296 Mathematics Semester 3 STPM Answers 5. (a) m = 72 (b) 1 3 (c) 8 9 6. s 2 3 4 5 6 7 8 P(S = s) 1 16 2 16 3 16 4 16 3 16 2 16 1 16 7. x 1 2 3 P(X = x) 5 7 5 21 1 21 8. x 2 3 4 , 3 5 P(X = x) 1 5 2 5 2 5 9. (a) s 1 2 3 P(S = s) 2 3 11 36 1 36 (b) P(S=s) s 1 36— 11 36— 24 36— 1 2 3 11. (a) 1 4 (b) 189 1 024 12. x 0 1 2 3 P(X = x) 27 125 54 125 36 125 8 125 13. Sample space = {WWW, WWL, WLW, LWW, WWD, WDW, DWW, DDD, DDL, DLD, LDD, DDW, DWD, WDD, LLL, LLW, LWL, WLL, LLD, LDL, DLL, WLD, WDL, LWD, LDW, DLW, DWL} w 0 1 2 1 3 2 2 5 2 3 P(W = w) 1 27 3 27 6 27 7 27 6 27 3 27 1 27 14. Sample space = {(+ + +), (+ + –), (+ – +), (– + +), (+ – –), (– + –), (– – +), (– – –)} 15. x 1 2 3 4 , 8 9 P(X = x) 2 3 2 9 2 27 1 27 Exercise 3.2 1. (a) x 0 1 2 F(x) 1 4 3 4 1 (b) x 0 1 2 3 F(x) 125 216 200 216 215 216 1 2. x 4 5 6 7 8 F(x) 0.1 0.2 0.5 0.9 1 3. X = x 5 6 7 8 9 P(X = x) 0.05 0.11 0.26 0.34 0.24 4. (a) 0.35 (b) 0.50 (c) 0.28 5. (a) 1 16 (b) 9 16 (c) X = x 2 3 4 P(X = x) 4 16 5 16 7 16 Exercise 3.3 1. 2 1 3 2. (a) 0.2 (b) 16.4 3. –RM 5 6 4. 4 5 5. s 2 3 4 5 6 7 8 9 10 P(S) 1 25 2 25 3 25 4 25 5 25 4 25 3 25 2 25 1 25 6. x P(X = x) x – 3 (x – 3)2 (x – 3)2 P(X = x) 1 0.1 –2 4 0.4 2 0.2 –1 1 0.2 3 0.3 0 0 0 4 0.4 1 1 0.4 Var(X ) = 1 7. (a) 3.3 (b) 13 (c) 2.11 8. (a) 5 1 4 (b) 27 7 16 9. 1, 1 2 10. 0.75 11. 2 12. x 1 2 3 4 P(X = x) 2 5 3 10 1 5 1 10 Var(X ) = 1 Exercise 3.4 1. f(x) is a p.d.f. 2. (a) 2 2 15 44— 3 44— g(y) y (c) (i) 2 11 (ii) 7 11
297 Mathematics Semester 3 STPM Answers 3. (a) 9 16 (b) 27 64 4. 0 1 2 3 f(x) x —2 3 (a) 1 6 (b) 23 24 (c) 13 24 5. m = 2 15 1 4 f(x) x —8 15 —2 15 (a) 1 5 (b) 1 5 6. 1 8 (a) 19 24 (b) 1 6 7. c = 1, k = 4 8. a = 1 18 , b = 1 17 (a) 29 68 (b) 3 9. (a) 5 (b) 1.11, 3.92 10. (a) m = 7, median X = 7 (b) 2, 11 Exercise 3.5 1. (a) 0, x 1, F(x) = x 10 – 1 10 , 1 x 11, 1, x 11. 1 0 1 11 F(x) x (b) 3 5 2. (a) 0, y 0, y2 20 , 0 y 4, G(y) = 8 15 + y 15 , 4 y 7, 1, y 7. —4 5 4 1 0 7 G(y) y (b) 2 3 3. (b) 0, x 0, x 2 , 0 x 1 2 , F(x) = 6x – 3x 2 – 2, 1 2 x 1, 1, x 1. (c) 1 3 4. 0, x 0, x3 3 , 0 x 1, F(x) = 5x 12 – x2 24 – 1 24 , 1 x 5, 1, x 5. (a) 1 24 (b) 7 12 5. (a) 2 3 (b) 0, x 0, x2 3 , 0 x 1, F(x) = x – x2 6 – 1 2 , 1 x 3, 1, x 3. (c) 13 24 6. 0, y 3, y 10 – 3 10 , 3 y 7, G(y) = 2 5 , 7 y 10, y 5 – 8 5 , 10 y 13, 1, y 13. 1 10 7. 0, w 0, w2 12 , 0 w 2, H(w) = 1 3 , 2 w 4, w2 30 – 1 5 , 4 w 6, 1, w 6. 7 10
