Advanced Engineering Mathematics Exercises on Module 4: Probability and Statistics

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Advaced Egieerig Mathematics Eercises o Module 4: Probability ad Statistics. A survey of people i give regio showed that 5% drak regularly. The probability of death due to liver disease, give that a perso drak regularly, was 6 times the probability of death due to liver disease, give that a perso did ot drik regularly. The probability of death due to liver disease i the regio is 0.005. If a perso dies due to liver disease what is the probability that he/she drak regularly?. I a productio lie ICs are packed i vials of 5 ad set for ispectio. The probabilities that the umber of defectives i a vial is 0,,, are /, /4, /4, /6 respectively. Two ICs are draw at radom from a vial ad foud to be good. What is the probability that all ICs i this vial are good?. A isurace compay floats a isurace policy for a evetuality takig place with probability 0.05 over the period of policy. If the sum isured is Rs. 00000 the what should be the premium so that the epected earig of the isurace compay is Rs. 000 per policy sold? 4. Suppose is a discrete radom variable with pmf give by α + α P ( = ) =, P ( = 0) =, P ( = ) =, where α is a real umber. Fid the rage of α for which this is a valid pmf. Also determie the maimum ad miimum values of Var (). Write the cdf for α =. Hece show that the media is ot uique. 5. Let be a cotiuous radom variable with the pdf b f( ) = ( + ) 0, otherwise., e < < e, Fid bp, ( < <.5) ad V( ). Also fid the cdf ad quatiles. 6. Let be a radom variable with momet geeratig fuctio t e M () t =. t e Fid P(5 7) ad E ( ). 4 7. The average umber of accidets durig 9:00 a.m. to 9:00 p.m. is thrice the average umber durig 9:00 p.m. to 9:00 a.m. at a traffic juctio. Give o accidets are recorded durig a day (4 hour period), the coditioal probability that the recordig time was betwee 9:00 p.m. to 9:00 a.m. is thrice the

coditioal probability that the recordig time was betwee 9:00 a.m. to 9:00 p.m. What is average umber of accidets durig the day (4 hour period)? 8. Questios are asked to Girish i a quiz competitio oe by oe util he fails to aswer correctly. The probability of his aswerig correctly a questio is p. The probability that he will quit after aswerig a odd umber of questios is 0.9. Fid the value of p. 9. A missile ca successfully hit a target with probability 0.75. If three successful hits ca destroy the target completely, how may missiles must be fired so that the probability of the completely destroyig the target is ot less tha 0.95? 0. A parallel system has idepedet compoets. The lifetime i (i hours) of the i th i compoet is epoetially distributed with mea, for i =,,. What is the probability that the system is workig eve after 7 hours? If the system is workig after 7 hours, what is the probability that oly third compoet is workig?. The time to failure (i years),, of the electroic tubes produced at two maufacturig plats I ad II follows a gamma distributio. For the tubes produced at plat I, the mea is ad the variace 4, whereas for the tubes produced at plat II, the mea is 4 ad variace 8. Plat II produces four times as may tubes as plat I. The tubes are itermigled ad supplied. What is the probability that a tube selected at radom will work for at least 6 years?. The legth of a door hadle (i cm), maufactured by a factory, is ormally distributed with µ = 6.0 ad σ = 0.. The place to fi o the door ca allow error i legth up to 0. cm. What percetage of hadles maufactured i the factory will be defective? How much should be the value of σ, so that the umber of defectives i reduced to oly 5%?. The marks (out of 00) of studets i a class are ormally distributed. Grade A is awarded for more tha 80 marks, grade B betwee 60-80, grade C betwee 40-60 ad F below 40 marks. If 0% studets get A ad 5% get F, fid the percetage of studets gettig grade B. 4. Five percet of items produced by a compay are defective. Items are sold i boes of 00 ad a guaratee is give that a bo cotais ot more tha defectives. Usig Poisso approimatio to biomial distributio, determie the probability that a bo will fail to meet the guaratee? 5. The probability that a studet joiig a Ph.D. program i IIT Kharagpur will complete withi 4 years is 0.5. Fid the approimate probability that out of et 000 studets more tha 60 complete withi 4 years? 6. Fatal accidets are observed to occur at a stretch of highway accordig to a Poisso process at a rate of 5 per moth. Fid the approimate probability that i a year there are less tha 50 fatal accidets at that stretch.

