MEAN 6.0000 WHEN WE HAVE LARGE DATA SETS, WE GROUP THE DATA. IN THIS CASE OUR GROUPS WILL BE: STD DEV 2.17 IN A NORMAL DISTRIBUTION THE MEAN, MEDIAN AND MODE ARE ALL THE SAME NUMBER X-VALUES Z-VALUES GROUP NUMBER OF DATA POINTS IN IT 2 -1.84 2.0-2.9 2 2 -1.84 3.0-3.9 3 7 3 -1.38 4.0-4.9 4 6 3 -1.38 5.0-5.9 5 3 -1.38 6.0-6.9 6 5 4 -0.92 7.0-7.9 5 4 -0.92 8.0-8.9 4 4 4 -0.92 9.0-9.9 3 3 4 -0.92 10.0-10.9 2 5 -0.46 2 5 -0.46 5 -0.46 1 5 -0.46 5 -0.46 0 12 23 34 45 78 6 0.00 X-VALUES: 65 76 (3.83) 6 0.00 98 109 6 6 6 6 0.00 0.00 0.00 0.00 7 0.46 7 7 7 0.46 0.46 0.46 Z-VALUES ARE CALCULATED AND SHOW THE NUMBER OF STANDARD DEVIATIONS A DATA POINT (OR GROUP OF DATA POINTS) IS FROM THE MEAN Z = ( X - MEAN) / STD DEV OF COURSE IF YOU HAVE A Z-VALUE YOU CAN BACK CALCULATE TO GET THE X-VALUE: X = Z TIMES THE STD DEV + THE MEAN AS AN EXAMPLE USE THE Z-VALUE OF +0.46 TO SEE WHAT X-VALUE IT REPRESENTS: X = 0.46 x 2.17 + 6.0 = 7.0 7 8 8 8 8 9 9 9 0.46 0.92 0.92 0.92 0.92 1.38 1.38 1.38 SIMILARLY THE X-VALUE (OR GROUP) THAT IS -1 STANDARD DEVIATION BELOW THE MEAN (Z = -1) WOULD BE: X = 1 x 2.17 + 6.0 = 3.83 WHILE THE SHAPE OF OUR EARLIER GRAPH WAS RECTANGULAR, THIS GRAPH HAS A BELL SHAPE, REFERRED TO AS THE "NORMAL" DISTRIBUTION. AND, AS WITH THE RECTANGULAR GRAPH THE AREA SHOWN IN THIS GRAPH REPRESENTS 100 % OF OUR DATA AND ANY PORTION OF IT REPRESENTS THE PERCENTAGE OF OUR DATA IN THAT AREA AS WELL AS THE PROBABILITY THAT SPECIFIC DATA ARE IN THAT REGION. HERE IS A MORE TYPICAL BELL CURVE GRAPH: TO CALCULATE THE AREA IN ANY SECTION OF THIS BELL CURVE GRAPH WE NEED TO KNOW THE EQUATION FOR THE LINE, AND THEN WE USE INTEGRAL CALCULUS TO CALCULATE THE AREA UNDER THAT CURVE FOR THE AREA SELECTED. n 10 10 1.84 1.84 Z-VALUES: -2 -1 0 +1 +2 THE X-VALUES ARE THE DATA POINTS AND WE HAVE GROUPED THEM AS SHOWN. x= 1 FORTUNATELY FOR US, THESE AREAS HAVE BEEN CALCULATED AND PUT INTO TABLES. THE HORIZONAL AXIS HAS THE Z-VALUES ON IT, WHICH ARE THE STANDARD DEVIATIONS FROM THE MEAN AND THE TABLE GIVES US THE AREA TO THE LEFT OF THAT Z-VALUE. (SEE & PRINT THE ATTACHED TABLES) TO GET THE AREA TO THE RIGHT WE MUST SUBTRACT THE TABLE VALUE FROM 1.0000 SINCE THE SUM OF BOTH AREAS EQUALS 100% OR 1.00 AS A DECIMAL. HERE ARE SOME "WORKED-OUT" AREAS: 68.3% OF THE AREA (OUR DATA VALUES) ARE WITHIN -1 AND +1 STANDARD DEVIATIONS FROM THE MEAN 95.4% OF THE AREA (OUR DATA VALUES) ARE WITHIN -2 AND +2 STANDARD DEVIATIONS FROM THE MEAN 99.7% OF THE AREA (OUR DATA VALUES) ARE WITHIN -3 AND +3 STANDARD DEVIATIONS FROM THE MEAN AN "UNUSUAL" VALUE IS CONSIDERED ONE THAT IS MORE THAN TWO STANDARD DEVAIONS FROM THE MEAN (Z = + 2) MORE SPECIFICALLY, THIS "UNUSAL" OR RARE EVENT AREA CORRESPONDS TO 5% OF THE GRAPH AREA ACCORDINGLY, THESE CRITICAL Z-VALUES ARE +1.645 ON THE RIGHT END AND -1.645 ON THE LEFT END OF THE GRAPH. SEE IF YOU CAN FIND THE AREA IN THE TABLE (0.05) FOR THESE Z-VALUES. . Chapter 8 - Confidence Intervals #1-5 , 23-37, 38-42, 49-61 , 62- 63, 79-94, 97, 103, 106, 111 , 119, 123, 133 Use the following information to answer the nexc five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish lO consouct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. 1. Identify the following: a. - x = 'l 'k'+ / ~J)vi~ b. a = t.; / c n = ---r;O"/ }{ '- 2.Inwords,define therandoruvariablesXand X. , , 3. Whic distributio should you u e for this prob em? j wCMl IJ '>e- ( o .. e:- "'~ 1 It : \\) ~~~"'a r"<..- { I t~ [ .llfvA.?o (11 ~leell.~t- '{. ~ ~ ~ v,e, ~t o~f i\\.l"" ~~rn a lfY\\afl, I t th- 0 1 Ul ( V'ltMI\\_ - -1 .,._- '7 o. z;. (z.- r~ v ( 1 1 .-t<; ' 11 <;ct o ~ ) (;. . 3IR qt ;\\ 4.--Consouct a 95% confidence mterval for the populauon mean wet&_tit of new.b.o ek hants. State the confidence mterval, s- 1 sketch the raph, and calculate the error bound. z 'I '1--l 1!1l> ~50 ) ~/ Z- '+ t . 1'5 ~ ~ ~ A.! ( 5.