Innova Junior College H2 Mathematics JC2 Preliminary Examinations Paper 2 Solutions 0 (*)

Transcription

1 Soluion 3 x 4x3 x 3 x 0 4x3 x 4x3 x 4x3 x 4x3 x x 3x 3 4x3 x Innova Junior College H Mahemaics JC Preliminary Examinaions Paper Soluions 3x 3 4x 3x 0 4x 3 4x 3 0 (*) Hence x or x or x 4 For x 3 e, making use of he resul in above par, x x 4e 3 e x 3 3 x 3 < e or < e 4 (no solns since e x is always posiive) Hence, e x 3 3 x ln

2 Soluion (i) Hour Sar of hour End of hour + 3 = = = 64 4 n n n n Number of complee hours = 9 Alernaive Soluion n n Toal < > > Number of complee hours = 9 (ii) Day Sar of day End of day x x.0[ x] x x.0x.0 x 3 x A he end of day n, he number commens n x x x n (*)

3 n.0.0 n x 0.0 n n x.0 (iii) 30 x x x 8755 (o neares ineger) Day : no. of commens removed 5000 Day : no. of commens removed Day 3: no. of commens removed As n, no. of commens removed Sofware Y is unable o remove all he commens because evenually i is only able o remove commens.

4 3 Soluion (i) Mehod : d x d x(5 x) 0 (*) Doing parial fracions A B x(5 x) x 5 x A(5 x) B( x) x(5 x) A B d P d 5x 5(5 x) 0 ln ln 5 5 x x c 0 x ln c 5 x x Ae, where A e 5 x c Given, x when 0 x(4 e ) 5e 5 x e xe 4 4 5e x 4 e 0 Ae A 5 4

5 (i) Mehod : d x d (*) x(5 x) 0 (ii) 5 d x d ( ) x x 5 5 x ( ) ln c ( ) 0 Given x ln c 5 x x when x(4 e ) 5e When x =, x c Ae, where Ae 5 x 0 Ae A 5 4 0, 5 x e xe 4 4 5e 5 x or x 4 e 4e 5e 4 e 8 e 5e 3e 8 e ln 3 8 I akes ln 3 years. (iii) As, x 5. The populaion of wild boars will increase and sabilise a 500 evenually.

6 4 Soluion (i) l : r 0, where is a real parameer. 7 4 p : r sin ( dec pl) (ii) For he poin of inersecion beween l and p, The posiion vecor of poin of inersecion is Coordinaes of poin of inersecion are (,, 3). (iii) The line perpendicular o p passing hrough (,0, 7) is r 0 0,

7 6 3 3 ON (iv) Mehod : Le he coordinaes of A be (,0, 7). Le A be he refleced poin of A in p. Using raio heorem, OA OA' ON 5 OA' ON OA The refleced line conains he poin A and poin of inersecion beween l and p. The direcion vecor of he refleced line is 4 r, (iv) Mehod : Le he coordinaes of A be (,0, 7).Le A be he refleced poin of A in p. Le he coordinaes of B be (,, 3). BA BA' BN BA' BN BA r, 3

8 5 Soluion (i) Sysemaic sampling (ii) (slower, more difficul o collec) Sysemaic sampling is a more edious process o selec he employees, whereas quoa sampling is quick and easy. (iii) Anoher possible reason: migh miss ou a cerain group of people due o differen reporing imes. The inerviewer could consider ranspor mode of he employees as he sraum. A possible quoa for each sraum is as follows: By privae By public By walking ranspor ranspor The inerviewer can hen sand a he enrance of he building and selec he sample unil he above quoa is me. 6 Soluion X X E.93 Var.4 Since n is large, by Cenral Limi Theorem,.4 X ~ N.93, n approximaely. Given ha P X (*) Mehod : Using GC o se up able when 4 P X ( 0.4) n, n, n, when 43 when 44 P X ( 0.4) P X ( 0.4) leas n is 43. Mehod : Using algebraic mehod via sandardizaion P X P Z n

9 From GC, (**).4 n leas n is 43. n n.939 n 4.53

10 7 Soluion (i) Mehod : Required probabiliy 3 C4 C4 4 4 C4 C (3 sig fig) Mehod : Required probabiliy (3 sig fig) Mehod 3: Required probabiliy C C3 C C C3 C 4 C (3 sig fig) (ii) Mehod : Required probabiliy C C 4 C4 4!!! 0.99 (3 sig fig) Mehod : Required probabiliy (3 sig fig) 4 3

