Continuous Rndom Vrile : The continuous rndom vrile hs its vlues in n intervl, nd it hs proility distriution unction or proility density unction p.d. stisies:, 0 & d Which does men tht the totl re under the curve o is. And the proilities concerning re chieved y: d Which represents the re under the curve o nd etween,. We deine the mthemticl Epecttion Or the epected vlue or the men vlue o s: While the vrince o is deined s: µ = E = d V E d E where E d The stndrd devition o is the positive squre root o its vrince. Some Common Continuous Rndom Vriles We Will Consider: There re mny dierent continuous proility distriutions such s: The Norml distriution 3The Chi-Squre distriution The Student distriution 4 Fisher roility distriution
The Norml roility Distriution The rndom vrile is sid to hve Norml proility distriution with the two prmeters µ nd σ or, ~ N i its p.d. hving the orm : 0, ; nd e Clerly & 0, d. And this p.d. stisies the importnt properties: The curve o is ell shped nd SYMMETRIC round µ It my e proven tht: } { & E E V d E 3 s 0 The norml distriution depends on the two prmeters nd. Where µ determines the loction o the curve. But, determines the scle o the curve, i.e. the degree o ltness or eked ness o the curve. Moreover the norml vrile with, 0 is clled stndrd norml nd is given the symol 0, ~ N, then; & ; dz z e z z s z 0 } { & 0 E E V dz z z E z Now, to evlute ech o the proilities: d d d &
We my use the stndrd liner trnsormtion to ind the required proilities s ollows: k k Note tht: > 0 = < 0 = 0.5 I ~ N0, then: i- < = 0.977 is the re under the -curve to the let to. ii- -.55 < <.55 = 0.9946 0.0054 = 0.989 is the re etween -.55 nd.55. iii- -.74 < <.53 =0.9370 0.003 = 0.9339 is the re etween -.74 nd.53. iv- >.7= 0.9966 = 0.0034 is the re under the -curve to the right o.7. v- = 0.84 = 0 vi- It is ment y Atht A A. For instnce: 0.0.8, 0.05.645, 0.05.96, 0.0.33, 0. 005.575 Norml Distriution Applictions Emple: The Uptime is custom-mde light weight ttery-operted ctivity monitor tht records the mount o time n individul spend the upright position. In study o children ges 8 to 5 yers. The reserchers ound tht the mount o time children spend in the upright position ollowed norml distriution with men o 5.4 hours nd stndrd devition o.3.find i child selected t rndom, then: -The proility tht the child spend less thn 3 hours in the upright position 4-hour period 3
35.4 < 3 = <.3 = -.85 = 0.03 -The proility tht the child spend more thn 5 hours in the upright position 4-hour period > 5 = > 5 5.4.3 = - 0.308-0.3 = - < - 0.3 = - 0.350= 0.648 3-The proility tht the child spend ectly 6. hours in the upright position 4-hour period is = 6. = 0. 4-The proility tht the child spend rom 4.5 to 7.3 hours in the upright position 4-hour period 4.5 5.4 4.5 < < 7.3 =.3 < < 7.35.4.3 = -0.69 < <.46 = <.46 < -0.69 = 0.979 0.45 = 0.688. Emples: The rin weights o certin popultion o dults ollow pproimtely norml distriution with men,400 gm nd stndrd devition 00 gm. Wht the percentge o the rin weights re:,500 gm or less? c,35 or more? Between,35 nd,500 gm? d,475 gm or more? e Between,475 gm nd,600 gm? Solution: Assume rndom vrile = weights o certin popultion o dults; ~ N,400, 00 ; to ind ech o the required:,500,400,500 0. 859 00 And so the required percentge is 8.59 %. 4
