Mathacle. PSet Stats, Concepts In Statistics Level Number Name: Date:

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1 APPENDEX I. THE RAW ALGEBRA IN STATISTICS A I-1. THE INEQUALITY Exmple A I-1.1. Solve ech iequlity. Write the solutio i the itervl ottio..) 2 p - 6 p -8.) 2x- 3 < 5 Solutio:.). - 4 p -8 p³ 2 or pî[2, + )..) 2x- 3 < 5 is equivlet to ( 2x - 3) < 5 d - (2x- 3) < 5 xî( -1, 4). Exmple A I-1.2. Write i the iequlity i the solute form for vrile x. Solutio: The geerl solutio to the iequlity ove is + - x- < 2 2 For this exmple, the solutio is x - 6 < 8 2-m Exmple A I-1.3. Solve the iequlity 5 s i the itervl ottio d i iequlity form. Solutio:, for m, where s > 0. Write the swer 2-m 5 s 2-m 5s 2-5s m 2+ 5s or [2-5s, 2+ 5s ] 140

2 Solve ech iequlity. Write the solutio i the itervl ottio. 1.) 2 p - 4 p -2 2.) - 6 (1+ 2x) ³ 6(2x-1) + 2x 3.) 5x+ 10-4x-17 < 9-2x 4.) 2 x - 3< 11 or - 8x -10 < ) 2x + 11 < 5 141

3 6.) 8x + 3 > 11 Write the ech iequlity i the solute iequlity form for vrile x. 7.) 8.) 9.) 142

4 Solve the idicted vrile. Write the swer i the itervl ottio d i iequlity form. Assume tht s, m, pˆ, qˆ, z * re ll positive umers. 10.) 5 3 -m 1, for m 16-m 11.) ³ 3, for m s 10-m 12.) 1, for m s 5-m 13.) 2, for m s x-m 14.) z *, for m s ˆ 0 p- p 15.) z *, for p 0 pq ˆ ˆ 143

5 0.8(1-0.8) 16.) , for iteger 0.6(1-0.6) 17.) z * 0. 03, for 900 z * 18.) The solutio set for the iequlity 2x -m 3 is xî[1, 4]. Wht is the vlue of m? 19.) The iequlity ove c e descried y m d s? 1.96s x-m for vrile x. Wht re

6 Aswers 1.). [ 1, + ) 2.) (-, 0] 3.) (-13, - 3] 4.) (-, 7) È (9, + ) 5.) (- 8, - 3) 7 6.) (-,- ) È (1, ) 4 7.) x - 4 < 8 8.) 1 7 x - < ) x -1 > 2 10.) 2 m 8, or [2, 8] 145

7 11.) m 16-3s orm³ s, or (-,16-3 s ] È [16+ 3 s, + ) 12.) 10- s m 10 + s, or [10- s,10 + s ] 13.) 2s 2s é 2s 2s ù 5- m 5 +, or ê5 -, 5+ ú ë û s s é s s ù 14.) x- z * m x+ z *, or êx- z *, x+ z * ú ë û pq 15.) ˆ ˆ pq ˆ ˆ é pq ˆ * ˆ ˆ ˆ ˆ pq ˆ ˆ ˆ ù p- z p0 p+ z *, or ê p- z *, p+ z * ú ë û 16.) > ) z *< m- 3 m ) 5. x m = s - 3- (-13) 19.) m = =-8, = s =

8 A I-2 THE PROBLEM OF AREA UNDER THE NORMAL CURVE The stdrd orml proility desity fuctio is give y f (z) : f(z) = 1 e 2p -1 2 z 2 Exmple A I-2.1. Fid the f(z) for the give z vlues z f(z) Sketch the grph of the f(z) sed o the clculted dt ove: Solutio: z - 4 f(z)

9 [Ti-84] The re uder the curve f(z) for give itervl ( z, z ) c e clculted y The vrile z AREA ( z < z< z ) = ormlcdf ( z, z,0,1) vlue for give re A = AREA - < z< z ) c e clculted y ( z = ivnorm( A, 0,1) Note tht AREA (- < z<+ ) = 1, Tht is, the etire re uder the curve For the clcultio purposes, whe e used i the plce of ifiity. z or f(z) z re ifiity, lrge umer such s 1000 is oe. c Exmple A I-2.2. Fid the re uder the curve vlue of AREA (-2< z< 1.5)? f(z) whe - 2 < z <1. 5. i.e., wht is the Solutio: AREA (-2< z< 1.5) = ormlcdf (-2, 1.5, 0, 1) = Exmple A I-2.3. Fid 10 th percetile, z, where AREA (- < z< z ) = 0. 1, or the re tht is elow Solutio: Sice AREA (- < z< z ) = 0. 1, z = ivnorm ( 0.10, 0, 1) =

