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230
Seat IE # ________ Please read these directions. Closed book and notes. 60 minutes.
Name ___ < KEY > ___
Covers through the normal distribution, Section 4.6 of Montgomery and Runger, fourth edition. Cover page and four pages of exam. Pages 8 and 12 of the Concise Notes. A normal-distribution cdf table. No calculator. No need to simplify beyond probability concepts. For example, unsimplied factorials, integrals, sums, and algebra receive full credit. Throughout, f denotes probability mass function or probability density function and F denotes cumulative distribution function. For one point of credit, write your name neatly on this cover page. Circle your family name. For one point of credit, write your name on each of pages 1 through 4.
Score ___________________________
Exam #2, October 19, 2010
Schmeiser
IE 230 Probability & Statistics in Engineering I
Name _______________________
Closed book and notes. 60 minutes. 1. Suppose that the random variable X has the discrete uniform distribution over the set {0, 1,..., 10} and that the random variable Y has the continuous uniform distribution over the set [0, 10]. (a) (b) (c) (d) (3 pt) T (3 pts) T (3 pts) T (3 pts) T F F F F
E(X ) = E(Y ). V(X ) = V(Y ). FX (5) = FY (5). f X (5) = f Y (5).
2. (from Montgomery and Runger, 3113) In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over twenty years to show that the number of soldiers killed by horse kicks each year in each corps in the Prussian cavalry followed a Poisson distribution with a mean of 0.61. (a) (7 pts) For a particular corps over one year, determine the value of the probability of more than one death. ____________________________________________________________ Deaths occur with rate = 0.61 deaths per year per corps. Let N = "number of deaths in one year in a particular corps". Then N is Poisson with rate = t = (0.61)(1) = 0.61 deaths. Therefore, P(N > 1) = 1 [P(N = 0) + P(N = 1)]
=1 e 0!
0
+
e 1!
1
=1e
0.61
[1 + 0.61]
____________________________________________________________ (b) (7 pts) For a particular corps over ve years, determine the value of the probability of no death. ____________________________________________________________ Deaths still occur with rate = 0.61 deaths per year per corps. Now let N denote the number of deaths during ve years. Then N is Poisson with mean = t = (0.61)(5) = 3.05. Therefore, P(N = 0) =
e 0!
0
=e
3.05
____________________________________________________________
Exam #2, October 19, 2010
Page 1 of 4
Schmeiser
IE 230 Probability & Statistics in Engineering I
Name _______________________
3. (from Montgomery and Runger, 461) The lifetime of a semiconductor laser at a constant power is normally distributed with a mean of 6000 hours and standard deviation of 500 hours. (a) (8 pts) Sketch the corresponding normal pdf. Label and scale both axes. ____________________________________________________________ Sketch two axes. Label the horizontal axis with x and the vertical with f X (x ). Scale the horizontal axis with at least two numbers. The easiest is to place the mean 6000 at the center. Maybe place 5500 and 6500 at the points of inection. Maybe place 4500 and 7500 where the bell curve disappears. Scale the vertical axis. Placing zero at the bottom of the bell curve is an easy number. The height at the mode is 1 / (500 2). Or sketch a rectangle to determine a height. ____________________________________________________________ (b) (8 pts) Determine the value of the probability that a randomly selected laser fails between 4000 and 5000 hours. Provide at least three digits of precision. ____________________________________________________________
P(4000 < X < 5000) = P(7000 < X < 8000) = FX (8000) FX (7000) = FZ (4)FZ (2) 0.999968 0.977250 0.0227
Notice that including, or excluding, 4000 and 5000 makes no difference. ____________________________________________________________ (c) (8 pts) Let q denote the correct answer to Part (b). If three lasers are used in a product, and all have independent lifetimes, determine the value of the probability that all three lasers fail between 4000 and 5000 hours. ____________________________________________________________ Let Li denote that laser i has a lifetime in (4000, 5000). Then P(Li ) = q 0.0227. Then L 1, L 2, and L 3 are Bernoulli trials, with probability of success q . Let X denote the number of trials that succeed. So X is binomial with n = 3 and p = q . From Page 8 of the Concise Notes,
P(N = 3) = f X (3) = 33 33 3 q (1 q ) =q 3
____________________________________________________________
Exam #2, October 19, 2010
Page 2 of 4
Schmeiser
IE 230 Probability & Statistics in Engineering I
Name _______________________
4. Result. For c = 1, 2,..., the geometric cdf is FX (c ) = 1 (1 p )c , where p denotes the probability of success on each Bernoulli trial. (3 pts each) For each of the six lines (af), state why the corresponding equality is true. Each blank requires one reason; reasons may be reused. Choose the reasons from this list: (i) (ii) (iii) (iv) (v) (vi) Events partition the sample space. Events are complementary. Events are mutually exclusive. Events are independent. Events are the same. Denition of conditional probability.
