GAYAZA HIGH SCHOOL MATHS SEMINAR- APPLIED MATHS SOLUTIONS

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1 PROBABILITY AND STATISTICS. (a) Let X be a r.v number of games won. X~B(6, ) (i) Expectation, E(X) = np 6x = 4 (ii) P(X ) = P(X < ) = (P(X = ) + P(= 0)) = C x x C0x x (b) Let X be a r.v number outside the tolerance limit. Since n >00, then X~N(np, npq) μ = np = 00x0.5 = 0 and σ = npq = 00x0.5x0.85 =.5 (i) P(X > ) = P(X.5) = P (Z >.5 0 ) = P(Z >.79) = (.79).5 = = (ii) P(0 X 0) = P(9.5 < X < 0.5) = P ( = P(.4 < Z < 0.05) = (0.05) + (.4) = = < Z < ) Marks Frequency Mid fx Cumulative fx (f) point(x) frequency(c.f) 5-< < < < < < < < < f=5 fx=59 fx =006 COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page

2 (a) (i) Var ( x) Mode 40 x0 4 (ii) (b) (i) From the graph below, 68 th percentile= x5 th 85 value = 5.5 th value (ii) Number of students who scored above 47% = (5-65) =60 students COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page

3 GAYAZA HIGH SCHOOL MATHS SEMINAR- APPLIED MATHS SOLUTIONS COMPILED BY: THEODE NIYIRINDA. TEL: / Page

4 . (a) (i) P(AuB) = P(A) + P(B) P(AnB); but P(AnB)= P(A)xP(A/B) = x (0.) 0.6 = x y - (0.)x x y 6 8x 0y 4x 5y...() (ii) P(BuC) = P(B) + P(C) 0.9 y ( x y) 0.9 y x 9 0y 0x...( ) Solving equation () and () simultaneously gives; x 0.5, y 0. For independent events P(AnB) = P(A) x P(B) P(AnB) = 0.5x0. = 0. and P(A)xP(B) = 0.5x0. = 0.. Hence A and B are independent events. (b) (i) P(AuB) = P(A) + P(B) P(AnB) P( B). P( B) P( B) 6 P( AnB) P( A) xp( B) P A P( B) B P( A) P( A) (ii) P B P ( B na) P( B ) (iii) A P( A) P ( B) 4 x x d 4. (i) f ( x) dx allx ax ( d x) dx a d a a( 4d )...( i) E ( X ) x. f ( x) dx 0.6 x. ax ( d x) dx allx x x 0.6 a d 0.6 a d 4 5 a(5d 4)...( ii) Divide equation (ii) by equation (i) a(5d 4) 5d 4 4d d From equation (i) a a(4d ) (b) P(0.9 x ) 0.9 x 4 x x ( x) dx COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 4

5 5. (a) (i) P(X = x) X P(X = x) = k X X 4 P(x=x) k k k k But k + K + K + K = k 44 k = (ii) Var(x) = Var(X) = [E(X ) (E(X) )] But E(X) = ( k) + ( k ) + 4 ( k) + (4 k ) = 5k = 60 and k k k E( x ) k 4 4k Hence Var (x) 94x b) P(X = x) = all x K + k+ k+..+nk = K(+++ +n) = K( n (n+)) = K = n(n+).(i) Also E ( x) x. P( X x) allx K + (K) + (K) +...+n(nk) = 5 K( n ) = 5 K( n(n+)(n+) ) = 5 6 Substituting K from equation (i) n(n+) x(n(n+)(n+) 6 ) = 5 4n + = 0, 4n = 8, n = 7 Hence k =, k = 7(8) 8 COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 5

6 ii) Var( x ) Var( x) 4E ( x ) E( x) x P(X=x) x X.P(X=x) X. P( X x) 8 Hence Var ( x) (a) -<x<0; f(-) = k(-+)=k, f(0) = k(0+) = k 0 <x<; f(0)=f() = k (5 ) (5 ) <x<; f ( ) k k ; f ( ) k k f(x) f(x)=k(x+) k f(x)= k f(x)=k (5- x) k - 0 x Area under the graph = ( kx 4) k(4 ) (b) t F ( x) f ( x) dx k COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 6

7 F( t) x 0; t F ( x) 0 x ; ( x ) dx x 4x 4 F t) F(0) dx 0 t x x t F( 0) t t 4x ( F( t) 4t F ( x) 4x F( ) 7 t 0 4t t 4t x ; x F t F 5 ( ) () dx F( t) 7 t x 5x t 7 F ( x) 0x x 5 F() 6 Hence t 5t 5 6 0t t 5 F(x) = x 0 4x 0x x 5 6 { 4x (c) P0.5 x P(0.5 X nx x X = 6 X<- -<X<0 <X< X> 0<X< ) P( x ) F() F() P( x ) F() 0x 5 4x 4x COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 7

