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Things that you should know before starting these exercises 1. X - N(p,o) refers to a random variable X with a mean of p and a standard deviation of o. - N(3.1 , 1.23!.refers to a random variable X whose mean is 3.1and whose standard deviation is 1.23.1 [Thus, X ,..the normal with mean=0 and standard deviation=1". [That is, Z - N(0 , 1) ] STflNDAN NqRrnflL 3. e Z-scores are in standard deviations from 0 (the mean). [For example,Z=L.5 means 1.5 standard deviations from 0 (the mean].1 I^:o t= 1.5 iS l,S sfq..&*^t &q*'rqf i*rr {rm^ $\* il.er^ (0) 4. The standard normal table ALWAYS gives the area under the normal curve to the left of the given z value. 6.y6tnp{.,-: 7 The area under a normal curve represents a probability- *\^s G.r'rsu i s-.-0.X\ l] , (t^sn^g 9r* shdq"r nsr#*\ 1"W*) t=l for example , PIZ < 1.2) is the probability that the standard normal random variable Z is LESS THAN 1.2. u,qq\l ({t'' ULlu) t: l,I So [email protected]{ l.})= St\9 ,:- ttsg 7. /-a---- 6. To use the standard normaltable to calculate a probability for a GREATER THAN, you must use the fact that P(Z > a) = 1- P{Z <a}. for example, P(Z > 1] = 1- P(Z < 1) ...and you can look up P(Z < 1) in the standard normal table.l [So, (, **'F* t^t'Q*, P(e'i)= O't't lJ So 7(z,l)= t - o,tYl3 : 0,rstl 7. To calculate the probability that a standard normal random variable is between two numbers, you {a} look up the larger number in the standard normal table, (b) look up the smaller number and (c) subtract the smaller number from the larger. P(t';?s6) f6Th: zLr "rb)=?(z:1,'d") , l*rlcwp .tlr tc.}.-ks-, o,ttcs B. - o') I I*3 :O-t5 : To calculate a probability related to a normal random I with mean p and standard deviation o N(p, o] l, it must first be converted to a standard variable * IX ti X normal random variable Z using , -f.. Once converted, thg probability is calculated N ( 1$-? l$ri \n' N( e-,-.pQa .. L'd. Ld \n,' Exa"-'pQ*'. T 3.xz tl): f (ry o.S =f; G 9. L? z- c.5 ) - 3 \ *\ (-t-4- L+(+\ =\ r(\: LZ r( e,,L."*( (Z ':o'5) o"6?t5 $c"'.-.\ttsl) lf you must find a z-score corresponding to a certain probability, I z 1_ ,/ s) ){ i5)* : Q"5\J \).Jr,'3t) $s 0\f^*.r;eul)*r) #' \)o ./ -a-\ you have to find the number in the body of the standard normal table that is closest to the given probability and then cross- r ,1 t \ $t Yka- T'>Tdi g1(tr I f ,ni t\* Z-,1?ltJ*:\'|t-'+ f;a*6rqo w\'r---Lsi"9**o'{1 E-"' Is*11 r fi$\* ^\ '"CS''t.o\ yL\ Ln b K"")"^^ referencethatnumberbacktothecorrespondipgzvalue. 1R h* . 