12Continuous ONLINE PAGE PROOFS. probability distributions Kick off with CAS

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1 Continuous proility distriutions. Kick off with CAS. Continuous rndom vriles nd proility functions. The continuous proility density function.4 Mesures of centre nd spred.5 Liner trnsformtions.6 Review

2 . Kick off with CAS Eploring proility density functions Continuous dt is dt tht is mesured, such s heights nd weights, nd my ssume ny vlue within given rnge. Often the dt is recorded in intervls, such s heights etween 5 nd 6 cm, nd the proility of vlue flling within prticulr intervl cn e clculted. These proilities cn e sketched vi histogrm, nd the midpoints of ech column on the histogrm cn e joined to form grph. Topic 8 covered res under curves; recll tht s the intervls get smller nd smller, the grph joining the midpoints pproches smooth curve. Some curves cn e modelled y specific functions. If the re under function is ectly, nd if f() for ll -vlues, then this function cn e clled proility density function. These two conditions must e met. Sketch the grph of f() = ( ) over y the domin. Wht is the re under the curve for the given domin? Is f()? c Is this grph proility distriution function? Sketch the grph of f() = 5e 5 over the domin. Wht is the re under the curve for the given domin? Is f()? c Is this grph proility distriution function? Sketch the grph of f() = sin() over the domin π. Wht is the re under the curve for the given domin? Is f()? c Is this grph proility distriution function? y = f() (, ) 4 If time permits, design your own proility density function, ensuring tht the two key conditions re met. Plese refer to the Resources t in the Prelims section of your ebookplus for comprehensive step-y-step guide on how to use your CAS technology.

3 . Interctivity Continuous rndom vriles int-64 Continuous rndom vriles nd proility functions Continuous rndom vriles Discrete dt is dt tht is finite or countle, such s the numer of soft-centred chocoltes in o of soft- nd hrd-centred chocoltes. A continuous rndom vrile ssumes n uncountle or infinite numer of possile outcomes etween two vlues. Tht is, the vrile cn ssume ny vlue within given rnge. For emple, the irth weights of ies nd the numer of millimetres of rin tht flls in night re continuous rndom vriles. In these emples, the mesurements come from n intervl of possile outcomes. If neworn oy is weighed t 4.46 kilogrms, tht is just wht the weight scle s output sid. In relity, he my hve weighed kilogrms. Therefore, possile rnge of outcomes is vlid, within n intervl tht depends on the precision of the scle. Consider n Austrlin helth study tht ws conducted. The study trgeted young people ged 5 to 7 yers old. They were sked to estimte the verge numer of hours of physicl ctivity they prticipted in ech week. The results of this study re shown in the following histogrm. Frequency y Physicl ctivity Hours Rememer, continuous dt hs no limit to the ccurcy with which it is mesured. In this cse, for emple, < mens from seconds to 59 minutes nd 59 seconds, nd so on, ecuse is not restricted to integer vlues. In the physicl ctivity study, tking on prticulr vlue is equivlent to tking on vlue in n pproprite intervl. For instnce, Pr(X =.5) = Pr( X < ) Pr(X =.5) = Pr( X < ) nd so on. From the histogrm, Pr(X =.5) = Pr( X < ) 56 = ( ) = In nother study, the nose lengths, X millimetres, of 75 dults were mesured. This dt is continuous ecuse the results re mesurements. The result of the study is shown in the tle nd ccompnying histogrm. 454 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

4 Nose length Frequency 7.5 < X.5.5 < X < X < X < X < X < X < X < X 7.5 Frequency y Nose length Length in mm It is possile to use the histogrm to find the numer of people who hve nose length of less thn 47.5 mm Pr(nose length is < 47.5) = 75 = = 5 It is worth noting tht we cnnot find the proility tht person hs nose length which is less thn 45 mm, s this is not the end point of ny intervl. However, if we hd mthemticl formul to pproimte the shpe of the grph, then the formul could give us the nswer to this importnt question. In the histogrm, the midpoints t the top of ech r hve een connected y line segments. If the clss intervls were much smller, sy mm or even less, these line segments would tke on the ppernce of smooth curve. This smooth curve is of considerle importnce for continuous rndom vriles, ecuse it represents the proility density function for the continuous dt. This prolem for continuous rndom vrile cn e ddressed y using clculus. Topic Continuous proility distriutions 455

5 For ny continuous rndom vrile, X, the proility density function is such tht Pr( < X < ) = f()d which is the re under the curve from = to =. Units & 4 AOS 4 Topic Concept Proility density functions Concept summry Prctice questions Interctivity Proility density functions int-644 f() A proility density function must stisfy the following conditions: f() for ll [, ] f()d = ; this is solutely criticl. Other properties re: Pr(X = ) =, where [, ] Pr( < X < ) = P( X < ) = Pr( < X ) = Pr( X ) = f()d Pr(X < c) = Pr(X c) = Proility density functions In theory, the domin of continuous proility density function is R, so tht f()d =. However, if we must ddress the condition tht f()d =, then the function must e zero everywhere else. c f()d when, nd < c <. 456 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

6 WOrKeD exmple Sketch the grph of ech of the following functions nd stte whether ech function is proility density function. ( ), f() =, elsewhere f() =.5, 4, elsewhere c f() = e,, elsewhere think Sketch the grph of f() = ( ) over the domin, giving n -intercept of nd n end point of (, ). Mke sure to include the horizontl lines for y = either side of this grph. Note: This function is known s tringulr proility function ecuse of its shpe. Inspect the grph to determine if the function is lwys positive or zero, tht is, f() for ll [, ]. Clculte the re of the shded region to determine if ( )d =. WritE/drW f() f() = ( ) (, ) (, ) (, ) Yes, f() for ll -vlues. Method : Using the re of tringles Are of shded region = se height = = Method : Using clculus Are of shded region = = ( )d ( )d = = ( ()) ( ()) = + = 4 Interpret the results. f() for ll vlues, nd the re under the curve =. Therefore, this is proility density function. topic COntInuOus proility DIstrIutIOns 457