- ) 1 2 ~ 'l . ~ 4 Z. ~ Dl/z_ :. (. C -::. I 5 f\\ :. S o r; -- z '\\l. ~ ~~ J 5. What wil happen to the confidence interv obtained, if 500 newborn elephants are weighed instead of 50? Why? ~1:t.S I --rhe ciJfl&1Jr,..c.e ,vo-+c'rv-~'~~1 v)l! Jacrl,.J( a'LJ fxCQr.-~ mor t pre-cr.;e-,.k'-lil.-~t{d.... 1s~o ( o~ fctfl.d /Jivt--1, Iv~. O"'- r et<~""flt. tk' f. BfV\\ \\(V,'fl Jocrt"-~ fr;/1 4-, 1$ ~ +o /. )I()) S' o o._,.) u.... r ef{ct \\:),M, ) w~lf (11>'1.~[ fr;IV'\\ J..'t"L-,6., -lo l't-~.$ 1q, -l,H~I,'l6a 1 (S' \\1"'t ~}'\\.t4~ tC;) 1 z.'l-'6.'~ (\\ -_1-~ Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall ter m was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean a e for the sam le was 30.4. We are interested in the oue mean age for Wimer Foothill College srudents. Let X = the age of a Winter Foothill College student. y -: . ~ 0 , : :: 25. n~= ~"c ~~ . = b ~ =15 26. In words, define the random variable - + 'i . . TN.. X.1-- () "'-1 ovt" 'tN-Ctfl. 1 ({ c..(( Fo-rltt~ C I( Nr 5 lv ,1.1- t') 2vr Cl r c, k -te r ~"' ~~ s ~~. .... rfe I ~.,.t/ 27. What is_ x estimating. .I t / 1 y.. ,., ec;, ~.~ .l'":7 ~ ~Jf"'"''o./' ({ILQ..,.' 28. Is a known? e~ , r+''J 1 ~, / 29. As a result of your answer to Exercise 8.26, state the exact distributio n to use when calculating the confidence y :t l"fz. interval. Q ..r(,V\\!A vSt ( .... 6:h,..,~ ,,...lt>,.~l (j, ''"'u (Y) .J.t.... o ,., i?M.vfl 1 IJ7',._l +~ 4r"",J tt 6 Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises. 0 30. How much area is in both tails (combined)? a = 31. How much area is in each tail? ~ ) ;. l. v = 2-.<:;-/o 2 5" I J ( ?,b 28'" Chapter 8- Confidence Intervals #1-5, 23-37, 38-42 , 49-61 ' 62- 63, 79-94, 97, 103, 106, 111 ' 11 9, 123, 133 32. Identify the following spedfications: a. lower limit 2- 't, s 2 b. c. upper limit error bound 36' '2 r S , ~ 8' 33. The 95% confidence interval is: 7.-'t. t:; 2- ) 6' 2 8" 34. Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean. ~ C.L. = ~~~.. (/ l. S/- 36,Z-[) Figure 8.8 35. In one complete sentence, explain what the interval means. \\JQ... eG,f.fi'oJt:e, vv:-f-l;, '1S'7'J eo~~e~JCt.~ ~~ tk ~>'\\ S-{~1., o.re... ~vV:r- z...!\\-.S2 cx~d -.!!!j!' ~~ D \\)or {jl \\..{! V';" r 0 -witt~{ Q1l j Co llvu 1 >6 . 2--~ 'r()rf ofJ. 36. Using the same mean, standard deviation, and level of confidence, suppose that n ere 69 instead of 25. Would the h. error bound become larger or mailer? How do you know? ~., .,4L ~ ,..., tnpr(. :, 4rf" ,.f-,'P -to f"\\P.k. fhr L-Y""'~ ~ t"'-Vr~ ~ '-rr>r V,""'~ fll"'- tl ?fC<'"'e.. ~./"' ((' , V C 011 ,,cy'- 1>,...1 ~rl-.6vrt-5- f"Ailtr e.rrur b"" j, 37. Using the same mean, stand d deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why? 1 4 Atl el<;t tk ~a.rN.., -.;IW\\ ofVL lowvr +k 1d leve 1_, Q:, (,, w t- i.V"'"'JJ ~ ~ 5~1 11r {J'I-'l.f ,._. ~ r""e1J (' V."'I,tr'7 \\\\.t\\v&iP txw~ ~ (rC.. ~JI( -to Mel~ a ("1-vft frl-~;~ 1"-~5 k ..f~ ~ ~ -tk+ w(. L /.( ~ llfC 1,.. r lv'\\ _/ '1J L I (lq>J~-)JAQ'-\\ II. l ~ ( ~" , , v "'' fl .. c gtt<-J #.. "- ~r- ) a-rte.. ~.1-u ,A,._ { ( ,__1' 52. Is ax known? No 53. As a result of your answer to Exercise 8.52, state the exact distribution to use when calculating the confidence interval. <, 1\\-a0 1M' "' f- I,, 0 11 11 ~I),M-'t,1 r vVa PDf!.. I~+IWI ~ vvo)J. l-tavl- {o vc;J!. .+k .::: v it (\\ t r~ ~{ ) J Construct a 95% confidence interval for the true mean number of colors on national flags. . 1 1"' 1 "S-f'.I'V'-M-t.o Chapter 8- Confidence Intervals #1-5, 23-37, 38-42, 49-61 , 62- 63, 79-94, 97, 103, 106, 111 , 119, 123, 133 54. How much area is in both tails (combined)? 55. How much area is in each tail? 56. Calculate the followi ng: a. lower limit z . q 1... 6 I b. upperlimit 3 . "' G6 c. error bound 0 . ~ b 0) 2. . t1_.z.-6 f 57. The 95% confidence interval is__ 3 . S'6 ' C 58. Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean. C.L. = Ctc;(R Figure 8.9 nr.. (i ( ~!I t--f ,c? ~tv~ (~ M +rv..(..