1; 6J% il#"'&'os\ +. c"lP9 n ---- ar,,O,0S Q-OL oJ\@",0'* Normal Distribution Examples HoME t-l I NEXT I Some simple problems to work in class (use table as needed): 1. Find the percent of area (round to nearest percent) under the normal curve (a) between the mean and 0.35 standard deviations from the mean. (b) between the mean and -0.35 standard deviations from the mean. 2. Find the percent of the total area under the standard normal curve between the z-scores: z = -1.1 and z = 0.8. 3. Find the z-score such that 1 8% of the total area is to the (a) left and (b) right of z. 4.P(-1.1 <Z<2.A1=7 5. lf X - N(4.2,1.1), (a) P(X < a) (b) P(x > 5) (clP(4<x<5) calculate the following: HOME Some simple problems to work in class (use table as needed): 1. Find the percent of area {round to nearest percent) under the normal cun/e N"k'. f*',s rnmsurej i"., :t*..tn,",[ dq{,rli.o.s (a) between the mean and 0,Q€ standard deviations from the mean. Su\-uq,Z=Olts %;$*o,o\o' A tb*A \*&.fs: o,ir)- F(a^.il bd'"**. b { b,\5 =?(acTt | r* luk..,1: 'in it6\,frui "*\'htlt L ::1"o" Soco - = e"bt6t e=os;*-*d*oq*t = ! A"f".^, a:kt$.,ktfi"" A f G.\, l\;tr\$rt = t lo:[i,.p r I , t I ' I o"SocD n I fi \l-otJ, $ fof r n*,1 t '/r tt 0rtr,' t, €=*l.l t:o"E lo"ky.; {G i^ si,i. o;\Ki = o.b5t1 Kr.it* G*t 13,6t % n"f"l +o!l= c.i : 3 bKuqr.t s1n p1fu;u o"3C3a: P(.1'.\)-P(t;i, l) u,1 I \otn t**{J,* I be1,t:t-66 *l,i{ q lr ]qlr/q_ 2. Find the percent of the total area under the standard normal curve scores: z= -1.1 and z = 0.8. .rl I nt"n*lc.urv( P(q") - P(i.;o,ls) h:1":11,. o,, b*.*Y*,,..-4,3t 4O = t:=(}-lt *:CI 7o ir c ({5+ddn= t) (b) between the mean and -0.35 standard deviations from the mean. q6,.t, ( A,39 o, -r-'t, ? 39 35 Gr.cj kr-.*r, Z.= -?: --Ao rt'=' 3.1't sJ 65.Nolo r Normal Distribution Examples t-NEXrl Some simple problems to work in class (use table as needed): (trr irn-r!\*p !>* k""1,u'J5,*y}-.,i \*1 k*ror', fiq t4ittq N.4u :. &*", \r$ia *d.v*,l- r,+ ru-$ +',* $; *,J T\1* Z se,;rt T\r* {1'vripntqs $\"Ls Q i t c,. 3. Find the z-score such that 18o/o of the total area is to the (a) lefl f$** (r) (t; E=i sr,*o;nr;F[t"t: ..$-+trqi,l, t^Li*" a..,& $,,r,$ F"e S t\or**t,t(ou, | \ur,r,o\ fru.'- A- gr'*Q' tn!"lq+ -]lj n.lt\t .8,'.1\ *€.- :],1* t,1 ( l*rsjt** )l',it -+, t* r;-ff**"*" ck:-* 4.P(-1.1 <Z<2.01=7 AG,. r;r) f f '),SlSt rllbi1b " *.1{t\ "*'t er:i\. * iP{<, *l.i) .,,*.r,t,*, \.u* '** :CJ)li * il,'t\ i) P,i351 t,\I.r, Normal Distribution Examples Some simple problems to work in class (use table as bI\ ?= vf (a) P(X < 4) f,(xzY) =tr(*Aq*4) =fl(r.- *):fl(e.q,\t|b T**^.\^Ut)Jt,r^"Y'.','-' J-r*-.$J^n *uil.^S\.!{ 9-" rqqflVlnA0,\Jl?{ }:\ q) \ (b) P(x ' .,i-1, f,,rta \"93ISU/cvlt"'^) sl 1"!,&"5 "tE " -+.)-\ (zaH3 ?(x'i=TP(x.$ = \*? T"l--l --l P(e. *)= \- f (ea o")rl)- P(a. oJ 3) : - t"l tl, [email protected] I ^ r \ \ ^ J \ I I \ {t*^\Lfu- (c)P(4<X<5) cz. 5* n/ e0,l t P(24--o,tn) : L?.o^l) t'G o,\x =f,f t{ *"Hu- + P(q. xas)=P(t#& \.\ l'1 @ -\J n ;GI ] 61.". c.Y1x6