7. Let (, ) have the joit desity fuctio give by a + by f (, y) =, y 0, y, = 0, elsewhere. If E ( ) eists, the fid a ad b. For these values of a ad b fid the margial ad coditioal distributios of ad. Are ad idepedet? Also fid E ( ), E ( ), Var( ), Var( ) ad Cov(, ). Also fid P ( 0.5), P ( ), P ( 0.5 = ), P ( > = 0.5) P ( / ), P ( + < ) ad P ( > ). 8. From a store cotaiig defective, partially defective ad good computers, a radom sample of 4 computers is selected. Fid the epected umber of defective ad partially defective computers i the sample. What will be the covariace betwee the umber of defective ad umber of partially defective computers? 9. Let (, ) have bivariate ormal distributio with desity fuctio 9 ( ) ( ) ( y ) ( y ) 6 + f( y, ) = e, < y, <. 4π Fid the correlatio coefficiet betwee ad, P (< < = ), ad P 4< + < 6. ( ) 0. Let,, be idepedet epoetial radom variables with the probability desity f( ) = λe λ, > 0. Defie radom variables, ad as + = + +, =, =. + + + Fid the joit ad margial desities of, ad. Are they idepedet?. Suppose that for a certai idividual, calorie itake at breakfast is radom variable with mea 500 ad stadard deviatio (s.d.) 50, calorie itake at luch is a radom variable with mea 900 ad s.d. 00, ad calorie itake at dier is a radom variable with mea 000 ad s.d. 80. Assumig that itakes at differet meals are idepedet of oe aother, what is the approimate probability that the average calorie itake per day over the et year (65 days ) is at most 40?. Let,,...,, + be idepedet N(0,) radom variables ad Fid the distributio of / +. i i = =.. Let,,..., be idepedet N(0,) radom variables ad W = ( ) i i. What is the distributio of W? i=

4. Legths of pis (i mm) produced by a machie follow a N ( µσ, ) distributio. Fid the maimum likelihood estimators of µ ad σ based o a radom sample of size 0 with observatios: 7., 7., 7.0, 6.95, 6.89, 6.97, 6.99, 6.9, 7.05, 7.0. 5. Let,,..., be idepedet N ( µσ, ) radom variables. Fid ubiased estimator of µ. Is it cosistet? 6. Let,,..., be idepedet Ep ( µσ, ) radom variables. Fid the method of momets ad maimum likelihood estimators of µ ad σ. Compare them by evaluatig their mea squared errors. 7. Breakig stregth (i kg) of the frot part of a ew vehicle is ormally distributed. I 0 trials the breakig stregths were foud to be 578, 57, 570, 568, 57, 570, 570, 57, 596, 584. Fid a 95% cofidece iterval for the mea breakig stregth. Ca we say that the mea breakig stregth is sigificatly less tha 570 at % level of sigificace? Further test the hypothesis that the variace is less tha 8 sq kg. At 5% level of sigificace. 8. Carbo emissios o 8 radomly selected vehicles of brad A were recorded as 50, 50, 40, 80, 90, 0, 0, 80, whereas those of 0 radomly selected of brad B were recorded as 40, 0, 70, 90, 70, 00, 50, 00, 90, 70. Set up 90% cofidece itervals for the differece i the meas ad ratio of variaces of the two populatios. Also test the hypothesis that the variaces of the two populatios are equal (at 0% level of sigificace). Based o the result of this test, coduct a appropriate test for the hypothesis that the average emissio from vehicles of brad B is less tha the average emissio from vehicles of brad A (at 5% level of sigificace). 9. A eperimet was coducted to compare the recovery time (i days) of patiets from a serious disease usig two differet medicatios. The first medicie was give to a radom sample of 5 patiets ad the sample mea ad sample variace of the recovery time were observed to be 6 ad.4 respectively. The secod medicie was admiistered to a radom sample of 9 patiets ad sample mea ad sample variace of the recovery time were observed to be 0 ad.0 respectively. Test the hypothesis that the average recovery time usig the first medicie is sigificatly less tha the oe by usig the secod (at 5% level). Assume the two populatios to be ormal with equal variaces. 0. I a radom sample of 00 families watchig televisio i Bombay at ay give time, it was foud that 45 were watchig Network A. At the same time, i a radom sample of 0 families watchig televisio i New Delhi, it was foud that were watchig Network A. Test the hypothesis that Network A is equally popular i both states (at this time) at % level of sigificace.

