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Math 110 EXAM QUESTIONS AND ANSWERS | LATEST 2024/25 | RATED A+
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Math 110 EXAM QUESTIONS AND ANSWERS | LATEST 2024/25 | RATED A+
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Math 110 EXAM QUESTIONS AND ANSWERS | LATEST 2024/25 | RATED A+ Probability IDensity ICurve I- I I I Ithe Icurve Iused Ito Idescribe Ithe Idistribution Iof Ia Icontinuous Irandom Ivariable Normal ICurve I- I I I Ithe Ibell-shaped Icurve Ithat Idescribes Ithe Idistribution Iof Imany Iphysical Iand Ipsychological Iattributes. IMost Iscores Ifall Inear Ithe Iaverage, Iand Ifewer Iand Ifewer Iscores Ilie Inear Ithe Iextremes Z-Score I- I I I Ia Imeasure Iof Ihow Imany Istandard Ideviations Iyou Iare Iaway Ifrom Ithe Inorm I(average Ior Imean) Example: IA Iwoman Iwhose Iheight Iis Ix=67 Iinches Ifrom Ia Inormal Ipopulation Iwith Imean Iμ=64 Iand Iσ=3 Iinches. I What Iis Ithe IZ-score? I- I I I IThe IZ-score Iis Iz I= I(x I- Iμ)/σ I= I(67-64)/3 I= I1 z Iscore Iformula I- I I I Iz I= I(x I- Iμ)/σ finding Ithe Iarea Iunder Ithe Inormal Icurve I- I I I Inormalcdf(lower, Iupper, Iμ, Iσ) To Iget Iit: I2nd+Var I-> I2: Inorm Example: IFind Ithe Iarea Ito Ithe Ileft Iof Iz I= I1.26 Iin Ia Inormal Icurve I- I I I Inormalcdf( -1E99, I1.26, I0, I1) Example: IA Istudy Ireported Ithat Ithe Ilength Iof Ipregnancy Ifrom Iconception Ito Ibirth Iis Iapproximately Inormally Idistributed Iwith Imean Iμ I= I272 Idays Iand Istandard Ideviation Iσ I= I9 Idays. IWhat Iproportion Iof Ipregnancies Ilast Iless Ithan I259 Idays? I- I I I IStep I1: IFind Ithe IZ-Score. I(259-272)/9 I= I-1.444 Step I2: IUse Ithe IZ-Score Ito Ifind Iarea Ileft Iof IZ I= I-1.444 normalcdf( -1E99, I-1.444, I0, I1) Finding IX Ifrom Iz-score I- I I I IinvNorm(Area Ito Ithe Ileft, Iμ, Iσ) To Iget Ito Iit: I2nd+Vars I-> I3:invNorm Mensa Iis Ian Iorganization Iwhose Imembers Ipossess IIQs Iin Ithe Itop I2% Iof Ithe Ipopulation. IIf Ithe IIQs Iare Inormally Idistributed Iwith Ia Imean Iof I100 Iand Ia Istandard Ideviation Iof I15, Iwhat Iis Ithe Iminimum IIQ Inecessary Ifor Iadmission? I- I I I IFirst: IFind Ithe Iz-score Ithat Icorresponds Ito Ithe Igiven Iarea Iusing IinvNorm . z I= IinvNorm(0.98, I0, I1) x I= Iμ I+ Iz I* Iσ I= I100 I+ I2.0537(15) x I= I130.8 x I≈ I131 mean Iof Isampling Idistribution Iof Ix̄ I- I I I Iμx̄ I= Iμ The Imean Iof Ithe Isampling Idistribution Iis Idenoted Iby Iμx̄ Iand Iis Iequal Ito Ithe Imean Iof Ithe Ipopulation Iμ. standard Ideviation Iof Isampling Idistribution Iof Ix̄ I- I I I Iσx̄ I= Iσ/√n The Istandard Ideviation Iof Ithe Isampling Idistribution I(Standard IError) Iis Idenoted Iby Iσx̄ Iand Iequals Ithe Istandard Ideviation Iof Ithe Ipopulation Idivided Iby Ithe Isquare Iroot Iof Ithe Isample Isize. Sampling Idistribution Iof Ia Iskewed Ipopulation I- I I I Iif Ia Ipopulation Iis Iskewed, Ia Ilarger Isample Isize Iis Inecessary Ifor Ithe Isampling Idistribution Iof Ix̄ Ito Ibe Iapproximately Inormal. The ICentral ILimit ITheorem I- I I I IThe Itheory Ithat, Ias Isample Isize Iincreases, Ithe Idistribution Iof Isample Imeans Iof Isize In, Irandomly Iselected, Iapproaches Ia Inormal Idistribution. In Ipractice, Ia Isample Isize Iof In>30 Iis Ienough. Example: IBased Ion Idata Ifrom Ithe IU.S. ICensus, Ithe Imean Iage Iof Icollege Istudents Iin I2011 Iwas Iμ I= I25 Iyears, Iwith Ia Istandard Ideviation Iof Iσ I= I9.5 Iyears. IA Isimple Irandom Isample Iof I125 Istudents Iis Idrawn. IWhat Iis Ithe Iprobability Ithat Ithe Isample Imean Iage Iof Ithe Istudents Iis Igreater Ithan I26 Iyears? I- I I I IStep I1: IConfirm Ithe Iassumptions . I n=135>30 Thus, Iwe Imay Iuse Ithe Inormal Icurve. Step I2: IThen, Icompute Iμx̄ Iand Iσx̄. μx̄ I= Iμ I= I25 σx̄ I= I9.5/√125 I= I0.85 normalcdf(lower, Iupper, Iμx̄, Iσx̄) normalcdf(26, I1E99, I25, I0.85) I= I0.1197
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