IE 230 Seat # Closed book ad otes. No calculators. 60 miutes, but essetially ulimited time. Cover page, four pages of exam, ad Pages 8 ad 12 of the Cocise Notes. This test covers through Sectio 4.7 of Motgomery ad Ruger, fourth editio. Remider: A statemet is true oly if it is always true. Score Exam #2, Sprig 2007 Schmeiser
IE 230 Probability & Statistics i Egieerig I 1. True or false. (For each: three poits if correct, zero poits if blak or wrog.) (a) T F A radom variable is a fuctio that assigs a real umber to every outcome i the sample space. (b) T F If X is a discrete radom variable, the f X (c )=P(X = c ) for every real umber c. (c) T F If X is a cotiuous radom variable, the f X (c )=P(X = c ) for every real umber c. (d) T F If X is ormally distributed with mea 6 ad stadard deviatio 2.3, the P(X 6)=0.5. (e) T F If X is (cotiuous) uiformly distributed with mea 6 ad stadard deviatio 2.3, the P(X 6) = 0.5. (f) T F Always, the expected value of a radom variable is oe of its possible values. (g) T F If X is a geometric radom variable, the P(X = 3) = P(2.5 X 3.5). (h) T F If X is biomially distributed with mea µ X = 10 ad = 25, the the probability of success is p = 10 / 25. (i) T F Beroulli trials are idepedet of each other. 2. (8 poits) (Motgomery ad Ruger, 3 109) Suppose that X has a Poisso distributio with P(X = 0)=0.06. Determie the value of E(X ). The Poisso pmf is f X (x )=e µ µ x / x! for x = 0, 1,... Therefore, P(X = 0)=e µ µ 0 / 0!=e µ. The coditio that P(X = 0)=0.06 yields µ= l(0.06) 2.81 Exam #2, Sprig 2007 Page 1 of 4 Schmeiser
IE 230 Probability & Statistics i Egieerig I 3. (Motgomery ad Ruger, 3 87) Assume that each of your calls to a popular radio statio has a probability of 0.07 of coectig, that is, of ot obtaiig a busy sigal. Assume that your calls are idepedet. (a) (6 poits) What is the experimet for Part (b)? Call the statio util a call is coected. (b) (6 poits) Determie the probability that your first call that coects is the seveth. Let X deote the umber of calls util the first coected call. The X is geometric with probability of success p = 0.07. Therefore, P(X = 7)= f X (7)=p (1 p ) 6 = (0.07) (0.93) 6 0.0453 (c) (6 poits) Determie the expected umber of calls util your first coectio. Because X is geometric, E(X )=1 / p = 1 / (0.07) 14.3 Exam #2, Sprig 2007 Page 2 of 4 Schmeiser
IE 230 Probability & Statistics i Egieerig I 4. (7 poits) (Motgomery ad Ruger, 4 13) The gap width is a importat property of a magetic recordig head. Suppose that, i coded uits, the width is a cotiuous radom variable havig the rage 0 < x < 2 with f X (x )=x / 2. Determie the cdf F X. For every real umber x, the cdf is F X (x )=P(X x ). Therefore, F X (x )=0if x < 0 F X (x )= 0 x (c / 2)dc = x 2 / 4 if 0 < x < 2 F X (x )=1if x > 2 5. (Motgomery ad Ruger, 4 54) The fill volume of a automated fillig machie used to fill cas of carboated beverage is ormally distributed with a mea of 12.4 fluid ouces ad a stadard deviatio of 0.1. (a) (4 poits) What are the uits of the stadard deviatio? fluid ouces (b) (6 poits) Sketch the pdf of X. Label both axes. Scale the horizotal axis. Sketch a horizotal ad a vertical axis. Sketch the bell curve, which should be symmetric, with the bell disappearig just beyod three times the poits of iflectio. Label the horizotal axis with a dummy variable, such as x. Label the vertical axis with f X (x ). Scale the horizotal axis with at least two poits. Probably oe of the poits is the mea. The poits of iflectio should be at 12.3 ad 12.5 fluid ouces. The bell should disappear at about 12.1 ad 12.7 fluid ouces. (c) (6 poits) O your sketch, idicate the probability that the fill volume is more tha 12 fluid ouces. Shade uder the bell curve everythig to the right of 12 ouces. All (or almost all) of the visible area should be shaded. (d) (6 poits) Determie the value (approximately) of the probability that the fill volume is more tha 12 fluid ouces. 12.0 ouces is four stadard deviatios below the mea. We kow that the area withi three stadard deviatios from the mea is about 0.997. Therefore, the area greater tha 12.1 ouces is about 0.9985. Therefore, the area greater tha 12.0 ouces is betwee 0.9985 ad 1. A reasoable guess might be P(X < 12.0)=0.9999 (The correct value, from Table III or MSexcel or..., is closer to 0.99997.) Exam #2, Sprig 2007 Page 3 of 4 Schmeiser