,35,400,500,400,35,500.75 00 00 = 0.859 0.040 0. 7758,35,500 c,35.75.75 0. 9599 00 The required percentge will e 95.99%,475,400 d,475 0.75 0.75 0.7734 0. 66 00 The required percentge is.66% e,475,400,600,400,475,600 0.75 00 00 EERCISE 0.977 0.7734 0.038 The required percentge is 0.38%. ---------------------------------------------------------------- Q. A Suppose tht is distriuted ccording to the stndrd norml distriution. The re under the curve to the let o z. 43 is: A 0.0764 B 0.936 C 0 D 0.833 The re under the curve to the let o z =.39 is: A 0.768 B 0.977 C.73 D 0.083 3 The re under the curve to the right o z 0. 89 is: A 0. 785 B 0.833 C 0.867 D 0.054 4 The re under the curve etween z. 6 nd z 0. 65 is: A 0.7576 B 0.8665 C 0.054 D 0.44 5 The vlue o k such tht 0.93 k 0. 047 is: 5
A 0.8665 B. C. D.00 B Suppose tht is distriuted ccording to the stndrd norml distriution. Find: < 3.9 > 4.5 3 < 3.7 4 > 4. Q. The inished inside dimeter o piston ring is normlly distriuted with men o centimeters c.m nd stndrd devition o 0.03 centimeter. Then, The proportion o rings tht will hve inside dimeter less thn.05 c.m is: A 0.0475 B 0.955 C 0.757 D 0.843 The proportion o rings tht will hve inside dimeter eceeding.97 c.m is: A 0.0475 B 0.843 C 0.587 D 0.454 3 The proility tht piston ring will hve n inside dimeter etween.95 nd.05 c.m is: A 0.905 B 0.905 C 0.454 D 0.757 Q3. The verge lie o certin type o smll motor is 0 yers with stndrd devition o yers. Assume the live o the motor is normlly distriuted. The mnucturer replces ree ll motors tht il while under gurntee. I he is willing to replce only.5% o the motors tht il, then he should give gurntee o: A 0.03 yers B 8 yers C 5.66 yers D 3 yers Q4. A mchine mkes olts tht re used in the construction o n electric trnsormer. It produces olts with dimeters ollowing norml distriution with men o 0.060 inches nd stndrd devition o 0.00 inches. Any olt with dimeter less thn 0.058 inches or greter thn 0.06 inches must e scrpped. Then The proportion o olts tht must e scrpped is equl to 6
A 0.0456 B 0.08 C 0.977 D 0.3333 E 0.667 I >= 0.949, then equls to: A 0.069 B 0.0659 C 0.0649 D 0.0669 E 0.0609 Q5. The dimeters o ll erings mnuctured y n industril process re normlly distriuted with men = 3.0 cm nd stndrd devition = 0.005 cm. All ll erings with dimeters not within the speciictions d cm d > 0 will e scrpped. Determine the vlue o d such tht 90% o ll erings mnuctured y this process will not e scrpped. I d = 0.005, wht is the percentge o mnuctured ll erings tht will e scrped? Q6. The weight o lrge numer o t persons is nicely modeled with norml distriution with men o 8 kg nd stndrd devition o 9 kg. The percentge o t persons with weights t most 0 kg is A 0.09 % B 90.3 % C 99.8 % D.8 % The percentge o t persons with weights more thn 49 kg is A 0.09 % B 0.99 % C 9.7 % D 99.8 % 3 The weight ove which 86% o those persons will e A 8.8 B 8.8 C 54.8 D 8.8 4 The weight elow which 50% o those persons will e A 0.8 B 8 C 54.8 D 8 Q7. The rndom vrile, representing the liespn o certin electronic device, is normlly distriuted with men o 40 months nd stndrd devition o months. Find 7
<38. 38<<40. 3=38. 4The vlue o such tht <=0.734. 4.4 Q8. I the rndom vrile hs norml distriution with the men nd the vrince, then <+ equls to A 0.877 B 0.477 C 0.577 D 0.777 E 0.977 Q9. I the rndom vrile hs norml distriution with the men nd the vrince, nd i <3=0.877, then equls to A 3.84 B.84 C.84 D 4.84 E 8.84 Q0. Suppose tht the mrks o the students in certin course re distriuted ccording to norml distriution with the men 70 nd the vrince 5. I it is known tht 33% o the student iled the em, then the pssing mrk is A 67.8 B 60.8 C 57.8 D 50.8 E 70.8 Q. I the rndom vrile hs norml distriution with the men 0 nd the vrince 36, then. The vlue o ove which n re o 0.96 lie is A 4.44 B 6.44 C 0.44 D 8.44 E.44. The proility tht the vlue o is greter thn 6 is A 0.9587 B 0.587 C 0.7587 D 0.0587 E 0.5587 8