(vii) Denition of cumulative distribution function. (viii) Denition of probability mass function. (ix) (x) (xi) Multiplication Rule. Total Probability. Substitute known values.
Proof. Let c be a positive integer. Let Ai denote success on the i th trial.
FX (c ) = P(X c ) = 1 P(X > c ) = 1 P(A 1 A 2 . . . A c ) = 1 P(A 1) P(A 2) . . . P(A c ) = 1 [1 P(A 1)] [1 P(A 2)] . . . [1 P(Ac )] = 1 (1 p )
c
(a) ___ < vii > ___ (b) ___ < ii > ___ (c) ___ < v > ___ (d) ___ < iv > ___ (e) ___ < ii > ___ ___ (f) < xi > ___
Exam #2, October 19, 2010
Page 3 of 4
Schmeiser
IE 230 Probability & Statistics in Engineering I
Name _______________________
5. For (ae), provide the name of the corresponding family of distributions. (a) (3 pts) The number of coin ips until the fourth "head". ____________________________________________________________ negative binomial ____________________________________________________________ (b) (3 pts) When drawing, without replacement, ve cards from a standard card deck, the number of aces drawn. ____________________________________________________________ hypergeometric ____________________________________________________________ (c) (3 pts) When drawing, with replacement, ve cards from a standard card deck, the number of aces drawn. ____________________________________________________________ binomial ____________________________________________________________ (d) (3 pts) The number of insect parts in a candy bar. ____________________________________________________________ Poisson ____________________________________________________________ (e) (3 pts) When playing Keno, the number of matched numbers. ____________________________________________________________ hypergeometric ____________________________________________________________ 6. Short answer. (a) (3 pts) Evaluate
6 , the number of ways to choose two items from six items. 2
____________________________________________________________
6! 6 = = 15 2 2! 4!
____________________________________________________________ (b) (3 pts) What is the numerical value of 0! ? ____________________________________________________________ ____________________________________________________________ (c) (3 pts) T
F P( < X 2) = FX (2) FX (). 1! = 1
(d) (3 pts) The mean of the binomial distribution is n p , the number of trials multiplied by the probability of success. The mean is in units of "trials". What are the units of the variance? ____________________________________________________________ "trials" squared ____________________________________________________________ (e) (3 pts) The mean of a distribution corresponds to the center of gravity. The variance of a distribution corresponds to... ____________________________________________________________ the moment of inertia. ____________________________________________________________
Exam #2, October 19, 2010
Page 4 of 4
Schmeiser
IE 230 Probability & Statistics in Engineering I
Discrete Distributions: Summary Table random variable X distribution name general x 1, x 2, . . . , xn range probability
Name _______________________
expected value
n
variance
mass function P(X = x ) = f (x ) = f X (x )
n
xi f (xi )
i =1
(xi )
i =1 2
2
f (xi )
= = X = E(X )
n
= = X = V(X )
2
2
= E(X )
n
2 2
X
discrete uniform
x 1, x 2, . . . , xn
1/n
xi / n
i =1
[ xi / n ]
2 i =1
X "# successes in 1 Bernoulli trial" "# successes in n Bernoulli trials" "# successes in a sample of size n from a population of size N containing K successes" "# Bernoulli trials until 1st success" "# Bernoulli trials until r th success" "# of counts in time t from a Poisson process with rate "
equal-space uniform indicator variable binomial
x = a ,a +c ,...,b where x = 0, 1 x = 0, 1,..., n
1/n n = (b a +c ) / c x 1x p (1p )
n Cx
a +b 2 p where
c (n 1) 12 p (1p ) p = P("success") np (1p ) p = P("success") (N n ) np (1p ) (N 1) p =K /N
2
2
p (1p )
x
n x
np where
hypergeometric (sampling without replacement) geometric
x= (n (N K )) , ..., min{K , n } and integer x = 1, 2,... x = r , r +1,...