7 Sketch the grph of f() =.5 for 4. This gives horizontl line, with end points of (,.5) nd (4,.5). Mke sure to include the horizontl lines for y = on either side of this grph. Note: This function is known s uniform or rectngulr proility density function ecuse of its rectngulr shpe. Inspect the grph to determine if the function is lwys positive or zero, tht is, f() for ll [, ]. Clculte the re of the shded region to determine if 4.5d =. f().5 f() =.5 (,.5) (4,.5) (, ) (4, ) Yes, f() for ll -vlues. Agin, it is not necessry to use clculus to find the re. Method : Are of shded region = length width =.5 = Method : Are of shded region = 4.5d =.5 4 =.5(4).5() = = 4 Interpret the results. f() for ll vlues, nd the re under the curve =. Therefore, this is proility density function. c Sketch the grph of f() = e for. End points will e (, ) nd (, e ). Mke sure to include the horizontl lines for y = on either side of this grph. c f() (, ) f() = e (, e ) (, ) (, ) 458 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

8 Inspect the grph to determine if the function is lwys positive or zero, tht is, f() for ll [, ]. Yes, f() for ll -vlues. Clculte the re of the shded region think to determine if WOrKeD exmple Given tht the functions elow re proility density functions, find the vlue of in ech function. f() = ( ), 4, elsewhere As the function hs lredy een defined s proility density function, this mens tht the re under the grph is definitely. WritE Remove from the integrl, s it is constnt. Antidifferentite nd sustitute in the terminls. 4 Solve for. e d =. e d = 4 e d = e = ( e + e ) = ( e + ) =.79 4 Interpret the results. f() for ll vlues. However, the re under the curve. Therefore this is not proility density function. 4 f() = e 4, >, elsewhere f()d = ( ) d = 4 4 ( ) d = ( ) d = ( ) ( ) 4 = = 9 + = 8 = = 8 topic COntInuOus proility DIstrIutIOns 459

9 As the function hs lredy een defined s proility density function, this mens tht the re under the grph is definitely. f()d = e 4 d = Remove from the integrl, s it is constnt. To evlute n integrl contining infinity s one of the terminls, we find the pproprite limit. Eercise. PRctise Work without CAS lim Continuous rndom vriles nd proility functions WE Sketch ech of the following functions nd determine whether ech one is proility density function. f() = 4 e, log e.5, f() =, elsewhere, elsewhere Sketch ech of the following functions nd determine whether ech one is proility density function. f() = cos(), π π f() =, 4, elsewhere, elsewhere k k 4 Antidifferentite nd sustitute in the terminls. lim k 5 Solve for. Rememer tht numer divided y n etremely lrge numer is effectively zero, so lim =. k e 4k lim lim k lim k k k e 4 d = e 4 d = e 4 d = 4 e 4 e 4k e 4k k = = = lim k 4e + 4k 4 = + 4 = 4 = = 4 46 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

10 Consolidte Apply the most pproprite mthemticl processes nd tools WE Given tht the function is proility density function, find the vlue of n. f() = n( ),, elsewhere 4 Given tht the function is proility density function, find the vlue of. f() =, <,, elsewhere 5 A smll cr-hire firm keeps note of the ge nd kilometres covered y ech of the crs in their fleet. Generlly, crs re no longer used once they hve either covered 5 kilometres or re more thn five yers old. The following informtion descries the ges of the crs in their current fleet. Age Frequency < < 6 < 8 < 4 4 < 5 5 < < 7 Frequency y Age of rentl cr 4 5 Age in yers Determine: i Pr(X ) ii Pr(X > 4). Determine: i Pr( < X 4) ii Pr(X > X 4). 6 The ttery life for tteries in television remote controls ws investigted in study. Hours of life Frequency < < < 45 Frequency y Remote ttery life 45 < < Bttery life in hours How mny remote control tteries were included in the study? Wht is the proility tht ttery will lst more thn 45 hours? c Wht is the proility tht ttery will lst etween 5 nd 6 hours? d A new ttery producer is dvocting tht their tteries hve long life of 6+ hours. If it is known tht this is just dvertising hype ecuse these tteries re no different from the tteries in the study, wht is the proility tht these new tteries will hve life of 6+ hours? Topic Continuous proility distriutions 46

11 7 A numer of eperienced shot-putters were sked to im for line metres wy. After ech of them put their shot, its distnce from the -metre line ws mesured. All of the shots were on or etween the 8- nd -metre lines. The results of the mesurements re shown, where X is the distnce in metres from the -metre line. Shot puts y 8 Metres Frequency 7 6 < < 6 4 < < Distnce in metres How mny shot-put throws were mesured? Clculte: i Pr(X >.5) ii Pr( < X ) c A guest shot-putter is visiting the thletics clu where the mesurements re eing conducted. His shot-putting ility is equivlent to the ilities of the clu memers. Find the proility tht he puts the shot within 5 cm of the -metre line if it is known tht he put the shot within metre of the -metre line. 8 Sketch ech of the following functions nd determine whether ech function is proility density function. Note: Use CAS where pproprite. f() = π, e cos() +, f() = 4 π 4, elsewhere, elsewhere c f() = sin(), π, elsewhere Frequency d f() =, <, elsewhere 9 The rectngulr function, f, is defined y the rule c,.5 < <.65 f() =., elsewhere Find the vlue of the constnt c, given tht f is proility density function. 46 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