10 Give the fuctio f (z), fid the re or the idicted vriles for ech of the prolems elow. Lel the domi (itervl) d shde the proility re. 1.) AREA (-1< z< 1) 2.) AREA ( z < 2) 3.) AREA ( z-1 < 1) 4.) AREA (- < z< -2) 5.) Fid z, where AREA (- < z< z ) = 0. 05, or the re tht is elow 5 th percetile, 149

11 6.) Fid z, where AREA ( z < z<+ ) = 0. 2, or the re tht is ove 80 th percetile. 7.) Fid z *, where AREA ( z < z*) = ) Fid z *, where AREA ( z > z*) = æ ö 9.) Fid the iteger such tht AREA ç z < = è ø 1.64s 10.) Fid s such tht AREA ( z < ) =

12 Aswers 1.) AREA (-1< z< 1) = ormlcdf (-1, 1, 0, 1) = ) AREA ( z < 2) = AREA (-2< z< 2)» ) AREA ( z-1 < 1) = AREA (0< z< 2)» ) The c e pproximted y reltive lrge umer such s Therefore, AREA (- < z<-2)» AREA (-100 < z<-2) = ormlcdf (-100, - 2, 0, 1) = ) z = Sice AREA (- < z< z ) = 0. 05, z = ivnorm ( 0.05, 0, 1) = ) z = The re o the left side is 1-0.2= 0. 8, z = ivnorm( 0.8,0,1) =

13 7.) z*= Sice - z * < z< z *, z = -z * d z = z * = 0.975, z* = z = ivnorm (0.975,0,1) = The re o the left side is 8.). Sice - z * > z or z> z *. The re o the left side for clcultig the = 0.95, z * = ivnorm (0.95,0,1) = z*= 1.64 z * is ) = 400. Sice = 1.96 = s 10.) s = Sice = 1.64 s =

14 A I-3. THE Z TRANSFORMATION The geerl ell curve is descried y the fuctio p (x) : 1 p( x) = e s 2p 2-1æ x-mö ç 2 è s ø xî (-, ) I Ti-84, the re udereth the curve p(x) c e clculted y AREA ( x < x< x ) = ormlcdf ( x, x, m, s ) The fuctio f(z) is ssocited with fuctio p(x) through the lier trsformtio: z = x-m s where m d s re some costts. The re uder f (z) i the itervl is the sme s the re uder p(x) i CI, i.e., AREA ( x < x< x ) = AREA ( z < z< z ) d whe I = z, z ] is give, CI = [ x, x ] = [ m+ z s, m+ z s]. [ I 153

15 Exmple A I-3.1. Give the re uder the f (z) is AREA (- < z< z ) = 0. 95, m = 10 d, fid d d the re uder. Tht is, wht is s = 2 x x p(x) AREA ( x < x< x )? Verify tht AREA - < z< z ) = AREA ( x < x< x ). Solutio: From the give, z = - (, = ivnorm( 0.95, 0, 1) = Becuse z z= x-m x = m- zs = 10- (2) =-, d = m+ z * s = (2) = Sice, choose lrge umer such s s The re uder AREA (- < x< 13.29) = ormlcdf (-1000, 13.29, 10, 2)» z = - x z p(x) s, is the, Exmple A I-3.2. Give AREA ( z < z*) = 0. 95, Verify tht AREA ( x x< x ) = < m = 10 d s = 2, fid x d x. Solutio: Sice - z * < z< z *, this implies z = -z *, z = z *, d AREA (- < z< z*) = z * = ivnorm (0.975, 0, 1) = 1.96, x = m- z * s = (2) = 6. 08, x = m+ z * s = (2) = Ad the re uder is AREA ( 6.08< x< 13.93) = ormlcdf (6.08, 13.95, 10, 2)» p(x) Exmple A I-3.3. AREA ( z < z*) = 0. 95, x = 0.5, s = Decide if m could e Solutio: 0.5-m z = z* = ivnorm (0.975, 0,1) = 1.960, < < Þ 0.01< m < Yes, m could e Exmple A I-3.4. AREA ( z < z*) = p, m = 4, s = Wht is the miimum z * such tht Î [ x, x ]? Wht is p? 5 Solutio: Sice - z * < z< z * d = x-4 x- 4 z, the - z * < < z * or 4-0.5z * < x< z * This implies tht z * ³ 5 or z *³ 2. p = AREA ( z 2) =