~ V\\-M-W "(}Co ort;. O~o.." "" ', ~.qz.6( "'..) ~.c,~6&, 60. Using the same x , sx, and level of confidence, suppose that n were 69 instead of 39. Would the error bound becor 59. In one complete sentence, expl what the in erval means. .,;t'= \\).h.. C 7"f.~ .. tli) Yo (Qfl larger or smaller? How do you know? . 1 _ \\) k ~(..t,... ~ c~~ V" Jr c ; eo:+-"' ,.vo.).<7( (?C.{fp...J 1 1 1~ Yl o!Jtr. b~ ..!'A"rf1J ~ql ~ e.-tY/1/ V/0~ lfo/t--vV? 0 ~,../d. ~ 1"'-o~ f>~C,)<. f (/ 1 ~' e'>-' ..,,_,~ 61. Using the same x , sx, and n = 39, how would the error bound c ange if the confidence level were reduced to goq Why? k W/l.e,-Ju.-.a. le"t.f v/rl oJr.-e~-J 'to ~0~"' .sli~ ~ ~ J a.!l(..J V7to 111.()..~ fN-arf "" M ""' (If((,'~{ b""i ,,~~ A.((CAF (;.k Q !. c. error bounur v-. o 1 .!-J , rr4 0 ' 00 I '-._-r ll_,.A-::- 7:: VI(, 1 .__ c~tV\\ :::.. J ~ 't -;;-- ' o 1 21 1 - c 0 ~ f 0 = \\ I rl'rdl,_;,l '7 1 I' 1- 'S"P ';1-- 1 (0' ~ ) ( 0 .1.) 1 of o. o1$- 3 89. The 92% confidence interval is ""'\\ n q__. '~ , .__ ~ 1 / '6 O 0, g . o, o1 ~ 3 ~ - EetV\\ D , 1 S ~ 90. Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion. Chapter 8- Confidence Intervals # 1-5, 23-37, 38-42, 49-61 ' 62- 63, 79-94, 97, 103, 106, 111 ' 119, 123, 133 C.L. =~2 Figure 8.10 91. In o_n,e complete , s~ntenc explajn wha 'I'll-- ~ltv'~ v-1 "fl/o CA"- (p-.tN" i7 ~-t~ r. e interval meflilS. ..e.,-..A{ Of a n,r.J--11/1. J 1\\ ttl~ ';{I f"" ,. . j 0. g 1 ~ '. .. ~t -f\\u-. -\\-f,.A-- P. I { I ~ J , -:p.1-:J ~ 92. Using the same p' and level of confidence, suppose that n were creased to 100. Would the error und become larger or smaller? How do )IOU know? c ~cc~)(. '1\\/t... "'-''"'( J. ~~t J c)- A- _." fVl() v ~ HI ~rl pre c: <::.C. '\\~ err~"r brv.:J: wr..-l~ (}~ ~ 93. sing the ame 1p' and =80, how w uld the error bound chaqge if the confid nee levelt..w - ere increased to 98%?1 ~y? \\ (uur IIA"tl Nll"'-\\ ,f\\(.f (<,(_ ' ~ (U tOO,,ll \\1-'-tl -y.l,(L irCdtt~ fv q 8 10 plcc~l.- VJ W&\\-tiQ l'ltiv<. f;) ,ll(r t'') 94. If you decreased the allowable error ound, why would the minimum sample size increase (keeping the same level of 4 r ( p.,-, c- (. 1 conf 1de nce)?. I! " ~'-', ~<-1'-s- , , /\\ 1 ~ e r,fV'vvtJ /11 cve u ;e tv Of'! /11 t/f{c:;/ 11 (!trtt'fti\\ .... .l k') t 0., h. I I. 5crn-e1 derr!af ~ , --1 ~ ,r.(..re. ~k. 'v" "'-"" I r '"led ~~ (11 ({Yt MtJ+hb"-,Jr''"'''J, ;I\\ .fht -@rrfYiwla E:l\\ rv1 " --:z ..l }! '$-' s. ~ \\ef'A().;(I ,.. ? "~ ( v' ,I,. ~ ,1 los-+~ ~-' P(A. !v..f-M rv 1/IA ( ,wet!/ Chapter 8- Confidence Intervals #1-5 , 23-37, 38-42, 49-61 ' 62- 63, 79-94, 97, 103, 106, 111 ' 119, 123, 133 r {Jb" 9.x ~Jcn<(' tt 97. Suppose that an accounting firm does a study to determine the time needed to com lete one erson's tax forms. I randomly surveys J..O.Q._p.eople. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. Th1 population distribution is assumed to be normal. 1. in a "''""rv 0 rfte~~ ) ~~ - / i. X = J__ ~, (, ii. a = 71. o / iii. n = \\OO - X -= r;. +-w-.e ,or~+,'~ . _ (i. ~ 4' I Co""f~~ 1~ ~,..f/c-ttl ()e. r:;J- r r ~!)(!I~ fA'( '~ ( In words, define the random variables X and X . ; _, , "'"JJ .D c. -::::=- . ,..= r.e,.~ ~~= , s ~ .,.., 1 Wllich distribution should you use for this problem? ExplcilifYour choice. IJb r f"'. ~ "' r Construct a 90% confidence interval for the population mean time to complete the tax form~ s. 1"' i. State the confidence interval. I 6ttc; ( ~ ) -:. ~ f5 tV\\ -;: 1 , 1~ 1 c;; c l.. 1190 11toi:~ ii. Sketch the graph. \\ \\ iii. Calculate the error bound. C--rf f J _ ------....._ :2. ~- b ! _.6"'" - '\\~g-- 1 , It; 1s-l ~t -l~l'> e . If the firm wished to increase its level of confidence and keep the error bound the same b tailili'g another survey, whatchangesshould itmake? lkL ~Vff\\ s~oUirlc.vea.~ ik f q, ... ~,'-zc tYI -14 tTl r $w-\\u:..