12 The grph of function, f, is shown. y (, z) (, ) (5, ) If f is known to e proility density function, show tht the vlue of z is. Find the vlue of the constnt m in ech of the following if ech function is proility density function. f() = m(6 ),, elsewhere c f() = me, log e, elsewhere f() = me,, elsewhere Let X e continuous rndom vrile with the proility density function Show tht the vlue of k is f() = + k +,., elsewhere 9. X is continuous rndom vrile such tht f() = nd log e,, elsewhere f()d =. The grph of this function is shown. Find the vlue of the constnt. 4 X is continuous rndom vrile such tht, < f() = where is constnt.,, elsewhere Y is nother continuous rndom vrile such tht f(y) = y, y e., elsewhere f() log e() (, ) (, log e ()) (, ) Topic Continuous proility distriutions 46

13 Mster. Units & 4 AOS 4 Topic Concept Clculting proilities Concept summry Prctice questions Sketch the grph of the function for X nd find Sketch the grph of the function for Y nd find c Find the vlue of the constnt if f()d = 5 X is continuous rndom vrile such tht f() = e e f()d. f(y)dy. f(y)dy. n sin() cos(), < < π., elsewhere If f is known to e proility density function, find the vlue of the constnt, n. 6 A function f is defined y the rule If f() = log e (), >, elsewhere. f()d =, find the vlue of the rel constnt. Does this function define proility density function? The continuous proility density function As stted in section., if X is continuous rndom vrile, then Pr( X ) = f()d. In other words, y finding the re etween the curve of the continuous proility function, the -is, the line = nd the line =, providing f(), then we re finding Pr( X ). It is worth noting tht ecuse we re deling with continuous rndom vrile, Pr(X = ) =, nd consequently: Pr( X ) = Pr( < X ) = Pr( X < ) = Pr( < X < ) Pr( X ) = Pr( X c) + Pr(c < X ), where < c < f() This property is prticulrly helpful when the proility density function is hyrid function nd the required proility encompsses two functions. 464 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

14 WOrKeD exmple A continuous rndom vrile, Y, hs proility density function, f, defined y y, y where is constnt. f(y) = y, < y, elsewhere Sketch the grph of f. Find the vlue of the constnt,. c Determine Pr( Y ). d Determine Pr(Y < Y > ) think WritE/drW The hyrid function contins three sections. f( ) = nd f() = The first grph, f(y) = y, is stright f(y) line with end points of (, ) nd (, ). (, ) (, ) The second grph is lso stright line nd hs end points of (, ) nd (, ). Don t forget to include the f(y) = lines for > nd <. (, ) (, ) (, ) Use the fct tht for. f(y)dy = to solve c Identify the prt of the function tht the required y-vlues sit within: the vlues Y re within the region where f(y) = y. 9 f(y)dy = Using the re of tringle, we find: + = = c Pr( Y ) = = 9 = = 9 f(y)dy 9 y dy = y 8 = 8 () 8 () = 8 8 = 4 9 Note: The method of finding the re of trpezium could lso e used. y topic COntInuOus proility DIstrIutIOns 465

15 d Stte the rule for the conditionl proility. Find Pr( < Y < ). As the intervl is cross two functions, the intervl needs to e split. To find the proilities we need to find the res under the curve. 4 Antidifferentite nd evlute fter sustituting the terminls. 5 Find Pr(Y > ). As the intervl is cross two functions, the intervl needs to e split. 6 To find the proilities we need to find the res under the curve. As Pr( Y ) covers ectly hlf the re under the curve, Pr( Y ) =. (The entire re under the curve is lwys for proility density function.) 7 Antidifferentite nd evlute fter sustituting the terminls. 8 Now sustitute into the formul to find Pr( < Y < ) Pr(Y < Y > ) =. Pr(Y > ) d Pr(Y < Y > ) = = Pr(Y < Y > ) Pr(Y > ) Pr( < Y < ) Pr(Y > ) Pr( < Y < ) = Pr( < Y < ) + Pr( Y < ) = 9 ydy + 9 = = ydy ydy ydy 8 y + 8 y = 8 () 8 ( ) + 8 () 8 () = = 5 8 Pr(Y > ) = Pr( < Y < ) + Pr( Y ) = 9 ydy + = = 9 ydy y + ydy = 8 () 8 ( ) + = = 8 = 5 9 Pr( < Y < ) Pr(Y < Y > ) = Pr(Y > ) = = = 466 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

16 Eercise. PRctise Work without CAS Consolidte Apply the most pproprite mthemticl processes nd tools The continuous proility density function WE The continuous rndom vrile Z hs proility density function given y f(z) = Sketch the grph of f. Find Pr(Z <.75). c Find Pr(Z >.5). z +, z < z, z., elsewhere The continuous rndom vrile X hs proility density function given y where is constnt. f() = 4,, elsewhere Find the vlue of the constnt. Sketch the grph of f. c Find Pr(.5 X ). Let X e continuous rndom vrile with proility density function defined y f() = Sketch the grph of f. Find Pr π 4 < X < π 4. c Find Pr X > π 4 X < π 4. sin(), π, elsewhere. 4 A proility density function is defined y the rule f() = k( + ), < k( ),, elsewhere where X is continuous rndom vrile nd k is constnt. Sketch the grph of f. Show tht the vlue of k is 4. c Find Pr( X ). d Find Pr(X X ). Topic Continuous proility distriutions 467

17 5 The mount of petrol sold dily y usy service sttion is uniformly distriuted proility density function. A minimum of 8 litres nd mimum of litres re sold on ny given dy. The grph of the function is shown. Find the vlue of the constnt k. Find the proility tht etween nd 5 litres of petrol re sold (8, k) (, k) on given dy. c Find the proility tht s much s 6 litres of petrol were sold on prticulr dy, given tht it ws known tht t lest litres were sold Petrol sold (thousnds of litres) 6 The continuous rndom vrile X hs uniform rectngulr proility density function defined y f() = 5, 6., elsewhere Sketch the grph of f. Determine Pr( X 5). 7 The continuous rndom vrile Z hs proility density function defined y f(z) = z, z e., elsewhere e Sketch the grph of f nd shde the re tht represents f(z)dz. Find e f(z)dz. Eplin your result. The continuous rndom vrile U hs proility function defined y f(u) = e4u, u, elsewhere. c Sketch the grph of f nd shde the re tht represents constnt. Frequency d Find the ect vlue of the constnt if k e f(z)dz is equl to f(u)du, where is f(u)du. 8 The continuous rndom vrile Z hs proility density function defined y f(z) = cos(z), π z π., elsewhere 468 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

