project A
and not project B?
P (AB̄|A ∪ B) =
P (AB̄)
P (B̄|A)P (A)
(1 − P (B|A))P (A)
0.85 × 0.5
=
=
=
= 17/29 = 0.5862
P (A ∪ B)
P (A ∪ B)
P (A ∪ B)
0.725
(c) If Company E is awarded only one project, what is the probability that it will be project A?
P (AB̄)
P (A)P (B̄|A)
P (A)P (B̄|A)
=
=
P (AB̄ ∪ ĀB)
P (AB̄) + P (ĀB)
P (B̄|A)P (A) + P (Ā|B)P (B)
P (A)(1 − P (B|A))
0.85 × 0.5
=
=
= 17/26 = 0.6538
(1 − P (B|A))P (A) + (1 − P (A|B))P (B)
0.85 × 0.5 + 0.75 × 0.3
P (A|AB̄ ∪ ĀB) =
(d) On the basis of past performance, it is estimated that the probability of Company E completing
project A within a target time is 75%, whereas, if another company is awarded the project, the
probability of on-time completion of project A is only 50%. What is the probability that project
A will be completely in time?
The probability that E wins A was given P (A) = 0.5 and we know on-time work P (T |A) = 0.75
and P (T |Ā) = 0.5
P (T ) = P (T |A)P (A) + P (T |Ā)(1 − P (A)) = 0.75 × 0.5 + 0.5 × 0.5 = 5/8 = 0.625
(e) If the project is completed in time, what is the probability that it was done by Company E?
P (A|T ) =
P (T |A)P (A)
3/4 × 1/2
=
= 3/5 = 0.6
P (T )
5/8
(7) Mechanical failures of a certain type of aircraft constitute a Poisson process. On the average,
based on past records, an aircraft will have a failure once every 5000 hr of flight time.
We are given ν = 1/5000 failures/hr
(a) If such aircrafts are scheduled for inspection and maintenance after every 2500 flight hours,
what is the probability of mechanical failure of an aircraft between inspections?
We let ν × (t = 2500) = 1/2. Then we say
P (Xt ≥ 1) = 1 − P (Xt = 0) = 1 − e−νt = 1 − e−1/2 = 0.3935
(b) In a fleet of ten aircrafts of the same type, what is the probability that not more than two
will have mechanical failures within the above scheduled inspection/maintenance interval? Assume
that failures between aircraft are statistically independent.
The number of airplanes with mechanical failures Y with p = 1 − e−1/2 is given
P (Y10 ≤ 2) = P (Y10 = 0) + P (Y10 = 1) + P (Y10 = 2)
10 2
10
9
= (1 − p) + 10p(1 − p) +
p (1 − p)8
2
= e−5 + 10(1 − e−1/2 )e−9/2 + 45(1 − e−1/2 )2 e−4 = 0.1781
(c) As the engineer in charge of aircraft safety, you wish to ensure an acceptable probability of
failure of no more than 5% for each aircraft. How should the inspection/maintenance schedule in
part (a) be revised? That is, what should be the revised inspection/maintenance (in flight hours)?
We want
0
P (X ≥ 1) = 1 − P (X = 0) = 1 − e−νt ≤ 0.05
⇒ t0 ≤ − ln(19/20)/ν = 256.5
There should be an inspection every 256.5 hours or less.
(8) The diameter of the dot produced by a printer is normally distributed with a mean diameter
of 0.002 inch and a standard deviation of 0.0004 inch.
We are given µ = 0.002 and σ = 0.0004.
(a) What is the probability that the diameter of a dot exceeds 0.0026 inch?
P (X > 0.0026) = 1 − P (X ≤ 0.0026) = 1 − Φ
0.0026 − µ
σ
= 1 − Φ (3/2) = 1 − 0.9332 = 0.0668
(b) What is the probability that a diameter is between 0.0014 and 0.0026 inch?
0.0026 − µ
0.0014 − µ
P (0.0026 > X > 0.0014) = Φ
−Φ
σ
σ
= Φ (3/2) − Φ (−3/2) = 2Φ (3/2) − 1 = 2 × 0.9332 − 1 = 0.8664
(c) What standard deviation of diameters is needed so that the probability in part (b) is 99.5%?
2Φ − 1 = 0.995
⇒
0.0006
= Φ−1 (0.9975) = 2.81
σ
⇒ σ = 0.0002135
(9) We wish to estimate the mean, µ, and standard deviation, σ, for the normal distribution we
think a random variable X is following. We have a data sample of size n from this random variable:
{x1 , x2 , ..., xn }. Using the method of maximum likelihood derive theoretically the MLE estimate
for the mean (µ̂) and the standard deviation (σ̂). You should get an explicit formula for both µ̂
and σ̂ as a function of n and the sample {x1 , x2 , ..., xn }. How do these estimates compare with the
point estimates x̄ and s given by the method of moments?
We are given X = N (µ, σ) there likelihood function
"
#
n
Y
1
1 xi − µ 2
√
L=
exp −
2
σ
2πσ
i=1
"
#
n
1
1 X
= √
exp − 2
(xi − µ)2
2σ
( 2πσ)n
i=1
n
√
1 X
⇒ ln L = −n ln( 2π) − n ln σ − 2
(xi − µ)2
2σ
i=1
Maximizing ln L(µ, σ) we want
⇒
1
2σ̂ 2
∂ ln L =0
∂µ µ=µ̂,σ=σ̂
!
n
X
2nµ̂ −
xi = 0
i=1
n
⇒ µ̂ =
1X
xi
n
i=1
and
∂ ln L =0
∂σ µ=µ̂,σ=σ̂
⇒−
n
n
1 X
+ 3
(xi − µ̂)2 = 0
σ σ
i=1
v
u n
u1 X
(xi − µ̂)2
⇒ σ̂ = t
n
i=1
We see then that µ̂ = x̄ but σ̂ 2 =
n−1 2
n s .
(10) A physical process is described by a continuous random variable X ≥ 0 whose cumulative
distribution function (CDF) is given by F (x) = 1 − e−2x , (x ≥ 0).
(a) Verify that the CDF satisfies all the appropriate limits on either side of the domain.
F (0) = 1 − 1 = 0
lim F (x) = 1 − lim e−2x = 1
x→∞
x→∞
(b) What is the probability distribution function (PDF) for X?
df (x)
= 2e−2x for x ∈ [0, ∞)
dx
fX (x) =
(c) What is the mean value of X?
Z
∞
E(X) =
Z
∞
xfX (x)dx = 2
0
Z
=
∞
xe−2x dx
0
e−2x dx = 1/2
0
where we have used
d
−2x x)
dx (−e
= 2xe−2x − e−2x
(d) What is the median of X?
F (xm ) = 1 − e−2xm = 1/2
⇒ xm = −1/2 ln(1/2) = 0.3466
(e) If you know, from a measurement, that X is above 1, what is the probability that is it not
above 2?
P (2 ≥ X|X > 1) =
F (2) − F (1)
−e−4 + e−2
P (2 ≥ X > 1)
=
=
= 1 − e−2 = 0.8647
P (X > 1)
1 − F (1)
e−2