Advaced Egieerig Mathematics Hits to Eercises o Module 4: Probability ad Statistics. Let A deote the evet that the perso driks regularly, ad let B deote the evet that the death is due to liver disease. Give C PA ( ) = 0.5, PB ( ) = 0.005, PB ( A) = 6 PB ( A) = α, say. By theorem of total probability C C PB ( ) = PB ( APA ) ( ) + PB ( A) PA ( ), OR α α 8 0.005 = +. α = 0.005 4 4 6 PB ( APA ) ( ) So PAB ( ) = = PB ( ). Let B j deote the evet that a vial has j defective ICs, for j = 0,,,. Let A deote the evet that two ICs draw at radom from a vial are good. It is give that PB ( 0) =, PB ( ) =, PB ( ) =, PB ( ) =. Also we fid that 4 4 6 4 5 5 PAB ( 0) =, PAB ( ) = =, PAB ( ) = =, 5 0 5 PAB ( ) = =. 0 Usig Bayes Theorem, PA ( B0) PB ( 0) 40 PB ( 0 A) = =. 69 PA ( B) PB ( ) j= 0 j j. Let P be the premium ad let deote the earig of the isurace compay per policy. The P, if there is o evetuality durig period of the policy = P -00000, if there is evetuality durig period of the policy Now E ( ) =.95 P+.05 ( P 00000) = 000 yields P = 6000. So Rs. 6000 should be the premium. 4. We must have 0 α ad 0 + α. This yields α. Also 4α V( ) =. This has miimum value 9 9 ad maimum value. For α =, the cdf is described by 0, <, 6, < 0, F ( ) =, 0 <,,. Let M be the media, the 0 M.

5. We should have f ( ) d =. It gives b =. Now ( < <.5) = ( ) = log e(7 6). P f d e E( ) = f ( ) d = d + e ( e e ) ( e e ) = + ta ta. ( ) = ( ) = e E f d d + e e d ( e e ). + e = = ( ) ( ) V = e e e e e e The cdf is give by ( ) ( ) 4 ta ta. 0, < e, F e e, e. ( ) = log e( + ), <, Q = e, M = e, Q = e. 5/4 / 7/4 6. This is mgf of NB(4, /) distributio. So the pmf of is give by k 4 4 k P ( = k) =, 4,5,... k= 76 4 P(5 7) = 0.69. E ( ) = = 6. 87 / 7. Let be the umber of accidets recorded durig a hour period. Let D deote the period betwee 9:00 a.m. to 9:00 p.m. ad N deote the period betwee 9:00 p.m. to 9: a.m. The it is give that /D ~P( λ ), /N ~P( λ ), for some λ > 0. λ P ( = 0 NPN ) ( ) e / PN ( = 0) = = λ λ P ( = 0 NPN ) ( ) + P ( = 0 DPD ) ( ) e / + e / λ = ( + e ) = α, say λ P ( = 0 NPD ) ( ) e / λ λ ( = 0 ) ( ) + ( = 0 ) ( ) / + / PD ( = 0) = = P NPN P DPD e e λ = ( + e ) = β, say

Give α = β λ = loge 0.549. So the average umber of accidets i a 4 hour period is 4λ = log.97. e 8. Let deote the umber of questios asked to Girish (the last oe he aswers wrogly or fails to aswer). P k p q k k ( = ) =, =,,... q Give P ( = k+ ) = 0.9 = 0.9 p=. p 9 k = 0 9. Let be the umber of successful hits. Suppose missiles are fired. The ~ Bi(,0.75). We wat so that P ( ) 0.95. This is equivalet to P ( = i) 0.05, or, +.. + 0.05, i= 0 4 4 4 or, 4 4 0 (9 + ) 4. The miimum value of for which this is satisfied is = 6. 0. Let deote the system life. The P ( > 7) = P ( 7) = P ( 9 7) = ( e )( e )( e ). i= P ( < 7, < 7, 7 7) P ( < 7, < 7, 7) = P ( 7) 9 ( e )( e ) e =. 9 ( e )( e ) ( e ) r r. Give E( I) = =, Var( I) 4 α = = α. So r =, α =. r r Similarly E( II) = 4 =, Var( I) 8 α = = α. So r =, α =. / Therefore P( > 6 I) = e d = e ad 6 By theorem of total probability 7 P ( > 6) = P ( > 6 IPI ) ( ) + P ( > 6 IIPII ) ( ) = e 0.69. 6. P( hadle is defective ) = P(6 0. < < 6 + 0.) = P(.5 < <.5) 0. = Φ (.5) = 0.0667 = 0.4 So the percetage of defective hadles is.4%. If the percetage of defectives is to be oly 5%, the we must have 0. P(6 0. < < 6 + 0.) = 0.05, or, Φ = 0.05, σ 0. 0. or, Φ = 0.05 =.96 σ = 0.5 σ σ i