18 Sketch the grph of f nd verify tht y = f(z) is proility density function. Find Pr π 6 Z π 4. 9 The continuous rndom vrile U hs proility density function defined y where is constnt. Find: f(u) = 4 (u u ), u, elsewhere the vlue of the constnt Pr(U <.75) c Pr(. < U <.5) d Pr(U =.8). The continuous rndom vrile X hs proility density function defined y Find: Pr(X >.) f () = 8,., elsewhere Pr(X > X >.5), correct to 4 deciml plces c the vlue of n such tht Pr(X n) =.75. The continuous rndom vrile Z hs proility density function defined y f(z) = z e where is constnt. Find:, z, elsewhere the vlue of the constnt such tht f(z)dz = Pr( < Z <.7) c Pr(Z <.7 Z >.), correct to 4 deciml plces d the vlue of α, correct to deciml plces, such tht Pr(Z α) =.54. The continuous rndom vrile X hs proility density function given s f() = e,, elsewhere. Sketch the grph of f. Find Pr( X ), correct to 4 deciml plces. c Find Pr(X > ), correct to 4 deciml plces. Mster The continuous rndom vrile X hs proility density function defined y f() = log e ( ),, elsewhere. Find, correct to 4 deciml plces: the vlue of the constnt if Pr(.5 X ). f()d = Topic Continuous proility distriutions 469

19 .4 Units & 4 AOS 4 Topic Concept Men nd medin Concept summry Prctice questions Interctivity Men int The grph of the proility function is shown. f(z) = π(z + ) Find, correct to 4 deciml plces, Pr(.5 < Z <.5). Suppose nother proility density function is defined s f() = Find the vlue of the constnt. +,., elsewhere Mesures of centre nd spred The commonly used mesures of centrl tendency nd spred in sttistics re the men, medin, vrince, stndrd devition nd rnge. These sme mesurements re pproprite for continuous proility functions. Mesures of centrl tendency The men Rememer tht for discrete rndom vrile, E(X) = μ = = n = n Pr(X = n ). This definition cn lso e pplied to continuous rndom vrile. We define E(X) = μ = f()d. f(z) If f() = everywhere ecept for [, ], where the function is defined, then E(X) = μ = (, π ) f(z) = f()d. π(z + ) z 47 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

20 Consider the continuous rndom vrile, X, which hs proility density function defined y f() =,, elsewhere For this function, E(X) = μ = = = = 4 = 4 = 4 f()d ( )d d 4 4 Similrly, if the continuous rndom vrile X hs proility density function of then f() = 7e 7,, elsewhere, E(X) = μ = = lim f()d k k 7e 7 d =.49 where CAS technology is required to determine the integrl. The men of function of X is similrly found. The function of X, g(), hs men defined y: E(g()) = μ = g()f()d. So if we gin consider f() =,, elsewhere Topic Continuous proility distriutions 47

21 then E(X ) = = f()d 4 d Interctivity Medin nd percentiles int-646 = 5 5 = 5 5 = 5 This definition is importnt when we investigte the vrince of continuous rndom vrile. Medin nd percentiles The medin is lso known s the 5th percentile, Q, the hlfwy mrk or the middle vlue of the distriution. For continuous rndom vrile, X, defined y the proility function f, the medin cn e found y solving m f ()d =.5. Other percentiles tht re frequently clculted re the 5th percentile or lower qurtile, Q, nd the 75th percentile or upper qurtile, Q. The interqurtile rnge is clculted s: IQR = Q Q Consider continuous rndom vrile, X, tht hs proility density function of, 5 f() =.e, elsewhere. To find the medin, m, we solve for m s follows: m.e d =.5 The re under the curve is equted to.5, giving hlf of the totl re nd hence the 5th percentile. Solving vi CAS, the result is tht m = This cn e seen on grph s shown. f().5 f() =.e = 5 47 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

22 Consider the continuous rndom vrile X, which hs proility density function of f() = 4, The medin is given y Pr( m) =.5: m m 4 d = m = 6 = m 4 = 8 m = ± 4 8, elsewhere. m =.688 ( m ) To find the lower qurtile, we mke the re under the curve equl to.5. Thus the lower qurtile is given y Pr( ) =.5: 4 4 d = = 4 6 = 4 4 = 4 = ± 4 4 = Q =.44 ( m) Similrly, to find the upper qurtile, we mke the re under the curve equl to.75. Thus the upper qurtile is given y Pr( n) =.75: n d = n = 4 n 4 6 = 4 n 4 = n = ± 4 n = Q =.86 (m ) Topic Continuous proility distriutions 47

23 WOrKeD exmple 4 think So the interqurtile rnge is given y Q Q = These vlues re shown on the following grph. f() (, ) f() = 4 Upper qurtile =.86 Lower qurtile =.44 =.447. (, ) Medin =.688 (, ) A continuous rndom vrile, Y, hs proility density function, f, defined y f( y) = ky, y, elsewhere where k is constnt. Sketch the grph of f. Find the vlue of the constnt k. c Find: i the men of Y ii the medin of Y. d Find the interqurtile rnge of Y. The grph f(y) = ky is stright line with end points t (, ) nd (, k). Rememer to include the lines f(y) = for y > nd y <. WritE/drW f(y) k (, k) (, ) (, ) y 474 Mths Quest MtheMtICL MethODs VCe units nd 4