- 1 , f. If the firm did another survey, kept the error ound the san1e, and,o y surve ed 49 people, what would happe to 'd ? Wh ? ( (#118/...r. <.r v/GM-lc:~ ;e. c E~~ ~ z~ G\\ ~J z_ . ~ (~} ,. \\1\\ ~(~ ~ \\~ 6131"'1 JJ(re'y~ a . . ) t~ ,1\\c r c~'5, 103. The average height of young adult males has a normal distribution with standard deviation of 2.5 inches. You want t< estimate the mean hei@t of students at_your college or university to within one inch with 93% confidence. How many mal1 students must you measure? r r;;~ J. ):_y\\ l /b I"- l, 1~ \\'1~ 0 'J \\)"' (V'W\\ ~.J} ;_, 4-, ~ 2 Ct 1- 2-0,\\\\~-q-r 8' ~ (1\\Ut! ,..,.L f-1'' I) ;J> \\. I II I '1 6' 2 I ("'..A It Chapter 8 - Confidence Intervals #1-5, 23-37, 38-42, 49-61, 62- 63, 79-94, 97, 103, 106, 111 , 119, 123, 133 106. Suppose that a committee is studying whether or not there is waste of time in our judidal system. It is interested in the mean amount of time individuals waste at the courthouse waitin to be called for jury duty. The committee randomly surveye 81 peo le who recently served as jurors. The sample mean wait time was eig t ours with a sample standard deviation of four hours. i. ji. X Sx = --'~=--- =_ 't_,____ iii. n = _~_f_ _ iv. n - 1 =---'~'-0_ __ ~ ul. fi ~ -e. Tlt.r Pj"" lw.f :J~ ,.,v:.-. ~~()'<""' .... ~ u vJth~ I r.t "".r ..y-t c~~~ dof\\-r ~,._.,.)JI " fwO Y" Define the random variables X and X in words. -1-o ~ C0ll.U 8rr l''l M 1' Ic Which distribution should you use for this problem? Explain your choice. 'i-::. 1ht. (J-.,vt.q'+ " ,. ""(ll ",.. 1, >' 0 V 0 - I &"" e . Explain in a complete sentence what the confidence interval means. u @ ~ ' I 1,q1 g,~ l,~'ls' f 3' &.6 64r 1-.IIS(, \\)Je.. evf,~tlc v-~1l ~fV\\.L tf,_}.l"~~vls .vlly+e. ~ ~-\\: vvtef\\- li Cf?l'. Ql .tk r.t~ ~,__-f ik_ (}..J4Pf Cllv..r~# wet;.frf1f 1- , ( t; 6 (J!J %' .~b .1.4 , 1 (,( .; /.J 2. ~(o ~'W" Ie-h.91t f'IL a. . /!M <\\~,.._4 ,. ~ c,...i tr} ~~~ J"/i ,~, '-6 0- Chapter 8 - Confidence Intervals #1- 5, 23-37, 38-42, 49-61 ' 62- 63, 79-94, 97, 103 , 106, 111 ' 119, 123, 133 c+ l11. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccu ied seats r fli ht over the ast ~ To accomplish this, the records of 225 flights are random! selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard fYiationis.A...Leats. - a ,r lr. \\} ~ t '\\ ..;J- w'" ._..Jt-. .10~ . 1''<1 ru~u'i~o"v<" l. x = \\).~ '-;'4l"'~~~~rll7~'')to- 'l" x~>._..-Jl' /" 0.~ ../' (, It, s X = 4' . I iii. n = 1-7 c; .. II. - I, 'Sf- . j ..f / 1.-- I 1':(/l.,tvf- fl'k a.... (\\, UW\\ l:Jt:r "\\]' VI' tJ cc. CA. I 11 ., (' /1 IJ ~~I, iv. n-1 = 7..2~ llr -tk n~~'-0' '1;tr. ). ... ;.....ov .. I (\\U 9t, "'(, o\\.Y\\17CC1Ap1-L- ~"' / ~ ~ I M- o'rer fk ftlVI 'ftlV lr 't (J'-J k . b. Define the random variables X and X in words. 'f-. "" Tk p.,rc r"'jt~ol ~rv'--c. Which distribution should you use for this problem? Explain your choice. ~ l.l ~ s~plt. /"' d. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight. i. State the confidence interval. f. :i -t.."'.f. \\}1 I, ~~,, " 2 ii. Sketch the graph. iii. Calculate the error bound. G 0 vv! -:: ~ .<)(.('fl..,.\\ . 0-'?qo _ 'f.. -= .=_=--::....__ rk 7o /~ ~-- ($) ~ \\l,b ~ \\1 . ~ t 1.1tbt ( t O.t~~t ~) ~ 2z..s ~ l '"' ~0> ~-~ V~\\\\it~ ~ ~ ~ ~ )('IJ{'- \\ f'Q~ ,~' kl~ 1\\.. \\ "~"" Q ~I) f"(J- \\}(,(, I >( ~ I", ~ ~,;"1 W/i-\\ 1' 'I' (f':(]"\\ . [.<.)( fl/ ~ ~-.~ e, 1\\ y l ~ '7 (}' 119. According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable ob. We are interested ? ' in the population proportion of people who feel the president is omg an accel?table job. ./. l -1 ~ p1 M ~....,~ 'r >;..t. a. Define the random variables X and P' in words. j :: tt~ t1 ~fIt- -t ~t fli'IV cJ (,p cv ~( ~ h ~b. c. '11 ).,,~ ? Which distribution should you use for this problem? Explain your choice. f~ .,/~ ~~r-UJ(I J fk. 5t:!""'rlz Construct a 90% confidence interval for the population proportion of people who feel e preside t 1s doin an ~ acceptable job. I 'J ~ ~I ~t er~: ~Mi. State the confidence interval. ~ 4lz_ 5 do, r c; F< r(e(;-1') 1 ii. Sketch the graph. f\\.r (" &l t 1 iii. Calculate the error bound. ,.. + t - rf.u.,k,l-. J-;-- ,.o/! ~ (\\-:. 