. Let be the marks out of 00. The ~ N ( µσ, ). 80 µ 80 µ P ( > 80) = 0. P Z> = 0. =.8, or, σ σ µ +.8σ = 80 () 40 µ 40 µ P ( < 40) = 0.5 Φ = 0.5 =.07, or, σ σ µ.07 σ = 40 () Solvig () ad (), we get µ = 57.905, σ = 7.67. Now P(grade B) = P(60 < 80) = P ( < 80) P ( < 60) 60 µ = 0.9 Φ = 0.9 Φ (0.) = 0.9 0.5478 = 0.5. σ So percetage of studets gettig grade B is 5.%. 4. Let deote the umber of defectives i a bo. The ~ Bi (00, 0.05). Sice = 00, p = 0.05 λ = p = 5. Therefore, P(bo will fail to meet the guaratee) = P ( > ) = P ( ) 5 5 5 5 7 5 = ( e + 5 e + e ) = e = 0.875. 5. Let deote the umber of studets joiig the Ph.D. program i IIT Kharagpur. The ~ Bi (000, 0.5). 50 59.5 50 So P ( > 60) P > PZ ( > 0.69) 87.5 87.5 =Φ ( 0.69) = 0.45. 6. Let deote the umber of fatal accidets over the year. 7. The ~ Ρ(80) Hece 80 59.5 80 P ( < 00) P PZ (.5) = 0.06 80 80 a b So y y 0 f ( y) = f (, y) d = +, y. a E( ) = dy + b dy < b = 0. y y Therefore, a f ( y) =, y. y Now f ( y) dy = a = 4. Hece, f ( y) =, y. Also y 4 f, ( y, ) =, 0, y. The margial pdf of is y f ( ) =, 0. Clearly, ad are idepedet. Other parts ca ow be aswered easily.

8. Let be the umber of defective computers i the sample ad let be the umber of partially defective computers i the sample. The joit pmf of (, ) is give by 0 p () 0 0 /70 9/70 /70 5/70 /70 8/70 8/70 /70 40/70 /70 9/70 /70 0 5/70 p (y) 5/70 0/70 0/70 5/70 Other parts ca ow be aswered easily. 9. µ =, µ =, σ =, σ =, ρ = / The coditioal distributio of = is N (4/,8/9). So P ( < < = ) = 0.979. Also ~ N(5,7). P 4 < + < 6 = 0.896. + So ( ) 0. Apply the Jacobia approach to get the required distributios. They are idepedet. The joit pdf of = (,, ) is λ y f ( y) = λ y e, y > 0,0< y <,0< y <. The margials ca be foud easily ow.. Let be the calorie itake at breakfast, be the calorie itake at luch ad be the calorie itake at dier. Let = + +. The µ = E( ) = 400, σ = Var( ) = 44900. Let i deote the itake o i-th day. Here = 65. Applyig Cetral Limit Theorem 65 L ( 400 ) Z ~ N(0,) 44900 So P ( 40) PZ ( 0.90) = 0.859.. The distributio of / + is t.. The distributio of W is χ. 4. ˆ µ = = ˆ σ = ML 7.006, ML 0.0054 5. A ubiased estimator for 6. µ is S ˆ µ σ, ˆ MM = i MM = i i= i= ˆ µ =, ˆ σ = ML () ML (). It is cosistet. The MSEs ca be calculated usig the form of the distributio. Problems 7-0: Solutios based o the formulae ad methods give i the lectures.