24 Solve ky dy = to find the vlue of k. ky dy = k y dy = c i Stte the rule for the men. c i μ = k y = k() = k = k = Using the re of tringle lso enles you to find the vlue of k. k = k = k = = y(y)dy y dy Antidifferentite nd simplify. = y ii Stte the rule for the medin. ii m m = () = f(y)dy =.5 Antidifferentite nd solve for m. Note tht m must e vlue within the domin of the function, so within y. ydy =.5 y m =.5 m =.5 m = ± m = ( < m < ) Topic Continuous proility distriutions 475

25 Units & 4 AOS 4 Topic Concept 4 Vrince nd stndrd devition Concept summry Prctice questions Write the nswer. Medin = d Stte the rule for the lower qurtile, Q. Mesures of spred Vrince, stndrd devition nd rnge The vrince nd stndrd devition re importnt mesures of spred in sttistics. From previous clcultions for discrete proility functions, we know tht Vr(X ) = E(X ) [E(X )] nd SD(X ) = Vr(X ) For continuous proility functions, Vr(X) = d ( μ) f() d f(y)dy =.5 ydy =.5 Antidifferentite nd solve for Q. y =.5 =.5 = ±.5 Stte the rule for the upper qurtile, Q. n n = Q =.5 < Q < f(y)dy =.75 ydy =.75 4 Antidifferentite nd solve for Q. y n =.75 n =.75 n = ±.75 n = Q = Stte the rule for the interqurtile rnge. IQR = Q Q 6 Sustitute the pproprite vlues nd simplify. = =.66 < Q < Interctivity Vrince, stndrd devition nd rnge int-647 = ( μ + μ )f()d 476 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

26 WOrKeD exmple 5 think = f()d = E(X ) μ f()μd + f()d + μ = E(X ) μ E(X) + μ = E(X ) μ + μ = E(X ) μ = E(X ) [E(X)] Two importnt fcts were used in this proof: Sustituting this result into SD(X) = μ f()d f()d f()d = nd Vr(X) gives us SD(X) = E(X ) E(X). f()d = μ = E(X). The rnge is clculted s the highest vlue minus the lowest vlue, so for the proility density function given y f() = 5, 6, the highest possile, elsewhere -vlue is 6 nd the lowest is. Therefore, the rnge for this function = 6 = 5. For continuous rndom vrile, X, with proility density function, f, defined y f() = +, 4 find:, elsewhere the men the medin c the vrince d the stndrd devition, correct to 4 deciml plces. WritE Stte the rule for the men nd simplify. μ = 4 f()d = = d + d topic COntInuOus proility DIstrIutIOns 477

27 Antidifferentite nd evlute. = 6 + Stte the rule for the medin. Antidifferentite nd solve for m. The qudrtic formul is needed s the qudrtic eqution formed cnnot e fctorised. Alterntively, use CAS to solve for m. m 4 4 = 6 ( ) + ( ) 6 ( 4) + ( 4) = = m 4 f()d =.5 + d =.5 So m = 8 ± (8) 4()(4) () m = 8 ± 8 = 4 ± m = 4 + s m 4,. Write the nswer. The medin is 4 +. c Write the rule for vrince. c Vr(X) = E(X ) [E(X)] Find E(X ) first. E(X ) = 4 + m = m + m ( 4) + ( 4) = m + m + 4 =.5 m + 8m + 6 = = = 4 f()d 4 + d + d = = 8 ( )4 + ( ) 8 ( 4)4 + ( 4) = 6 = + = m + 8m + 4 = Mths Quest MATHEMATICAL METHODS VCE Units nd 4

28 Sustitute E(X) nd E(X ) into the rule for vrince. Vr(X) = E(X ) E(X) = 8 = 64 9 = d Write the rule for stndrd devition. d SD(X) = Vr(X) Sustitute the vrince into the rule nd evlute. Eercise.4 PRctise Work without CAS Question Mesures of centre nd spred WE4 The continuous rndom vrile Z hs proility density function of z, z f(z) = where is constnt., elsewhere Find the vlue of the constnt. Find: i the men of Z ii the medin of Z. The continuous rndom vrile, Y, hs proility density function of f(y) = where is constnt. Find, correct to 4 deciml plces: the vlue of the constnt c the medin vlue of Y. y, y, elsewhere E(Y) WE5 For the continuous rndom vrile Z, the proility density function is log e (z), z e f(z) =, elsewhere. Find the men, medin, vrince nd stndrd devition, correct to 4 deciml plces. 4 The function = 9 = 9 =.474 f() = e,, elsewhere defines the proility density function for the continuous rndom vrile, X. Find the men, medin, vrince nd stndrd devition of X. Topic Continuous proility distriutions 479

29 Consolidte Apply the most pproprite mthemticl processes nd tools 5 Let X e continuous rndom vrile with proility density function of f() =,., elsewhere Prove tht f is proility density function. Find E(X). c Find the medin vlue of f. 6 The time in minutes tht n individul must wit in line to e served t the locl nk rnch is defined y f(t) = e t, t where T is continuous rndom vrile. Wht is the men witing time for customer in the queue, correct to deciml plce? Clculte the stndrd devition for the witing time in the queue, correct to deciml plce. c Determine the medin witing time in the queue, correct to deciml plces. 7 The continuous rndom vrile Y hs proility density function defined y y f(y) =, y 9., elsewhere Find, correct to 4 deciml plces: the epected vlue of Y the medin vlue of Y c the lower nd upper qurtiles of Y d the inter-qurtile rnge of Y. 8 The continuous rndom vrile Z hs proility density function defined y f(y) = z, z 8, elsewhere where is constnt. Find the vlue, correct to four deciml plces, of the constnt. Find E(Z) correct to 4 deciml plces. c Find Vr(Z) nd SD(Z). d Determine the interqurtile rnge for Z. e Determine the rnge for Z. 9 X is continuous rndom vrile. The grph of the f() proility density function f() = f() = (sin() + ) for π π (sin() + ) π π is shown. Show tht f() is proility density function. Clculte E(X) correct to 4 deciml plces. c Clculte, correct to 4 deciml plces: i Vr(X) ii SD(X). d Find the medin vlue of f correct to 4 deciml plces. (, π ) (π, π ) Mths Quest MATHEMATICAL METHODS VCE Units nd 4