12.00 -4 (. ~'-;g_bl qo !u q ~ff. ~q 0, S\\568 {),6\\ r l 64-S I t-z.-00 0 , 61 i l t~rJ\\ <- (; 0,0252 I :. o, oz.?z...J D, b I c:; /? -- ly o. 6 'J 32 , \\ . Chapter 8 - Confidence Intervals #1 -5, 23-37, 38-42, 49-61 ' 62- 63, 79-94, 97, 103, 106, 111' 119, 123, 133 123. A telephone poll ofl..QQO adult Americans was reported in an issue of Time Magazine. One of the questions asked was ''What is the main problem facing the country?" Twenty percent answered "crime." We are interested in the population \\ 1~ ro ortion who feel that crime is the main roblem. . ~; 11 .0 ~ Cl~ c: "'~ ..vl.o ~ (vJfJM 1/1- Ja . Definetherandom variablesXandP'inwords. X-::. p-t>pll-l_.,f,;'l\\. rr,pv~i IS~.(' M.ai"l V'Q~/1""-. k [o" ~'M" . Which distribution should you use for this problem? Explain your choice. ~I -=- Tk S" (;( . . . . f I~ p f l pur-tl "/1 r ' I\\ l,erv~-'( c. Construct a 95% confidence interval for the population proportion of adult Americans who feel that crime is the fl'l ( cIt.- .J I Q ~(~~ ~v-{btv main problem. I 1 ~ vJ"r- 0 flA~! en . . ...e.. \\)Jf A.. \\ \\4r& ;-0 ~{t( _i. State the confidence interval. f t ~ J~ , l~ ~"" L, .,tJ {1 "'1 , _"'v.f.Je,. i \\fr ~. ' 1i. Sketch the graph. - - 1S tN- fiL r ~ . ~ ~f" iii. Calculate the error bound. E P.IM '::. 0 o ~ ~~ ~ fl.if"' (/) 11 ~- Suppose we want to lower the sampling error. What is one way to accomplish that? (\\ _,( ~~~t e . The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is 3%. In one to three 1 ~ complete sentences, explain what the 3% represents. ---..... ,,1..., cJJ vR ,_)....( M- I 'i "() Jf ~~ ~ t\\ v' ~ f\\ (l ~ ~ 1 :: l to ,'tlu~;?<- ~ '7 &("'f k Slu. \\}.j e. c 0 \\J e--ls.:~ Io ,.-er ov-1 (_(!.(\\. e~ tfa,._< I "'11?r ....... I s ' error D. 2 o.a ~I-- Of\\.- wfM.j -\\:" \\<> 0 ,0 3 ) 133. You plan to conduct a survey on your college campus to learn about the political awareness of srudents. You want to estimate the true proportion of college srudems on your campus who voted in the 2012 residential electi with 95% confidence and a margm of error no ater than ive ercent. ow many srudents must you inteiView? - L c~..- -: t}S,;. <( ~- V' s I' .. (, o..;~ ::: } '<;" (_, ce-, -:.. I, 16 "2- s- "f.- \\.qb o.o') j --- n - 3 S 4J b t?<- ~ sr; q'-:- o.C: ~I ~ 0, ~ a" J f( ) Jo ;vl,tv7 ,11 i /f c 1 r f/ o r t~r {ka 1t. ( MEAN 6.0000 WHEN WE HAVE LARGE DATA SETS, WE GROUP THE DATA. IN THIS CASE OUR GROUPS WILL BE: STD DEV 2.17 IN A NORMAL DISTRIBUTION THE MEAN, MEDIAN AND MODE ARE ALL THE SAME NUMBER X-VALUES Z-VALUES GROUP NUMBER OF DATA POINTS IN IT 2 -1.84 2.0-2.9 2 2 -1.84 3.0-3.9 3 7 3 -1.38 4.0-4.9 4 6 3 -1.38 5.0-5.9 5 3 -1.38 6.0-6.9 6 5 4 -0.92 7.0-7.9 5 4 -0.92 8.0-8.9 4 4 4 -0.92 9.0-9.9 3 3 4 -0.92 10.0-10.9 2 5 -0.46 2 5 -0.46 5 -0.46 1 5 -0.46 5 -0.46 0 12 23 34 45 78 6 0.00 X-VALUES: 65 76 (3.83) 6 0.00 98 109 6 6 6 6 0.00 0.00 0.00 0.00 7 0.46 7 7 7 0.46 0.46 0.46 Z-VALUES ARE CALCULATED AND SHOW THE NUMBER OF STANDARD DEVIATIONS A DATA POINT (OR GROUP OF DATA POINTS) IS FROM THE MEAN Z = ( X - MEAN) / STD DEV OF COURSE IF YOU HAVE A Z-VALUE YOU CAN BACK CALCULATE TO GET THE X-VALUE: X = Z TIMES THE STD DEV + THE MEAN AS AN EXAMPLE USE THE Z-VALUE OF +0.46 TO SEE WHAT X-VALUE IT REPRESENTS: X = 0.46 x 2.17 + 6.0 = 7.0 7 8 8 8 8 9 9 9 0.46 0.92 0.92 0.92 0.92 1.38 1.38 1.38 SIMILARLY THE X-VALUE (OR GROUP) THAT IS -1 STANDARD DEVIATION BELOW THE MEAN (Z = -1) WOULD BE: X = 1 x 2.17 + 6.0 = 3.83 WHILE THE SHAPE OF OUR EARLIER GRAPH WAS RECTANGULAR, THIS GRAPH HAS A BELL SHAPE, REFERRED TO AS THE "NORMAL" DISTRIBUTION. AND, AS WITH THE RECTANGULAR GRAPH THE AREA SHOWN IN THIS GRAPH REPRESENTS 100 % OF OUR DATA AND ANY PORTION OF IT REPRESENTS THE PERCENTAGE OF OUR DATA IN THAT AREA AS WELL AS THE PROBABILITY THAT SPECIFIC DATA ARE IN THAT REGION. HERE IS A MORE TYPICAL BELL CURVE GRAPH: TO CALCULATE THE AREA IN ANY SECTION OF THIS BELL CURVE GRAPH WE NEED TO KNOW THE EQUATION FOR THE LINE, AND THEN WE USE INTEGRAL