+
Cx Cn x / Cn
K
N K
N
np where
p (1p )
x 1 Cr 1
x 1
1/p
x r
(1p ) / p
2
negative binomial Poisson
p (1p )
x
r
where r /p where
p = P("success") 2 r (1p ) / p p = P("success") = t
x = 0, 1,...
e
/ x!
where
x
Result. For x = 1, 2,..., the geometric cdf is FX (x ) = 1 (1 p ) . Result. The geometric distribution is the only discrete memoryless distribution. That is, P(X > x + c | X > x ) = P(X > c ). Result. The binomial distribution with p = K / N is a good approximation to the hypergeometric distribution when n is small compared to N .
Purdue University
5 of 22
B.W. Schmeiser
IE230
CONCISE NOTES
Revised August 25, 2008
Continuous Distributions: Summary Table random distribution range variable name cumulative distrib. func. probability density func. dF (y ) dy y =x = f (x ) = f X (x ) 1 b a expected value
variance
2
X
general
(, ) P(X x ) = F (x ) = FX (x ) x a b a
xf (x )dx (x )
f (x )dx
2
= = X = E(X ) a +b 2
= = X = V(X ) 2 2 = E(X ) (b a ) 12
2 2
2
X
continuous [a , b ] uniform
X
triangular
[a , b ]
2(x d ) a +m +b (b a ) (m a )(b m ) (x a ) f (x ) / 2 if x m , else (b a )(m d ) 3 18 1(b x ) f (x )/2 (d = a if x m , else d = b )
1 x
2
sum of random variables time to Poisson count 1 time to Poisson count r lifetime
normal (or Gaussian)
(, ) Table III
e
2
2
2
exponential [0, )
1e
x
e
x
1/
1/
2
Erlang
[0, )
k =r
e
x
(x )
k
x
r r 1 x
e
k!
(r 1)!
r /
r /
2
gamma
[0, ) [0, )
numerical 1e
(x /)
x x
r r 1 x
e
(r )
1 (x /)
r / (1+
r / 1
2
2
lifetime
Weibull
e
)
(1+
2
)
2
Denition. For any r > 0, the gamma function is (r ) = x
0
r 1 x
e
dx .
Result. (r ) = (r 1)(r 1). In particular, if r is a positive integer, then (r ) = (r 1)!. Result. The exponential distribution is the only continuous memoryless distribution. That is, P(X > x + c | X > x ) = P(X > c ). Denition. A lifetime distribution is continuous with range [0, ). Modeling lifetimes. Some useful lifetime distributions are the exponential, Erlang, gamma, and Weibull.
Purdue University
12 of 22
B.W. Schmeiser

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Jason Lim November 5, 2009 Intro to Philosophy Essay #1 and #2 SummaryIn Essay #1 of Metaphysics, Aristotle defines knowledge and how one obtains it. Humanity by its nature, Aristotle says, desires knowledge. Knowledge is gained through the senses, prima

Saint Louis - PHIL - 105

Meditation 1 1. Senses can deceive sometimes. -An object in the distance 2. It is not wise to trust something that has deceived us before. 3. If senses can deceive once, it is a possibility that senses are deceiving all the time. 4. Man sleeps, and in our

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Descartes and Locke The Senses and Perceiving the WorldJason Lim December 2, 2009Rene Descartes and John Locke are two of the most prominent figures in philosophy when it comes to dealing with the self. Descartes establishes the self through his Cogito