30 The continuous rndom vrile X hs proility density function defined y f() =,, elsewhere. Find the vlues of the constnts nd if E(X) =. The continuous rndom vrile, Z, hs proility density function of f(z) = z, z, elsewhere where is constnt. Show tht the vlue of is. Find the men vlue nd vrince of f correct to 4 deciml plces. c Find the medin nd interqurtile rnge of f. Find the derivtive of 4. Hence, find the men vlue of the proility density function defined y, f() = π 4., elsewhere Consider the continuous rndom vrile X with proility density function of f() = where h is constnt. h( ), h( ), < 4, elsewhere Find the vlue of the constnt h. Find E(X). c Find Vr(X). 4 Consider the continuous rndom vrile X with proility density function of f() = k,, elsewhere where, nd k re positive constnts. Sketch the grph of the function f. Show tht k =. c Find E(X) in terms of nd. d Find Vr(X) in terms of nd. Mster 5 The continuous rndom vrile Y hs proility density function f( y) =. log y e, y , elsewhere Verify tht f is proility density function. Find E(Y) correct to 4 deciml plces. c Find Vr(Y) nd SD(Y) correct to 4 deciml plces. d Find the medin vlue of Y correct to 4 deciml plces. e Stte the rnge. Topic Continuous proility distriutions 48

31 .5 6 The continuous rndom vrile Z hs proility density function f(z) = where is constnt. z, z, elsewhere Find the vlue of the constnt correct to 4 deciml plces. Determine, correct to 4 deciml plces: i E(Z) ii E(Z ) iii Vr(Z) Liner trnsformtions iv SD(Z). Sometimes it is necessry to pply trnsformtions to continuous rndom vrile. A trnsformtion is chnge tht is pplied to the rndom vrile. The chnge my consist of one or more opertions tht my involve dding or sutrcting constnt or multiplying or dividing the vrile y constnt. Suppose liner trnsformtion is pplied to the continuous rndom vrile X to crete new continuous rndom vrile, Y. For instnce Y = X + It cn e shown tht E(Y) = E(X + ) = E(X) + nd Vr(Y) = Vr(X + ) = Vr(X). First let us show tht E(Y) = E(X + ) = E(X) +. Since E(X) = then E(X + ) = f() d, ( + )f() d. Using the distriutive lw, it cn e shown tht this is equl to But E(X) = Also, E(X + ) = f() d, so f() d =, so = f() d + f() d + E(X + ) = E(X) + f() d f() d E(X + ) = E(X) +. f() d. Also note tht E(X) = E(X) nd E() =. Now let us show tht Vr(Y) = Vr(X + ) = Vr(X). 48 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

32 WOrKeD exmple 6 think Since Vr(X) = E(X ) E(X), then Vr(X + ) = E((X + ) ) E(X + ) = = ( + ) f() d (E(X) + ) ( + + )f() d E(X) + E(X) + Using the distriutive lw to seprte the first integrl, we hve Vr(X + ) = But E(X) = = density function. Thus, Thus, f() d + E(X) f() d + E(X) f() d, E(X ) = f() d + f() d nd f() d + f() d E(X) f() d E(X) f() d = for proility Vr(X + ) = E(X ) + E(X) + E(X) E(X) = E(X ) E(X) = (E(X ) E(X) ) = Vr(X) E(X + ) = E(X) + nd Vr(X + ) = Vr(X). A continuous rndom vrile, X, hs men of nd vrince of. Find: E(X + ) Vr(X + ) c E(X ) d E(X ) e E(X 5). WritE Use E(X + ) = E(X) + to find E(X + ). E(X + ) = E(X) + = () + = 7 Use Vr(X + ) = Vr(X) to find Vr(X + ). Vr(X + ) = Vr(X) = 4 = 8 topic COntInuOus proility DIstrIutIOns 48

33 c Use Vr(X) = E(X ) E(X) to find E(X ). c Vr(X) = E(X ) E(X) = E(X ) = E(X ) 9 E(X ) = d Use E(X ) = E(X ) to find E(X ). d E(X ) = E(X ) = = e Use E(X + ) = E(X ) + to find E(X 5). e E(X 5) = E(X 5 ) 5 = 5 = 6 WOrKeD exmple 7 think Solve It my lso e necessry to find the epected vlue nd vrince efore using the fcts tht E(X + ) = E(X) + nd Vr(X + ) = Vr(X). The grph of the proility density function for the continuous rndom vrile X is shown. The rule for the proility density function is given y f() = k, f() (, k) k, elsewhere where k is constnt. Find the vlue of the constnt k. Clculte E(X) nd Vr(X). c Find E(X ) nd Vr(X ). d Find E(X + ). k d = to find k, or lterntively use the formul for the re of tringle to find k. WritE Method : kd = k (, ) (, ) = k() = k = Method : k = k = k = k = 484 Mths Quest MtheMtICL MethODs VCe units nd 4