CALCULUS TO CALCULATE THE AREA UNDER THAT CURVE FOR THE AREA SELECTED. n 10 10 1.84 1.84 Z-VALUES: -2 -1 0 +1 +2 THE X-VALUES ARE THE DATA POINTS AND WE HAVE GROUPED THEM AS SHOWN. x= 1 FORTUNATELY FOR US, THESE AREAS HAVE BEEN CALCULATED AND PUT INTO TABLES. THE HORIZONAL AXIS HAS THE Z-VALUES ON IT, WHICH ARE THE STANDARD DEVIATIONS FROM THE MEAN AND THE TABLE GIVES US THE AREA TO THE LEFT OF THAT Z-VALUE. (SEE & PRINT THE ATTACHED TABLES) TO GET THE AREA TO THE RIGHT WE MUST SUBTRACT THE TABLE VALUE FROM 1.0000 SINCE THE SUM OF BOTH AREAS EQUALS 100% OR 1.00 AS A DECIMAL. HERE ARE SOME "WORKED-OUT" AREAS: 68.3% OF THE AREA (OUR DATA VALUES) ARE WITHIN -1 AND +1 STANDARD DEVIATIONS FROM THE MEAN 95.4% OF THE AREA (OUR DATA VALUES) ARE WITHIN -2 AND +2 STANDARD DEVIATIONS FROM THE MEAN 99.7% OF THE AREA (OUR DATA VALUES) ARE WITHIN -3 AND +3 STANDARD DEVIATIONS FROM THE MEAN AN "UNUSUAL" VALUE IS CONSIDERED ONE THAT IS MORE THAN TWO STANDARD DEVAIONS FROM THE MEAN (Z = + 2) MORE SPECIFICALLY, THIS "UNUSAL" OR RARE EVENT AREA CORRESPONDS TO 5% OF THE GRAPH AREA ACCORDINGLY, THESE CRITICAL Z-VALUES ARE +1.645 ON THE RIGHT END AND -1.645 ON THE LEFT END OF THE GRAPH. SEE IF YOU CAN FIND THE AREA IN THE TABLE (0.05) FOR THESE Z-VALUES. Wk4-HW4 Instructions WEEK 4 HOMEWORK: LANE CHAPTER 7 AND ILLOWSKY CHAPTERS 6 AND 7 THE NORMAL DISTRIBUTION Z-TABLES ARE INCLUDED IN THE COURSE RESOURCES AND YOU ARE TO USE TH TO SOLVE THESE PROBLEMS. THIS IS STRAIGHT FORWARD TABLE READING. THIS WEEK'S CONCEPTS ARE REALLY THE HEART OF OUR COURSE. PROBABILITY (FROM LAST WEEK) IS THE STATISTICS (AND THEORETICAL PHYSICS - WATCH THE PBS SHOW \"PARTICLE FEVER\"). THE NORMAL DISTRIBUTION AND THE AREAS UNDER PARTS OF IT ARE ALL WE ARE TRYING TO FIGURE OUT IN THE AREAS IN THE TABLE THAT CORRESPOND TO SPECIFIC (CALCULATED) Z-VALUES. OUR DATA ARE THE X-VA Z-VALUES FROM THEM. WE CAN ALSO BACK-CALCULATE X-VALUES FROM Z-VALUES. LONG STORY SHORT: THERE ARE SPECIFIC Z-VALUES OF INTEREST REFERRED TO AS \"CRITICAL VALUES\" AND CORRESPOND TO SMALL (RARE) AREAS IN ONE OR BOTH \"TAILS\" OF OUR NORMAL DISTRIBUTION: 1%, 5% OR THE NORMAL CURVE. THESE ALREADY SMALL AREAS CAN ALSO BE SPLIT BETWEEN BOTH ENDS LEAVING 0.5 THE PERCENTAGES ARE OUR \"SIGNIFICANCE LEVELS\" AND WE CHOSE ONE BASED ON HOW SURE WE WANT T 90%, 95% OR 99% (CORRESPONDING TO THE AREAS OF 10%, 5% AND 1%) NON-TEXT PROBLEM #1: LOOK THESE Z-VALUES UP FOR THESE AREAS: + 1%, + 5%, + 10% AND + 0.5%, + 2.5% THE + 5%) AND WRITE THEM DOWN AS THEY DON'T CHANGE. THE Z-VALUES ON THE LEFT END ARE NEGATIVE END ARE POSITIVE. HERE IS HOW WE USE THESE Z-VALUES TO SEE IF OUR DATA ARE IN THE \"RARE\" OR \"UNUSUAL\" AREAS TO THE NORMAL DISTRIBUTION. WHY DO WE CARE IF DATA ARE RARE? YOU WILL SEE. LET'S START WITH OUR QUANTITATIVE, CONTINUOUS DATA VALUES (REFERRED TO AS THE X-VALUES). WE NE OUR X-VALUE: [Z = (X - MEAN) / STD DEV ]. THIS Z-VALUE IS SIMPLY THE NUMBER OF STANDARD DEVIATIONS O MEAN. NOW, WE COMPARE OUR CALCULATED Z-VALUE (REFERRED TO AS THE \"TEST STATISTIC\") TO OUR CRITICAL V SIGNIFICANCE LEVEL WE CHOSE). IF THE TEST STATISTIC IS GREATER THAN THE POSITIVE (+) CRITICAL Z-VALUE WE ARE IN THE \"UNUSUAL\" OR R OF THE NORMAL DISTRIBUTION. IF IT IS LESS THAN THE NEGATIVE (-) CRITICAL VALUE IT ALSO IN THE RARE A THIS IS THE WAY WE TEST A HYPOTHESIS, AS YOU WILL LEARN IN LATER CHAPTERS. IF OUR TEST STATISTIC E REJECT OUR HYPOTHESIS, IF NOT WE ACCEPT IT. BUT, KEEP IN MIND THAT \"ACCEPTING' DOES NOT MEAN \"PR NOTHING BY ITSELF. IT SIMPLY ADDS SUPPORT TO OTHER INFORMATION. LET'S GET BACK TO THE NORMAL DISTRIBUTION. THE EXCEL PAGE ATTACHED PROVIDES AN EXAMPLE O SURE IT MAKES SENSE TO YOU. NOW, LET'S MOVE ON TO THIS WEEK'S HOMEWORK: LANE (C7) FIRST PROBLEM #8 THIS IS A STRAIGHT FORWARD Z CALCULATION. THIS Z-VALUE (SCORE) TELLS YOU HOW MANY S OF 65 IS FROM THE MEAN (IN THIS CASE IT'S A NEGATIVE DISTANCE, MEANING IT'S BELOW THE MEAN). USE TH SOFTWARE) TO FIND THIS AREA (NEGATIVE - Z-VALUES) AND REMEMBER TOO THAT THE TABLES GIVE THE ARE VALUE. TO GET THE AREAS TO THE RIGHT (BUT NOT NEEDED IN THIS PROBLEM) YOU SIMPLY SUBTRACT THE A SINCE THE AREA UNDER THE NORMAL CURVE ACCOUNTS FOR 100% OR 1.00 OF THE DATA. KEEP IN MIND THAT THESE \"AREAS\" ARE THE PERCENTAGE OF DATA BELOW YOUR SPECIFIC X-VALUE AND A (RELATIVE FREQUENCY) OF DATA BEING IN THAT AREA. THIS IS THE HEART OF IT. AS AN EXAMPLE, TO READ THE TABLE FOR A Z-SCORE OF +1.31 GO TO THE TABLE OF POSITIVE Z-SCORES AND AND THEN GO OVER TO THE 0.01 ON THE TOP AND GET THE READING FROM THE TABLE. IT'S 0.9049. IF YOU H THE OTHER TABLE. WHAT IF YOU HAVE 1.315? GET THE AREAS FOR 1.31 AND 1.32 AND IN OUR CASE SIMPLY A BETWEEN THEM. BACK TO THE AREA OF 0.9049, THIS AREA MEANS THAT 90.49% OF OUR DATA ARE BELOW THE DATA POINT THAT OF 1.31. IT ALSO MEANS THAT THERE IS A 90.49% PROBABILITY THAT A DATA POINT WILL BE IN THAT AREA BELO CONVERSELY, ABOUT 9.5% WILL BE ABOVE IT (100% - 90.49% = 9.5%). MAKE SENSE? PART B IN THIS PROBLEM WE USE AN EVEN SMALLER PERCENTAGE (LESS THAN 1%, BUT YOU NEED TO CALCU UNDER THE CURVE) PART C IS A LITTLE TRICKIER. NOTE THAT THE PROBLEM STATES THAT ONLY 10% BE OVER THE SPEED LIMIT. PERCENTAGE UNDER A SPECIFIC Z-VALUE. SO, IF 10% ARE OVER, THAN MEANS THAT 90% ARE UNDER. FIND T (+TABLE) CORRESPONDING (NEAREST NUMBER) TO 90%. ONCE YOU HAVE THAT Z-VALUE YOU CAN BACK-CAL VALUE, WHICH IS THE SPEED OF THE CAR. (THE FORMULA IS THE Z = (X-MEAN)/STD DEV SO USE ALGEBRA TO PART D - THE DISTRIBUTION (SHAPE OF THE CURVE) IS ACTUALLY SKEWED NOT BELL SHAPED, TELL ME IF IT'S SKEW. PROBLEM #11: ANOTHER TRICKY ONE. THE \"TOP 30%\" ARE THE ONES TO THE RIGHT SO WE NEED THE Z-VALU 70% OF THE AREA BEING BELOW IT. ONCE YOU HAVE FOUND THAT Z-SCORE, BACK-CALCULATE THE ACTUAL X NECESSARY SCORE ASKED FOR. USE THE SAME PROCEDURE FOR PART B (5% ABOVE MEANS 95% BELOW) PROBLEM #12: PARTS \"a\" IS FAIRLY STRAIGHT FORWARD, BUT PART \"b\" IS VERY TEDIOUS. THIS PROBLEM INVO DISTRIBUTION (THE BELL CURVE ) WHICH IS CONTINUOUS TO APPROXIMATE THE BINOMIAL WHICH IS DISCRET BINOMIAL HAS JUST TWO OPTIONS (LIKE HEADS/TAILS). THE NORMAL DEALS WITH CONTINUOUS DATA. WE WANT THE PROBABILITY OF GETTING 15 TO 18 HEADS OUT OF 25 COIN FLIPS. IF YOU LOOK AT THE HISTOG BELL SHAPE YOU SEE THAT EACH COLUMN IS CENTERED ON THE 1, 2, 3 ETC. THIS MEANS THAT FOR A VALUE RANGE OF 1.5 TO 2.5. DO YOU SEE THIS? OK SO HERE IS WHAT WE DO TO USE THE NORMAL TO APPROXIMAT FIRST: WE NEED TO CALCULATE THE MEAN, WHICH HERE EQUALS THE TOTAL NUMBER OF DATA POINTS (IN TH PROBABILITY OF GETTING A PARTICULAR RESULT (LIKE A HEAD), IN THIS CASE 50/50 OR 0.5. THE VARIANCE EQ DATA POINTS TIMES THE FIRST PROBABILITY (GETTING HEADS), WHICH IS 0.5 TIMES THE PROBABILITY OF NOT 0.5) YOU CAN SEE THAT WE COULD HAVE DIFFERENT PROBABILITIES, FOR EXAMPLE IF THE COIN WERE WEIG BUT THAT IS NOT THE CASE HERE. ONCE WE CALCULATE THIS VARIANCE, THE STANDARD DEVIATION IS SIMPL VARIANCE. WE NOW HAVE OUR SAMPLE MEAN AND STANDARD DEVIATION FOR OUR BINOMIAL. TO GET PROBABILITIES FROM OUR TABLES WE NEED Z-VALUES. WE WANT THE PROBABILITY OF GETTING 15 T TOSSES. BUT, SINCE THE NORMAL DISTRIBUTION IS CONTINUOUS, THE 15 GOES FROM 14.5 TO 15.5 AND 18 GO OUR RANGE WHEN USING THE NORMAL TO APPROXIMATE THIS BINOMIAL IS 14.5 UP TO ____? NOW, CALCULA TWO LIMITS. FROM THESE Z-VALUE WE GO TO THE TABLE AND FIND THE AREAS TO THE LEFT CORRESPONDING TO THEM. AREA BETWEEN THESE TWO LIMITS WE MUST SUBTRACT THE LOWER AREA FROM THE UPPER AREA. IS THIS C PICTURES TO SEE WHAT IS GOING ON. PART B: INVOLVES THE ACTUAL BINOMIAL EQUATION. THE BINOMIAL GRAPH IS NOT CONTINUOUS BUT WILL H POSSIBILITY, IN THIS CASE 15, 16, 17, AND 18 HEADS. WE NEED TO USE THE COMPLEX BINOMIAL FORMULA TH FOR PROBLEM LANE CHAPTER 5 #25. CHECK IT OUT. WE MUST USE THAT FORMULA TO GET THE PROBABILITY FOR EACH OF THESE FOUR POSSIBILITIES AND THEN TO DO THE MATH (CORRECTLY) ON THIS ONE FOR 0.5 BONUS POINTS, SHOW ME YOUR SET UPS