11 8 Soluion (i) Unbiased esimae of he populaion mean x (3 s.f.) 3 Unbiased esimae of he populaion variance 4. s (3 s.f.) (ii) The baery life of a PI-99 calculaor is assumed o be normally disribued. (iii) Le X be he r.v. denoing he baery life of a randomly chosen PI-99 calculaor. Le be he populaion mean baery life of he PI-99 calculaors. (iv) H H 0 : : where k k H 0 is he null hypohesis and To es a 5% level of significance. Under H 0 H X k, he es saisic is T ~ S. is he alernaive hypohesis. 3 Since he null hypohesis is no rejeced, -value falls ouside criical region..788 value x k s (*) 3 s s x.788 k x where x and s k The required se is k : 4.97 k 45.4

12 9 Soluion (i) The average number of fauls deeced by each sysem (for he rack and he rain) is consan from one day o anoher. (ii) Le X be he r.v. denoing he oal number of fauls deeced by he wo sysems in a periods of 0 days. (iii) X ~ Po ( ) 0, i.e. X ~ Po(4) P X (3 sig fig) Le Y be he r.v. denoing he oal number of fauls deeced by he wo sysems in a period of n days. Y ~ Po(0.4 n ) Given PY , Mehod : Algebraic mehod 0.4n 0.5n 0.5n (o.e. e 0.05 n e e 0.05 ) he smalles number of days required is 8. Mehod : GC able When 7 When 8 When 9 P Y ( 0.05) n, n, n, P Y ( 0.05) P Y ( 0.05) he smalles number of days required is 8. (iv) Le W and V be he r.v. denoing he number of fauls deeced on he rack and on he rack in a period of 0 days respecively. W ~ Po(.5) and V ~ Po(.5) Required probabiliy P W 3 V W 4 P W 3V W 4 P V W 4 P( W 3)P( V 0) P( W 3)P( V )+P( W 4)P( V 0) P V W 4 P( W 3)P( V ) P( W 4)P( V 0) P V W (3 sig fig)

13 0 Soluion (i) Ben s performance (i.e. wheher he loses or wins) in a game is independen of any oher games ha he plays wih Alex. (ii) Le X be he r.v. denoing he number of games ha Ben loses ou of 0 games. (iii) (iv) X ~ B 0,0.7 P( X 5) P( X 5) (3 sig fig) Le Y be he r.v. denoing he number of games ha Ben loses ou of n games. Y ~ B n,0.3 P( Y 8) 0.0 P( Y 8) 0.0 Using GC, When When When n 3, P( 8) ( 0.0) n 4 Y, P( Y 8) ( 0.0) n 5, P( 8) ( 0.0) Y he greaes value of n is 4. Le W be he r.v. denoing he number of games ha Ben loses ou of 50 games. W ~ B 50,0.3 As n = 50 is large, np 5 ( 5) and nq 35 ( 5), W ~ N5,0.5 approximaely P(0 W 0) P(9.5 W 0.5) (*) 0.90 (3 sig fig)

14 Soluion (i) m From he scaer diagram, a curvinlinear correlaion is observed beween m and (i.e. as increases, m decreases a a decreasing rae), and hence a linear model wih equaion of he form m a b canno be used o model he relaionship beween m and. (ii) Produc momen correlaion coefficien beween m and. (a) (b) Produc momen correlaion coefficien beween m and ln (iii) Since he absolue value of he correlaion coefficien beween m and (i.e. case (b)) is closer o, his indicaes ha he linear correlaion beween he variables m and is sronger as compared o ha beween he variables for case (a). ln case (b) is he beer model for he relaionship beween m and. m ln m79 3.ln (3 sig fig) (iv) When 50, m ln50.6 ( dec pl) The esimae obained is reliable, because he given value of = 50 lies wihin he given sample daa range for and he produc momen correlaion coefficien beween m and ln is very close o, hence indicaing a srong negaive linear correlaion beween he variables m and ln. ln

15 Soluion (i) X Y ~ N50, 3.65 X Y P (3 sig fig) (ii) E( X Y S) Var( X Y S) 39.4 X Y S ~ N, 39.4 P(mehod A is faser han mehod B by more han 5 mins) P S 5 ( X Y) 5 P( X Y S 0) (3 sig fig) (iii) Le A X Y and BS5 A ~ N50, 3.65 and B ~ N53,.6 A A A A B... B Le W E( W ) & Var( W ).76 0 W ~ N 5.8,.76 Required probabiliy P( W 50) (3 sig fig).