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exam2-key

Course Number: IE 230, Fall 2010

College/University: Purdue

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IE 230 Seat # ________ 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,...

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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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Purdue - IE - 230
IE 230Seat # _Name _Closed book and notes. 60 minutes. Cover page and four pages of exam. Pages 8 and 12 of the Concise Notes. No calculator. No need to simplify answers. This test is cumulative, with emphasis on Section 4.7 through Chapter 6 of Montgo
Purdue - IE - 230
IE 230Seat # _ Closed book and notes. 60 minutes. Cover page and four pages of exam. Pages 8 and 12 of the Concise Notes. No calculator. No need to simplify answers.Name _ &lt; KEY &gt; _This test is cumulative, with emphasis on Section 4.7 through Chapter 6
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Purdue - IE - 230
Quiz 1. September 1, 2010Seat # _Name: _ &lt; KEY &gt; _Closed book and notes. No calculators. Set theory. For all questions below, consider the universe U composed of persons in this room now. Let B denote the set of all students born in Indiana, M the set
Purdue - IE - 230
Quiz 2. September 8, 2010Seat # _Name: _Closed book and notes. No calculators. In probability, we always have an experiment. 1. (1 pt) The set of all outcomes is called the _. 2. (1 pt) Each replication of the experiment results in exactly one _. 3. (1
Purdue - IE - 230
Quiz 2. September 8, 2010Seat # _Name: _ &lt; KEY &gt; _Closed book and notes. No calculators. In probability, we always have an experiment. 1. (1 pt) The set of all outcomes is called the _ &lt; sample space &gt; _. 2. (1 pt) Each replication of the experiment re
Purdue - IE - 230
Quiz 3. September 15, 2010Seat # _Name: _Closed book and notes. No calculators. Remember. For T/F questions, a statement is true only if it is always true. Below, assume that all probabilities mentioned are not zero.1. (2 pts) The denition of conditio
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Quiz 3. September 15, 2010Seat # _Name: _Closed book and notes. No calculators. Remember. For T/F questions, a statement is true only if it is always true. Below, assume that all probabilities mentioned are not zero. 1. (2 pts) The denition of conditio
Purdue - IE - 230
Quiz 4. September 29, 2010Seat # _Name: _Closed book and notes. No calculators. Remember. You do not need to simplify answers. Consider ipping a coin twice, independently. For i = 1, 2, let Hi denote that ip i results in &quot;heads&quot; facing up. Let X denote
Purdue - IE - 230
Quiz 4. September 29, 2010Seat # _Name: _ &lt; KEY &gt; _Closed book and notes. No calculators. Remember. You do not need to simplify answers. Consider ipping a coin twice, independently. For i = 1, 2, let Hi denote that ip i results in &quot;heads&quot; facing up. Le
Purdue - IE - 230
Quiz 5. October 6, 2010Seat # _Name: _Closed book and notes. No calculator. For each question, provide the name of the corresponding family of distributions. 1. (1 pt) The 100 coin ips, the number that results in &quot;tails&quot;.2. (1 pt) The number of coin i
Purdue - IE - 230
Quiz 5. October 6, 2010Seat # _Name: _ &lt; KEY &gt; _Closed book and notes. No calculator. For each question, provide the name of the corresponding family of distributions. 1. (1 pt) The 100 coin ips, the number that results in &quot;tails&quot;. binomial 2. (1 pt) T
Purdue - IE - 230
Quiz 6. October 13, 2010Seat # _Name: _Closed book and notes. No calculator. Consider the probability density function f X (y ) = 0.1 for 0 y c and zero elsewhere. 1. (2 pt) Show that c = 10.2. (2 pt) Determine the value of f X (5.6).3. (2 pt) Determ
Purdue - IE - 230
Quiz 6. October 13, 2010Seat # _Name: _Closed book and notes. No calculator. Consider the probability density function f X (y ) = 0.1 for 0 y c and zero elsewhere. 1. (2 pt) Show that c = 10. _ Set 1 = f X (y ) dy = (0.1) dy = 0.1c and solve for c . 0
Purdue - IE - 230
Quiz 7. October 27, 2010Seat # _Name: _Closed book and notes. No calculator. For Questions 13, recall the following three statements. A binomial distribution concerns the number of successes in n Bernoulli trials, when p is the probability of success.
Purdue - IE - 230
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Purdue - IE - 230
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Purdue - IE - 230
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Purdue - IE - 230
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Purdue - IE - 230
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Purdue - MA - 170
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Purdue - MA - 170
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