IE 230 Probability & Statistics i Egieerig I 6. (Motgomery ad Ruger, 4 73) Suppose that the umber of asbestos particles i a sample of oe squared cetimeter of dust is a Poisso radom variable with a mea of 1000. (a) (6 poits) Determie the Poisso-process rate of the asbestos particles. (State the uits.) λ = 1000 (particles per squared cetimeter) (b) (6 poits) Determie the probability that o particles are foud i two square cetimeters of dust. Let X deote the umber of particles i two square cetimeters. The X is Poisso with mea µ = λ t = (1000) (2) = 2000 particles. Therefore, P(X = x )= f X (x )=e µ µ x / x! for x = 0, 1,... Therefore, P(X = 0)=e 2000 2000 0 / 0!=e 2000 (c) (6 poits) Cosider te squared cetimeters of dust. Determie the expected umber of asbestos particles. Let Y deote the umber of particles i te square cetimeters. The X is Poisso with mea µ = λ t = (1000) (10) = 10000 particles. Exam #2, Sprig 2007 Page 4 of 4 Schmeiser
IE 230 Probability & Statistics i Egieerig I Discrete Distributios: Summary Table radom distributio rage probability expected variace variable ame mass fuctio value X geeral x 1, x 2,..., x P(X = x ) x i f (x i ) (x i µ) 2 f (x i ) = f (x ) =µ=µ X =σ 2 2 =σ X = f X (x ) = E(X ) = V(X ) = E(X 2 ) µ 2 X discrete x 1, x 2,..., x 1 / x i / [ x i 2 / ] µ 2 uiform X equal-space x = a,a+c,...,b 1 / a+b c 2 ( 2 1) 2 12 uiform where = (b a+c ) / c idicator x = 0, 1 p x (1 p ) 1 x p p (1 p ) variable biomial x = 0, 1,..., C x p x (1 p ) x p p (1 p ) K N K hyper- x = C x C x / N C p p (1 p ) (N ) (N 1) geometric ( (N K )) +,..., mi{k, } where p = K / N ad iteger geometric x = 1, 2,... p (1 p ) x 1 1 / p (1 p ) / p 2 egative x = r, r+1,... x 1 C r 1 p r (1 p ) x r r / p r (1 p ) / p 2 biomial Poisso x = 0, 1,... e µ µ x / x! µ µ where µ=λt Result. For x = 1, 2,..., the geometric cdf is F X (x )=1 (1 p ) x. Result. The geometric distributio is the oly discrete memoryless distributio. That is, P(X > x + c X > x )=P(X > c ). Result. The biomial distributio with p = K / N is a good approximatio to the hypergeometric distributio whe is small compared to N. Purdue Uiversity 8 of 25 B.W. Schmeiser
IE230 CONCISE NOTES Revised August 25, 2004 Cotiuous Distributios: Summary Table radom distributio rage cumulative probability expected variace variable ame distrib. fuc. desity fuc. value X geeral (,) P(X x ) df (y ) dy y=x xf (x )dx (x µ) 2 f (x )dx = F (x ) = f (x ) =µ=µ X =σ 2 2 =σ X = F X (x ) = f X (x ) = E(X ) = V(X ) = E(X 2 ) µ 2 X cotiuous [a, b ] x a 1 a + b (b a ) 2 b a b a 2 12 uiform (x a )f (x ) / 2 X triagular [a, b ] 2(x d ) a+m+b (b a ) 2 (m a )(b m ) 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 µ 2 e 2 σ sum of ormal (,) Table III µ σ 2 2πσ radom (or variables Gaussia) time to expoetial [0,) 1 e λx λ e λx 1 /λ 1 /λ 2 Poisso cout 1 time to Erlag [0, ) k=r e λx (λx ) k λ r x r 1 e λx r /λ r /λ 2 k! (r 1)! Poisso cout r lifetime gamma [0, ) umerical λ r x r 1 e λx r /λ r /λ 2 Γ(r ) lifetime Weibull [0,) 1 e (x βx β 1 (x /δ)β e /δ)β δγ(1+ 1 ) δ 2 Γ(1+β 2 ) µ 2 δ β β Defiitio. For ay r > 0, the gamma fuctio isγ(r )= x r 1 e x dx. 0 Result. Γ(r ) = (r 1)Γ(r 1). I particular, if r is a positive iteger, the Γ(r ) = (r 1)!. Result. The expoetial distributio is the oly cotiuous memoryless distributio. That is, P(X > x + c X > x )=P(X > c ). Defiitio. A lifetime distributio is cotiuous with rage [0,). Modelig lifetimes. Some useful lifetime distributios are the expoetial, Erlag, gamma, ad Weibull. Purdue Uiversity 12 of 25 B.W. Schmeiser