AND CALCULA 0.2049). ANSWER THIS FOR PART B: HOW DOES THE TRUE BINOMIAL CALCULATION RESULT OF 0.2049 COMPARE TO TH THE BINOMIAL (FROM PART A)? BY WHAT PERCENTAGE IS THE APPROXIMATION OFF? ILLOWSKY CHAPTER 6 PROBLEM #60 THE EASY SOLUTION IS THAT SINCE THIS IS A NORMAL DISTRIBUTION, THE MEAN, MEDIAN AND NUMBER. SO, IF THE MEAN IS 5.3, THEN THE MEDIAN IS 5.3. BUT, I WANT YOU TO DO THIS USING THE Z-TABLES OF THE \"MEDIAN\". WHAT PERCENT OF OUR DATA POINTS ARE ABOVE AND BELOW THE MEDIAN? USING THE P THE AREA TO THE LEFT) GO TO THE TABLE AND GET THE Z-VALUE THAT CORRESPONDS TO THAT AREA. USING AND SD, BACK CALCULATE THE X-VALUE. 60. What is the median recovery time? a. 2.7 b. 5.3 c. 7.4 d. 2.1 PROBLEM #66: USE THE FORMULA FOR Z: Z = (X-u)/s AND EXPLAIN WHAT THE RESULT MEANS. 66. Height and weight are two measurements used to track a child's development. TheWorld Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean = 10.2 kg and standard deviation = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them. a. 11 kg b. 7.9 kg c. 12.2 kg PROBLEM #76: DO THE MATH AND READ THE TABLE NOT SOFTWARE. SKIP \"a\" (b) IT ASKS FOR \"LESS\" THAN SO IT'S THE VALUE (AREA) IN THE TABLE. CALCULATE THE Z-SCORE AND READ (WHICH IS THE PROBABILITY OR PERCENTAGE BELOW 220). (c) THE 80TH PERCENTILE SIMPLY MEANS 80% OF THE AREA (TO THE LEFT). THIS MEANS THAT 80% OF THE BAS THIS DISTANCE. FIND THE Z-VALUE CORRESPONDING TO 0.80 IN THE TABLE AND BACK-CALCULATE THE X-VAL SOFTWARE. 76. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. a. If X = distance in feet for a fly ball, then X ~ _____(_____,_____) b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement. PROBLEM #88: TRICKY. CALCULATE THE Z-SCORE FOR 30%. FIND THE AREA IN THE TABLE THAT CORRESPON LEAST 30\" MEANS THAT WE DON'T WANT THE AREA TO THE LEFT WHICH WOULD BE THOSE SCORES LESS THA TO THE RIGHT (THOSE SCORES GREATER THAN 30%). SO, SINCE OUR TABLES ONLY GIVE AREAS TO THE LEFT (OR 100%). PIECE OF CAKE. (b) 95TH PERCENTILE MEANS THAT 95% OF OUR DATA ARE IN THIS AREA. FIND THE Z-VALUE THAT CORRESPON TO A X-VALUE WHICH IN THIS PROBLEM IS ALSO A PERCENTAGE SO IT'S A LITTLE CONFUSING. 88. Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site. On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent. CHAPTER 6 | THE NORMAL DISTRIBUTION 367 a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30. b. Find the 95th percentile, and express it in a sentence ILLOWSKY CHAPTER 7 PAGE 375 DISCUSSES THE CENTRAL LIMIT THEOREM AND SHOWS THE FORMULAS WE WILL BE USING (NOT SO SAYS THAT IF WE TAKE THE MEANS OF MULTIPLE SAMPLES FROM THE SAME POPULATION, THOSE MEANS WIL REGARDLESS OF THE ACTUAL SHAPE OF THE ACTUAL POPULATION'S DISTRIBUTION.. PROBLEM #62: USE THE FORMULAS ON PAGE 400 (TEXT PAGE) FOR THIS ONE. IT'S THE SAME PROCEDURE AS FOR THE STD DEV, HENCE Z IS SLIGHTLY DIFFERENT, BUT THE SAME TABLES ARE USED. 62. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. a. If X = average distance in feet for 49 fly balls, then X ~ _______(_______,_______) b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X . Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of the average of 49 fly balls.. PROBLEM #70: PAY ATTENTION TO ALL THE CHOICES THAT ARE CORRECT AS WELL. REMEMBER THEM. 70. Which of the following is NOT TRUE about the distribution for averages? a. The mean, median, and mode are equal. b. The area under the curve is one. c. The curve never touches the x-axis. d. The curve is skewed to the right. PROBLEM #96: DO THE MATH, READ THE TABLES NOT SOFTWARE. SIMPLY CALCULATE THE Z-SCORES FOR 85 FORMULAS AND GET THE AREAS (PROBABILITIES/PERCENTAGES) FROM THE TABLES (THESE AREAS ARE VERY SUBTRACT THE AREA FOR 85 FROM THAT FOR 125 TO GET THE AREA BETWEEN THEM. THIS AREA IS THE PERC ASKED FOR. (IT'S BIG) 96. A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ tesst, what is the probability that the sample mean scores will be between 85 and 125 points? \f