8 7. Let X be a r.v weight of the cabbage X~N(50, 6 ) (a) (i) P( 48 X 57.4) P Z P 0. Z. 6 6 (b) P(48 <X <57.4) = (.) + (0.) = = (ii) P ( X 46.) P Z P( Z 0.65) 6 Hence P ( X 46.) 0.5 (0.65) y 50 P P ( Z Z) 0. where Z y 50 6Z 6 (c) X y 0. COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 8

9 From critical table, Z =.8 Hence y = 50 +6(.8) = g 8. (a) Let X be a r.v number of students with blue eyes. X~B (n, 5 ) P ( X 0) n 0 n 0 C x n 4 x log 5 n log(0.00) n Hence the least value is 0 log(0.00) n 4 log 5 (b) Let events A and B be John wins the first and second game respectively. P ( A) 0., P ( A ) 0. 7, P B 0. 6, A P B 0. 5 A A P( AnB) B P( B) P( AnB) P( A nb) P( A) xp B P( A ) xp B P( B) A A P but = 0.x x0.5 = x0.6 Hence P A B 0.85 (d) Let R and A be events Robot and aeroplane respectively; Dan 0 5 P ( RD ), P( AD ) 5 5 Emma 7 7 x P( RE ), P( AE ) x 7 x 4 x 4 7 P( RD xre ) 8 Hence 7 7 x ( x 4) x 8 x 4 8 COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 9

10 9. MECHANICS Weight α Area W=gA, where g is a constant Portion Area Weight Position of c.o.g from OA and OE Rectangle, OABE (4x6) =4cm 4g ( 7, 8) Triangle, CBD x x9 = 49.5cm 49.5g 0, Remainder (4 49.5) = 74.5g ( x, y ) 74.5cm Taking moments about O, x 7 x g 4g 49.5g y 8 y x y 8.78 Position of c.o.g is (5.980, 8.78) (b) COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 0

11 8.78 Tanθ= Θ= Tanα= 4 =48.8 Hence OB makes an angle of ( ) = 8.9 with the vertical 0. (a) Impulse = Change in momentum 6 4 F.t = m(v-u) x.5 v 9 8 v v Hence final velocity, v 0.5i 4.5 j. 5k (b) Let the body have mass, mkg, natural length, l and displaced by x equilibrium T l 0.5m x In equilibrium; e T = mg, but T l (0.5) mg... ( i) l mg COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page

12 (0.5 When displaced by x; restoring force is; mg T ma where T l ( 0.5) (0.5 x) x ma ma a x..(ii) l l l ml Compare eqn(ii) by a x mgl From equation (i) 0.5 g=0ms - Hence T 5 seconds 5. (a) Let the acceleration of the car be a ms - But T ml ml x) ml(0.5) T ; mgl Sinα= 0 Along the plane; 900 (400000g sin) 000a x0x a 0.5ms 0 000a 08x000 U 0ms V = 0 ms - 60x60 Using V = U +as 0 = 0 +(-0.5)S Hence distance travelled is S = 900m COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page

13 (b) Work done= F. AB but F = ( 5 4 ) + ( 0 7 ) + ( 0 ) = ( 4 5 ) N Distance, AB = ( 4) ( ) = ( 6) m Hence Work done =( ). ( 6) = (4x4 + x6 + 5x) = 8 Joules 5. Let the body make an angle, θ Distance = xr 60 (r ) xr Using V U as r V rg ( g)( h) But h r( cos60) r V rg ( g)( ) rg Hence V rgms (0.) rg Resolving along the radius, R 0x9.8( cos60) 49N R 0.gCos60 r COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page

14 . (a) Resolving along the plane; FCos 0 gsin FCos 0.5R 0 but R FSin0 gcos0 FSin0 gcos gsin x9.8(0.5cos0 Sin0) F N Cos0 0.5Sin0 0 5 Cos (b) (i) Resultant force, F Sin Taking moment about point O, G = 8x 4x = 8Nm Let the line of action of the resultant force act at a point (x, y) x y Momemnt of the resultant is G 8x 7 y 7 8 But G = G ; Hence 8x 7y = 8 (ii) The line crosses C at a point when x = 0. -7y = 8 8 y Hence it crosses at a 7 point 8 cm below point C 7 COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 4

15 4. Whole system Resolving vertically in equilibrium, R R W 5W 6W...( i) A C Taking moments about point A R C 4R R x4bcos45 C W 5W 6W C 4W WxbCos45 From equation (i) R A 6W 4W W Splitting the joint at B 5WxbCos45 COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 5