34 Write the rule for the men. E(X) = = f()d ( )d Antidifferentite nd evlute. = = ( )d = () Write the rule for the vrince. Vr(X) = E(X ) [E(X)] 4 Find E(X ). E(X ) = 5 Sustitute the pproprite vlues into the vrince formul. c Use the property E(X + ) = E(X) + to work out E(X ). Use the property Vr(X + ) = Vr(X) to clculte Vr(X ). = = f()d d = 4 = ()4 = Vr(X) = E(X ) [E(X)] = = 4 9 = = 8 c E(X ) = E(X) = = = Vr(X ) = Vr(X) = 9 8 = d Use the property E(X + ) = E(X ) + to clculte E(X + ). d E(X + ) = E(X ) + = + = 4 Topic Continuous proility distriutions 485

35 Eercise.5 PRctise Work without CAS Questions Consolidte Apply the most pproprite mthemticl processes nd tools Liner trnsformtions WE6 If the continuous rndom vrile Y hs men of 4 nd vrince of, find: E(Y ) Vr(Y ) c E(Y ) d E(Y(Y )). Two continuous rndom vriles, X nd Y, re relted such tht Y = X + 5 where is positive integer nd E(X + 5) = Vr(X + 5). The men of X is 9 nd the vrince of X is. Find the vlue of the constnt. WE7 The continuous rndom vrile X hs proility density function defined y k, f() = k, <, elsewhere where k is constnt. The grph of the function is shown. Find the vlue of the constnt k. Determine E(X) nd Vr(X). c Find E(5X + ) nd Vr(5X + ). d Find E((X ) ). Find E(Y) nd Vr(Y). 4 The continuous rndom vrile X hs proility density function defined y π cos (), f() = π., elsewhere Sketch the grph of f nd verify tht it is proility density function. Clculte E(X) nd Vr(X). c Clculte E(X + ) nd Vr(X + ). d Clculte E((X )(X )). 5 For continuous rndom vrile Z, where E(Z) = 5 nd Vr(Z) =, find: E(Z ) Vr(Z ) c E(Z ) d E Z. 6 The men of the continuous rndom vrile Y is known to e.5, nd its stndrd devition is.. Find: E( Y) E Y 7 The length of time it tkes for n electric kettle to come to the oil is continuous rndom vrile with men of.5 minutes nd stndrd devition of. minutes. If ech time the kettle is rought to the oil is n independent event nd the kettle is oiled five times dy, find the men nd stndrd devition of the totl time tken for the kettle to oil during dy. (, k) f() (, k) (, ) (, ) c Vr(Y) d Vr( Y) e Vr Y. k 486 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

36 8 The proility density function for the continuous rndom vrile X is f() = where m is constnt. Find: the vlue of the constnt m c E(5 X) nd Vr(5 X). m( ),, elsewhere E(X) nd Vr(X) 9 The continuous rndom vrile Z hs proility density function given y f(z) = z +, z, elsewhere where is constnt. Clculte, correct to 4 deciml plces: the vlue of the constnt the men nd vrince of Z c i E(Z + ) ii Vr(Z + ) iii E(Z + ). The continuous rndom vrile X is trnsformed so tht Y = X + where is positive integer. If E(X) = 5 nd Vr(X) =, find the vlue of the constnt, given tht E(Y) = Vr(Y). Then clculte oth E(Y) nd Vr(Y) to verify this sttement. The continuous rndom vrile Y is trnsformed so tht Z = Y where is positive integer. If E(Y) = 4 nd Vr(Y) =, find the vlue(s) of the constnt, given tht E(Z) = Vr(Z). Then clculte oth E(Z) nd Vr(Z) to verify this sttement. The continuous rndom vrile Z hs proility density function given y f(z) = z, z where is constnt., elsewhere Find the vlue of the constnt. Clculte the men nd vrince of Z correct to 4 deciml plces. c Find, correct to 4 deciml plces: i E(4 Z) ii Vr(4 Z). The dily rinfll, X mm, in prticulr Austrlin town hs proility density function defined y f() = kπ sin, π, elsewhere where k is constnt. Find the vlue of the constnt k. Wht is the epected dily rinfll, correct to deciml plces? c During the winter the dily rinfll is etter pproimted y W = X. Wht is the epected dily rinfll during winter, correct to deciml plces? Topic Continuous proility distriutions 487

37 Mster 4 The mss, Y kilogrms, of flour sold in gs lelled s kilogrm is known to hve proility density function given y f(y) = where k is constnt. k(y + ),.9 y.5, elsewhere Find the vlue of the constnt k. Find the epected mss of g of flour, correct to deciml plces. c On prticulr dy, the mchinery pckging the gs of flour needed to e reclirted nd produced tch which hd mss of Z kilogrms, where the proility density function for Z ws given y Z =.75Y Wht ws the epected mss of g of flour for this prticulr tch, correct to deciml plces? 5 The continuous rndom vrile Z hs proility density function defined y 5 log e (z), z f(z) = z, elsewhere where is constnt. Determine, correct to 4 deciml plces: the vlue of the constnt E(Z) nd Vr(Z) c E( Z) nd Vr( Z). 6 A continuous rndom vrile, X, is trnsformed so tht Y = X +, where is positive constnt. If E(X) = nd Vr(X) = 7, find the vlue of the constnt, given E(Y) = Vr(Y). Then clculte oth E(Y) nd Vr(Y) to verify this sttement. Give your nswers correct to 4 deciml plces. 488 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

38 ONLINE ONLY.6 Review the Mths Quest review is ville in customisle formt for you to demonstrte your knowledge of this topic. the review contins: short-nswer questions providing you with the opportunity to demonstrte the skills you hve developed to efficiently nswer questions without the use of CAS technology Multiple-choice questions providing you with the opportunity to prctise nswering questions using CAS technology ONLINE ONLY Activities to ccess ebookplus ctivities, log on to Interctivities A comprehensive set of relevnt interctivities to ring difficult mthemticl concepts to life cn e found in the Resources section of your ebookplus. Etended-response questions providing you with the opportunity to prctise em-style questions. summry of the key points covered in this topic is lso ville s digitl document. REVIEW QUESTIONS Downlod the Review questions document from the links found in the Resources section of your ebookplus. studyon is n interctive nd highly visul online tool tht helps you to clerly identify strengths nd weknesses prior to your ems. You cn then confidently trget res of gretest need, enling you to chieve your est results. topic COntInuOus proility DIstrIutIOns 489