16 Give the p (x), solve ech prolem. 1.) m = 5, s = 6 d x = 17, fid z. 2.) m = 20, AREA ( 30 < x< ) = or 45% of the re is ove x = 30. Fid s d z such tht AREA ( z < z< ) = ) m = 88, AREA (- < x< 50) = or 2% of the re is elow x = 50. Fid s d z such tht AREA (- < z< x ) = ) s = 5, AREA (- < x< 100) = 0. 8 or 80% of the re is elow x = 100. Fid d m. z 5.) s = 15.6, AREA ( 17.2< x< ) = 0. 1 or 10% of the re is ove x = d m.. Fid z 155

17 6.) AREA ( z < z< ) = 0. 05, m = 3, s = Fid x such tht AREA ( x < x< ) = Verify tht AREA ( x < x< ) = AREA ( z < z< ). 7.) AREA ( z < z*) = 0. 95, x = 3, s = 2. Fid z *, d m d mî [ m, m. m such tht ] 8.) AREA (- < z< z ) = 0. 95, x = 0.5, s = Decide if m could e AREA z < z * = 0., m = 0.4, s = 0. 25, wht re 9.) If ( ) z *d decide if x could e p( 1- p) AREA, x *= 0. 6, m = 0. 5, s = wht is the vlue of p? ) ( z < z *) = ) AREA ( z < z*) = 0. 95, m = 3, s = Decide if 5 is i the itervl CI tht cotis x. Tht is, Î [ x, x ]

18 12.) AREA ( z < z*) = p, m = 3, s = Wht is the miimum Î [ x, x ]? Wht is for the miimum 5 p z *? z * such tht 13.) AREA ( z < z*) = 0. 9, m = 3. Wht is the mximum s such tht Ï [ x, x ]? ) AREA ( z < z*) = , m = 0. 5, d s =, where is the miimum such tht x < 0. 53? is positive iteger. Wht 15.) Give two curves ove with H 0 : p0 ( x), m0 = -1, s 0 = 1 d H : p1 ( x), m1 = -3, s1 = 1. Aswer ech of the followig questios. 157

19 .) wht is to x * x * uder the tht AREA (- < x< x* H 0 ) = Tht is, the re from egtive ifiity H 0 curve is 0.1?.) wht is = AREA x* < x< + H )? ( c.) wht is PWR = 1-? 158

20 Aswers ). z = = 2 6 x -m ) m = 20, x = 30, z = ivnorm( 0.55,0,1) = , s = =» z ) m = 88, x = 50 x -m 50-88, z = , s = =» z ) s = 5, x = 100, z = , m = x - z s = (5) = ) s = 15.6, x = 17. 2, z = , m = x - z s = (15.6) = ) z = ivnorm( 0.95,0,1) = 1. 64, x = m+ zs = (0.5) = 3. 82, d to verify AREA ( 3.82 < x< ) = ormlcdf (3.82, 1000, 3, 0.5)» m 7.) z *= 1. 96, -1.96< z < 1. 96, -1.96< < m Î CI 2 = [ 3-2(1.96), 3+ 2(1.96)] = [-0.92, 6.92], 0.5-m 8.) z = ivnorm( 0.95, 0, 1) = 1. 64, < 1.64 Þ m > No x ) z * = ivnorm (0.975, 0,1) = 1. 96, -1.96< < 1.96 Þ -.09< x< Yes ( ) 10.) z * = ivnorm (0.985, 0,1) = 2. 24, = 2. 24, p ( 1- p) = , p = p(1- p) 50 or p =

21 x ) z * = ivnorm (0.97,0,1) = 1. 96, -1.96< z < 1. 96, -1.96< < 1. 96, 0.5 x Î CI = [ 3-0.5(1.96), (1.96)] = [-2.02, 3.98]. The swer is o. x ) - z * < z< z *, - z * < < z *, 3-0.5z * < x< z *, z * ³ 5 z* 4. p = AREA (-4< z< 4) = x ) z * = ivnorm (0.95, 0, 1) = 1. 64, -1.64< < 1. 64, s < x < s, s s < 5 Þ s < is t most s x ) z * = ivnorm (0.9875, 0, 1) = 2. 24, < < 2. 24, < x< , < æ1.12ö >ç» So the miimum ³ è 0.03 ø 15.).) x * = ivnorm (0.1, -1, 1) = ) = AREA ( x* < x<+ H ) = ormlcdf (-2.281,1000, -3,1) = c.) PWR = =