16 Considering rod BC; Taking moments at the hinge B; 4wxbCos45 8W T tan 45 TxbSin45 5W W T Resolving on rod BC; 5WxbCos 45 Vertically; Y+4W = 5W Y W Horizontally: X T W W W W Hence reaction is R X Y W 4 Let the direction of the reaction act at θ to the horizontal, 5. Y W Tan. 7 X W a m/s - T T T am/s - g 5g (i) Using F = ma For 5Kg mass; 5g-T=5a () For kg mass; T-g=a T g = a...() COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 6

17 Adding () and () 4g = 6a a g x ms (ii) From equation () T = 5x9.8 5x6.5 = 6.5N s ut at s 0 x.5 x6.5x(.5) 7. 50N (iii) 6. Usinθ U y θ z X Horizontal displacement; X = (Ucosθ)t Ucosθ t x U cos..() Vertical displacement; y U sin. t gt () Put t from. () in () x x y U sin. g U cos U cos Hence y = xtanθ gx u ( + tan θ) gx gx Sec y x tan y x tan U Cos U COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 7

18 (b) U = 0 g y 40m Ɵ 40m x For the particle to clear the tower; y x40 ( tan 40tan x tan 8( tan ) 40 40tan 8tan 48 0 tan 5tan 6 0 tan tan 0 ) 40 tan 0 or tan 0 tan 0r tan Hence tan NUMERICAL ANALYSIS 7. (a) x Sinx ln x From graph x. 5 root COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 8

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20 (b) let f ( x) Sinx ln x x ( n) x n f ( Sinxn ln x Cosxn xn x) Cosx x n Using x. 5 x x o Sin(.5) ln(.5).5.94 Cos(.5).5 Sin(.94) ln(.94).94.9 Cos(.94).94 x x Hence x. 9 root 8. (a) From the graph on the next page, there is a positive relationship between mock and final mark (b) Mock Final R m R F d d = 9.5 Spearman s Correlation coefficient, ρ = 6x9.5 7(7 ) = Comment: There is a high positive correlation between the Mock and Final mark COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 0

21 GAYAZA HIGH SCHOOL MATHS SEMINAR- APPLIED MATHS SOLUTIONS COMPILED BY: THEODE NIYIRINDA. TEL: / Page

22 9. True value= X Y; X=x± x and Y=y± y (X Y) max =X maxy max =(x+ x) (y+ y) =(x +x x+ x )(y+ y) = (x y+ xxy+ x y+ yx + x yx + x y) (X Y) min = X miny min = (x x) (y y) = (x x x + x )(y y) Error= = (x y xxy + x y yx + x yx x y) (Max limit min limit) = x y+ xxy+ x y+ yx + x yx+ x y x y+ xxy x y+ yx x yx+ x y = 4 xxy + yx + x y Error = xy x + x y + x y For maximum error; if x and y are very small then, x y 0 Error= xy x + x y Error = xy x + x y xy x + x y Hence Max error = xy x + x y Max limit= x max y max = ( ) ( ) =.484 Min limit=x min y min = ( ) ( ) =.0989 Limits are [.0989,.484] or (.0989 x y.484) 0. (a) COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page

23 A = pqsinα da dq = dp dα (psinα + q sin α + pqcosα ) dq dq Multiplying through by dq da = (p sin α dq + q sinα dp + pqcosα dα) da = (psinα dq + qsinα dp + pq cosα dα) da { psinα dq + qsinα dp + pqcosαdα } Maximum possible error = { psinα dq + qsinα dp + pqcosα dα } (ii) dp = 0.05 dq = 0.05 dα = 0.5 Error = π {4.5x sin 0x x sin 0x x8.4 cos 0x0.5x } 80 = Exact Area = sin 0 = 9.45 sq units %age error = = 6.4 (b) e (.679) = e (7.0) = 0.05 e () = 0.5 e (5.48) = Let.679 P P max = =.859 P min = 7.05 = The range is from 0.87 P.859 or 0.87,.859 COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page

24 . (a) h 0. 4 Let y x ln x 6 x, y5 y,..., y Sum y o 4 x ln xdx x x (b) Exact value; x x ln xdx ln x x. dx x x x x ln xdx ln x 9 x ln x x () ln () ln Hence x ln xdx Error = Therefore; % error = 0.05 x00 = Error can be reduced by increasing the number of sub-intervals or number of ordinates COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 4

25 . (i) START Read: a, b, c d = b 4ac Is d o? X = X = b i d a b i d a X = b d a X = b+ d a Print :X, X STOP (ii) d x x -6 i i For any inquiries feel free to drop me a line on contacts below COMPILED BY: THEODE NIYIRINDA. TEL: / niyirinda0@yahoo.com Page 5