Q. Suppose tht the mrks o the students in certin course re distriuted ccording to norml distriution with the men 65 nd the vrince 6. A student ils the em i he otins mrk less thn 60. Then the percentge o students who il the em is A 0.56% B 90.56% C 50.56% D 0.56% E40.56% Q3. The verge rinll in certin city or the month o Mrch is 9. centimeters. Assuming norml distriution with stndrd devition o.83 centimeters, then the proility tht net Mrch, this city will receive: Less thn.84 centimeters o rin is: A 0.838 B 0.76 C 0.5 D 0.08 More thn 5 centimeters ut less thn 7 centimeters o rin is: A 0.8504 B 0.496 C 0.650 D 0.34 3 More thn 3.8 centimeters o rin is: A 0.056 B 0.9474 C 0.30 D 0.4053 ---------------------------------------------------------------------- The Chi squre roility Distriution The rndom vrile = is sid to hve chi-squre proility distriution with the prmeter or the degrees o reedom d. υ i its p.d.. is: e, 0. Which does men tht there re so mny chi-squre distriutions depends on the d. =υ. The chisqure distriution stisies the ollowing properties: 0 0; 0 & d 9
This p.d.. stisies the importnt properties: The curve o is not ell shped ut it is skewed curve to the right depends on υ. We my prove tht the epected vlue nd the vrince o re: E & V We my denote y, to the criticl point tht stisies., Eventully, i rndom norml smple o size n nd represents the popultion vrince then n S is with n where S represents the smple vrince. Emples:. Determine the vlue o 5, 0.005 which stisies 0. 005. 5, 0.005 The required is tht tulted chi-squre vlue which leves n re o α =0.005 to the right with the d.= 5. From the tle we get: 5, = 3.80. 0.005. I 0 0. 99.So he sked us to ind the vlue o 0 which leves n re o 0.99 to the let with d. = 0. Or 0.0 37.766 3. 0,0.0 0 0,0.0 3. Find the proility 0, 0.99 0, 0.0. The required proility vlue equls 0.99 0.0= 0.98. The t- roility Distriution The Student distriution The Student distriution t with degree o reedom υ is ell shped nd symmetric s N 0, ut 0
is less peked in the center 0 nd higher tils t 0 depends on the υ where t t t, V T t >. The criticl point t, is T t, t dt. The distriution t pproches to N 0, s the numer o degrees o reedom υ pproches. But in prctice t is pproimted y N 0, s n 30. It cn e shown tht: - I ~ N 0, & Y ~ T / Y - I,,..., n ~ N, T n S / n Emples I T t = α where υ=7, α=0.975 then t. 3646, 7, 0.975 I T t = α where υ=4, α =0.995 thent. 3646, 4, 0.975 3 I T t 8, 0. 975 0. 975 then t 8,0. 975 = -.009 4 I T t 0. 99 So he sked us to ind the vlue o t such tht T t 0.99 nd so T t,0.0 0.0 i. e t, 0. 0 ----------------------------------------------------------.508 The F-distriution Or Fisher proility distriution F is sid to hve the F-distriution Or Fisher proility distriution with the two degrees υ nd υ i its proility density unction o the orm:
F F F / ; 0 F, Which stisies: The curve o F is not ell shped ut it is skewed curve to the right depends on the two prmeters υ nd υ. We my prove tht the epected vlue o the rndom vrile T is E F F F df & V F ; 4 4 We denote the criticl point F : F F F df nd we ccept,,,, F,, tht F,,. F,, 3 It cn e shown tht: / ~ & Y ~ F, ; Y / Emples I F F 0.05 F 3. 35. 0,8,0.05 0,8,0. 05 Find the F-vlue which stisies: F F 0 0. 05 :, 8 The required F vlue is tht criticl tulted F vlue which leves n re o 0.05 under the curve to the let. Tht is F vlue which leves n re o 0.95 to the right. From the F-le t α = 0.05, we cn sy tht: the required F vlue F 0,8,0.95 0.357 = 3.35 F 3.07 8, 0, 0.05
F 3 Find the tulted F-vlue o 0,8, 0. 99. We cn get the required F-vlue y using F 5.06 0,8, 0.99 F 8,0, 0.0 0.976 ---------------------------------------------------------------------- 3