22 A I- 4. THE COUNTING PROBLEMS [MATH] Here re some of the commoly used methods for coutig eqully likely evets. Fctoril of positive iteger is defied s the product of ( - 1) ( - 2) 2 1. It is deoted s!. Tht is! = ( - 1) ( - 2) 2 1 Note tht 0! 1s defied. The four mi methods for coutig re = Multiplictio Priciple ---- Suppose tht coutig of ojects is composed of t cosecutive opertios. If opertio 1 c e couted i those, opertio 2 c e couted i coutig c e performed i m1 m2 mt wys. m 2 m 1 wys d for ech of wys, d so forth. The the complete Additio Priciple ---- If the ojects to e couted re seprted ito cses with m1, m2,, mt s the umers of wys i ech cse, respectively. The totl umer of couts is the sum m1 + m2+ + mt of the vrious cses. Permuttio ---- For the ordered rrgemets of ojects. The formul to rrge r distict ojects less th or equl to the give distict ojects is! Pr = ( - r)! Note tht P =!. The formul cot e used to solve ll permuttio prolems. Comitio ---- For rrgemets of ojects with regrd to o order. The formul to select r distict ojects less th or equl to the give distict ojects is t C r! = r! ( - r)! Note tht C = C0 = 1. The sme s P r i permuttio prolems, the formul cot e used to solve ll comitiol prolems. 161

23 Solve ech prolem 1.) Evlute 4! = 2.) Evlute 9! 4! = 3.) Evlute 6! 4! 5! = 4.) Evlute ( + 2)! =! 5.) Solve ( + 2)! = 6! 6.) A rilwy hs 20 sttios. If the me of the poit of deprture d the destitio re prited o ech ticket, how my differet kids of sigle tickets must e prited? 7.) There re six sets ville i sed. I how my wys c six people e seted if oly three c drive? 162

24 8.) At THS, the school codes for clsses use the letters A, B, C, D, d E. Ech code is formed y usig o more th three of the five letters. How my codes re possile? 9.) Joh is plig to tke 4 differet AP clsses d 3 differet Hoors clsses. I how my wys c she tke the 7 clsses i row such tht t lest two AP clsses re ext to ech other? 10.) I primry electio, there re four cdidtes for myor, five cdidtes for city tresurer, d two cdidtes for coutry ttorey. I how my wys my voters mrk their llots..) if they vote i ll three of the rces?.) if they c exercise their rights ot to vote i y or ll of the rces? 11.) Ech of 5 me dced with ech of 5 wome, d the ech wom dced with ech of the other wome, How my dces were there? 163

25 12.) How my wys c 12 people e prtitioed ito group of four people d group of eight people? 13.) A TV director is schedulig certi sposor s commercils for upcomig rodcst. There re six slots ville for commercils. I how my wys my the director schedule the commercils i ech of the followig cses?.) If the sposor hs six differet commercils, ech to e show oce?.) If the sposor hs three differet commercils, ech to e show twice? c.) If the sposor hs two differet commercils, ech to e show three times? d.) If the sposor hs three differet commercils, the first of which is to e show three times, the secod two times, d the third oce? 164

26 Aswers 1.) 4! = 24 2.) 9! ! = 3.) 6! 6 5! 1 = = 4! 5! 4! 5! 4 4.) ( + 2)! ( + 2)( + 1) =! = ( + 1)( + 2)!! 5.) Solve ( + 2)! = 6 ( + 2)( + 1) = 6 = 1! 6.) 380. There re 20 sttios to strt d 19 sttios to ed. So 20 (19) = ) There re three wys to choose the driver. Ad oce the driver is chose, there re five sets to populr with five people. So, there re 3 5! = 360 wys to set those six people. 8.) The code c e oe, two or three letters. So, there re = 155 possile codes. 9.) There re totl 7!wys to tke the courses while there is o restrictio. The sequeces tht o two AP clsses re i row follow the ptter AHAHAHA. Or there re wys tht he cot tke two Hoors clsses i row. Therefore, there re 7! - 4! 3! = 4896 wys to tke t lest two AP clsses i row. 4! 3! 10.).) There re 4 5 2= 40..) There re 5 6 3= 90. For ech rce, there is choice for No Vote. 11.) There were totl of 5(5) + 5C2 = = 35 dces. 12.) 495. C = 165

27 13.).) 6 P 6 = 6! = 720.) There re to choose the 1 st commercil, to choose the 2 d d the 3 rd commercil. So, the totl umer of wys is 6C2 4C2 2C2 to choose C C C = st 2d 3rd c.) There re of wys is 6C3 to choose the 1 st commercil, 3C3 6C3 3C3 = 20 1st 2d d.) There re to choose the 1 st commercil, 3 rd commercil. So, the totl umer of wys is 6C3 3C2 to choose the 2 d. So, the totl umer to choose the 2 d d 1 1to choose the C C C C = st 2d 3rd 166