39 Answers Eercise. f() f() = e (log e, 4 4 ) ( ), 4 (, ) (log e, ) This is proility density function s the re is unit. (,.5) f() (, ) (, ) This is proility density function s the re is unit. (, ) f() (,.5) This is proility density function s the re is unit. This is not proility density function s the re is.99 units. n = 8 4 = f() =.5 (,.5) π π π π π π π π 4 4 f().7 (.5,.7) f() = f() =.5 cos() (4,.5) (.5, ) (4, ) (, ) 5 i 9 5 ii 4 5 i 7 5 ii i 5 8 ii c 46 8 c f() = ( e, e ) f() (, ) ( e, ) (, ) c 4 5 d This is proility density function s the re is unit. f() ( π 4,.7) f() = cos() + ( π,.9 4 ) ( π π 4, ), ( ) 4 π This is not proility density function s the re is π units. f() (, ) f() = sin() π, π ( ) (π, ) This is proility density function s the re is unit. 49 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

40 d f().5 f() = (,.5) f(y) e (, ) f(y) = y ( e, ) e 9 c = This is proility density function s the re is unit. f(z)dz = A tringle = h = 6 z = z = z = m = 8 m = c m = 4 ( + k + )d = + k + = () + k() + = 9 + 9k + = 9k + = 9k = k = = e 4 (, ) (, ) (, ) f() = (, ) f() 9 (, ) (, ) (, ) f() = e y dy = c = 5 n = 6 = e. As f() nd density function. Eercise. 5 5 c 8 = f() c 5 6 (, ) f(z) (, ) f(z) = z 4 (, ) (, ) e (e, ) π y f()d =, this is proility (, ) (, ) (, ) f() = 4 f(z) = z + y (, 4) (, ) z d + d = + y = sin() (, ) π (π, ) Topic Continuous proility distriutions 49

41 c c f(u) f(u) = e 4u 4 f() (, e 4 ) (, k) f() = k( + ) f() = k( ) (, ) A = h c 5 = 4 k = 4k k = 4 4 d 6 7 c f() f(z).5 e (, ) (, ) (, ) ( 6, ) 5 (, ) (6, ) (,.5) (, ) f() = 5 dz =. As f(z) nd z proility density function. e 5 e (, e ) f(z) = z f(z)dz =, this is (e, ) z d 8 π π e 4u du = 4 e 4 4 nd = 4 log e 5 cos(z)dz = π sin(z) π = sin π sin π = + = This is proility density function s the re under the curve is nd f(z) for ll vlues of z = 8 c.7 d 98 5 c 6 (, ) (, ) = log e f(z) (, ) (, ) π π, c.54 d.6 f(z) = cos(z) ( ) u z 49 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

42 f() (, ) (, ) f() = e y = 8 = c Vr(Z ) =.895, SD(Z ) =.957 d.75 e 7 9 π π (sin() + )d = π π (sin() + )d.95 c.5 = = tn Eercise.4 = i ii.565 or 5 c Medin = E(Z ) =.75, m =., Vr(z) =.44, SD(Z ) = E(X ) =, m =., Vr(X ) = 9, SD(X ) = 5 d = = = = d d ( ) = = As f() for ll -vlues, nd the re under the curve is, f() is proility density function. c m = min.5 min c m =.5 min 7.56 m =.65 c Q =.4, Q =.8899 d.5795 = π cos() + π = π cos(π) + π cos() + = π + π + = As f() for ll -vlues, nd the re under the curve is, f() is proility density function..78 c i.575 ii.7566 d m =.99 =, = 4 z dz = z dz = z = z = + = + = = = = E(Z ) =.64, Vr(Z ) =.4 c m = 6 8, interqurtile rnge = 5 4 π h = 4 c Topic Continuous proility distriutions 49

43 4 f() k f() = k (, k) (, k) π π ( cos())d = sin() π π = sin(π) + sin π = + = c d 5 k d = k = k k = k( ) = (, ) (, ) k = + ( ) f(y)dy = y. log e dy = c Vr(Y ) =.6, SD(Y ) =.4697 d m =.986 e i.786 ii.85 iii.76 iv.4 Eercise.5 5 c 9 d 5 = 5 E(Y ) = 5, Vr(Y ) = 5 k = 4 E(X ) =, Vr(X ) = 4 c E(5X + ) =, Vr(5X + ) = 5 d E((X ) ) = f() (π, ) As f() for ll -vlues nd the re under the curve is, f() is proility density function. E(X ) =.578, Vr(X ) =.46 c E(X + ) = 8.74, Vr(X + ) =.74 d E((X )(X )) = c 7 d c.44 d.44 e.6 7 E(5T ) = 7.5 minutes, SD(5T ) = 5.5 minutes 8 m = 4 E(X ) =, Vr(X ) =. c E(5 X ) =, Vr(5 X ) =.8 9 =.6487 E(Z ) =.974, Vr(Z ) =.49 c i.89 ii.4 iii.4 =, E(Y ) = 8, Vr(Y ) = 8 = or, E(Z ) = or 9, Vr(Z ) = or 9 = 49 6 E(Z ) =.759, Vr(Z ) =.9 c i.47 ii.978 k = mm c. mm 4 k = 4.8 kg c.6 kg 44 5 =.7755 E(Z ) =.49, Vr(Z ) =.6 c E( Z ) =.58, Vr( Z ) = =.5469, E(Y ) =.98, Vr(Y ) =.98 f() = cos() ( π, ) (π, ) 494 Mths Quest MATHEMATICAL METHODS VCE Units nd 4

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