GAYAZA HIGH SCHOOL MATHS SEMINAR- APPLIED MATHS SOLUTIONS

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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)) 5 0 6 6 = C x x C0x x 0. 98 (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) = 0.5 + (.79).5 = 0.5 + 0.465 = 0. 965 6. (ii) P(0 X 0) = P(9.5 < X < 0.5) = P ( 9.5 0 = P(.4 < Z < 0.05) = (0.05) + (.4) = 0.54 + 0.4865 = 0.8407.5 < Z < 0.5 0.5 ) Marks Frequency Mid fx Cumulative fx (f) point(x) frequency(c.f) 5-<9 6 7 6 6 474 9-<5 84 8 88 5-<40 7 7.5 0.5 45 7968.75 40-<50 0 45 50 75 60750 50-<55 8 5.5 945 9 496.5 55-<60 4 57.5 805 07 4687.5 60-<70 9 65 585 6 805 70-<75 4 7.5 90 0 05 75-<80 5 77.5 87.5 5 00.5 f=5 fx=59 fx =006 COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

006 59 (a) (i) Var ( x) 59. 686 5 5 Mode 40 x0 4 (ii) (b) (i) From the graph below, 68 th percentile= 68 00 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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

GAYAZA HIGH SCHOOL MATHS SEMINAR- APPLIED MATHS SOLUTIONS COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

. (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 0.6 0.8x 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 0 4 4 0 a( 4d )...( i) E ( X ) x. f ( x) dx 0.6 x. ax ( d x) dx allx 0 4 5 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 4 0.9 4 0.9 4 5 0 0.9 4 4 0.05 COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 4

5. (a) (i) P(X = x) X P(X = x) = k X X 4 P(x=x) k k k k 4 9 6 But k + K + K + K = 05 4 9 6 k 44 k = 44 05 (ii) Var(x) = Var(X) = [E(X ) (E(X) )] But E(X) = ( k) + ( k ) + 4 ( k) + (4 k ) = 5k = 60 and 9 6 4 k k k E( x ) k 4 4k 4 9 6 44 60 Hence Var (x) 94x 6. 04 05 4 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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 5

ii) Var( x ) Var( x) 4E ( x ) E( x) x 4 5 6 7 P(X=x) 4 5 6 7 x 8 8 4 8 9 8 6 8 5 8 6 8 49 X.P(X=x) 8 7 8 7 7 6 8 5 7 54 4 49 X. P( X x) 8 Hence Var ( x) 48 5 6. (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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 6

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 7 0.58 COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 7

7. Let X be a r.v weight of the cabbage X~N(50, 6 ) (a) (i) 48 50 57.4 50 P( 48 X 57.4) P Z P 0. Z. 6 6 (b) P(48 <X <57.4) = (.) + (0.) = 0.9 + 0.04 = 0.57 46. 50 (ii) P ( X 46.) P Z P( Z 0.65) 6 Hence P ( X 46.) 0.5 (0.65) 0.5 0.4 0. 74 y 50 P P ( Z Z) 0. where Z y 50 6Z 6 (c) X y 0. COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 8

From critical table, Z =.8 Hence y = 50 +6(.8) =57.69 57.7g 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 5 4 5 n 4 x 5 0.00 0.00 4 log 5 n log(0.00) n 0.95655 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.x0.6 + 0.7x0.5 = 0.85 0.x0.6 Hence P A 0. 66 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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 9

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 568 64.5 94.5 74.5g 4g 49.5g 0 74.5 y 8 y 79 0 46 x 5.980 y 8.78 Position of c.o.g is (5.980, 8.78) (b) COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 0

8.78 Tanθ= 5.980 Θ=57.7 6 Tanα= 4 =48.8 Hence OB makes an angle of (57.7 48.8 ) = 8.9 with the vertical 0. (a) Impulse = Change in momentum 6 4 F.t = m(v-u) x.5 v 9 8 v 6.5 4 0.5 v 4.5.5 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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

(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) 0.5 5 5 T ; mgl 0 00 0 Sinα= 0 Along the plane; 900 (400000g sin) 000a 500000x0x 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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

(b) Work done= F. AB but F = ( 5 4 ) + ( 0 7 ) + ( 0 ) = ( 4 5 ) N 8 4 4 Distance, AB = ( 4) ( ) = ( 6) m 5 4 4 Hence Work done =( ). ( 6) = (4x4 + x6 + 5x) = 8 Joules 5. Let the body make an angle, θ Distance = xr 60 (r ) xr 6 60 60 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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

. (a) Resolving along the plane; FCos 0 gsin FCos 0.5R 0 but R FSin0 gcos0 FSin0 gcos0 0 0 0.5 gsin x9.8(0.5cos0 Sin0) F 4. 707N Cos0 0.5Sin0 0 5 Cos45 4 0 7 (b) (i) Resultant force, F 0 8 5 Sin45 0 4 9 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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 4

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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 5

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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 6

Adding () and () 4g = 6a a g x9.8 6.5ms (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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 7

(b) U = 0 g y 40m Ɵ 40m x For the particle to clear the tower; y 40 9.8x40 ( tan 40tan x 0 9.8 40tan 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 0.5.0.5.0.5.0 Sinx ln x.76 0.844 0.590 0.6-0.78-0.9575 From graph x. 5 root COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 8

COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 9

(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.9.94 0.000 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 8 54 7 7 0 4 66 6 5.5 0.5 6 68 5 4 4 70 4 5 76 4 54 66 5.5.5 60 74 d = 9.5 Spearman s Correlation coefficient, ρ = 6x9.5 7(7 ) = 0.658 Comment: There is a high positive correlation between the Mock and Final mark COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 0

GAYAZA HIGH SCHOOL MATHS SEMINAR- APPLIED MATHS SOLUTIONS COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

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 = (.8 + 0.06) (.44 + 0.008) =.484 Min limit=x min y min = (.8 0.06) (.44 0.008) =.0989 Limits are [.0989,.484] or (.0989 x y.484) 0. (a) COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

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 0x0.05 + 8.4x sin 0x0.05 + 4.5x8.4 cos 0x0.5x } 80 = 0.608 Exact Area = 4.5 8.4 8 sin 0 = 9.45 sq units %age error = 0.608 9.45 00 = 6.4 (b) e (.679) = 0.0005 e (7.0) = 0.05 e () = 0.5 e (5.48) = 0.005 Let.679 P 7.0 5.48 P max =.6795 6.95.6785 =.859 P min = 7.05 = 0.87.5 5.485.5 5.475 The range is from 0.87 P.859 or 0.87,.859 COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page

. (a) h 0. 4 Let y x ln x 6 x, y5 y,..., y.0 0.0000.4 0.65.8.9044..86.6 6.459.0 9.8875 Sum 9.8875.56 y o 4 x ln xdx x.0.4 9.8875 x.56 6. 984 (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 6.8875 0. 6. 999 9 9 9 Error = 6.999 6.984 0. 05 Therefore; % error = 0.05 x00 = 0.4 6.999 Error can be reduced by increasing the number of sub-intervals or number of ordinates COMPILED BY: THEODE NIYIRINDA. TEL: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 4

. (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: 0776-8648/070-0048 Email: niyirinda0@yahoo.com Page 5

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