298 Mathematics Semester 3 STPM Answers 8. (a) 0, x 0, F(x) = x 3 , 0 x 1, 1, x 1. (b) 0.794 9. 0, x 0, F(x) = 1 – e–x , x 0. Median X = ln 2 10. (a) 0, x 1, F(x) = x2 24 – 1 24 , 1 x 5, 1, x 5. (b) 2.65, 4.36, 1.71 11. k = 1 3 (a) 0, x 0, 1 6 x 2 , 0 x 2, F(x) = 1 3 x, 2 x 3, 1, x 3. (b) Median X = 3 Exercise 3.6 1. 1 8 , 0 x 2, f(x) = 1 4 , 2 x 5, 0, otherwise. 1 4 , 3 8 2. (a) 2(x + 2) 9 , –2 x 1, f(x) = 0, otherwise. (b) 4 9 (c) –1 3. k = 1 5 , h = 1 15 1 10 x , 0 x 4 f(x) = 2 15 (x – 4), 4 x 7, 0 , other values of x. 4. (a) a = 1, b = – 1 27 (b) 0, x 0, F(x) = x3 27 , 0 x 3, 1, x 3. (c) x2 9 , 0 x 3, f(x) = 0, other values of x. 5. (b) 2x + 5 2 2 , 1 2 2 x 1 2 , f(x) = 0, otherwise. (c) 1 8 (12 – 5 2) 6. 2(x – 1), 1 x 2, f(x) = 0, other values of x. (a) 1 4 (b) 3 16 (c) 0.051 7. 1 16 x , 0 x 4, f(x) = 1 4 , 4 x 7, 0, otherwise. 8. (a) x – 1 6 , 1 x 3, f(x) = 7 – x 12 , 3 x 7, 0, otherwise. —1 3 10 3 7 f(x) x (b) 0.595 Exercise 3.7 1. (a) 6 (b) 5 1 3 (c) 133 1 3 (d) 133 1 3 2. (a) 4 (b) 19 (c) 3 (d) 3 16 (e) 3 3. 0, 1 2 4. a = 2 3 , b = 1 3 ; 7 18 5. a = 6 —1 a a 2a f(x) x E(X) = 3 2 a 6. a = 1 100 ; 10, 16 2 3 7. (a) 2x, 0 x 1, f(x) = 0, otherwise. (b) 2 3 (c) 1 18 (d) 2
299 Mathematics Semester 3 STPM Answers 8. (a) 1 3 (b) 1 3 , –1 x 0, f(x) = 2 3 , 0 x 1, 0, x > 1. (c) 1 6 (d) 11 6 (e) 11 18 9. (a) 3 20 (c) 27 44 10. 17 81 ; 2 1 9 , 17 81 11. (a) f(x) 0.2 0.4 0.6 0.8 1 0.1 1.6 0 x Mean X = 0.5 (b) 0.0473; E(Y ) = 2, Var(Y) = 0.1892 12. (a) 1, 2 (b) a = 1 2 , b = – 1 2 13. (a) –1 (b) 5 1 3 (c) 3 4 14. (a) 4.5, 4 1 12 (b) 3 7 (c) 5 7 1 8 0 —1 7 f(x) x 15. (a) 8 (b) 5 (c) 3 (d) 0.4 16. (a) –1.5 (b) 4 7 (c) 2 7 (d) 0.25 17. (a) 0, π 3 (b) 1 3 18. (b) (i) c = –1, d = 3 (ii) 1 4 (iii) 2 3 (3 – 3) 19. (a) 0.0907 (b) 0.3691 (c) 1 4 (d) 0.6321 (e) 1 4 ln 2 20. (a) 5 000 hours (b) (i) 0.7866 (ii) 0.1813 (c) 0.4651 (d) 0.0601 21. 1.65 Exercise 3.8 1. (a) 5 16 (b) 3 16 (c) 1 2 2. (a) 3 125 15 552 (b) 31 031 46 656 (c) 21 875 23 328 3. (a) 1 4 (b) 5 16 (c) 1 16 4. (a) 1 792 6 561 (b) 5 984 6 561 5. (a) 0.0369 (b) 0.0467 (c) 0.0410 6. 15 16 7. 1 2 8. 0.1852 9. (a) 0.0256 (b) 0.1808 (c) 0.8192 10. (a) 0.2150 (b) 0.0123 11. (a) 0.1678 (b) 0.00008448 12. (a) 0.0272 (b) 0.7021 13. (a) 10 243 (b) 64 81 14. (a) 0.0015 (b) 0.9769 15. 14 16. 6 Exercise 3.9 1. 10, 2.887 2. 8, 6.4 3. 8, 1.386 4. (a) 1 5 (b) 2 (c) 0.0264 5. 10, 5 6. 3, 1.643 7. 0.3, 0.297 8. (a) 0.8 (b) 0.0055 Exercise 3.10 1. 2 2. 8 3. 5, 0.2099 4. (a) 2 (b) 0.9936 5. (a) 3 (b) 3 (c) 0.3823 Exercise 3.11 1. (a) 0.2231 (b) 0.5578 (c) 0.0612 (d) 0.1913 (e) 1.5, 1 2 6 2. (a) 1.2 (b) 1.2 (c) 0.8795 (d) 0.0077 3. (a) 0.6703 (b) 0.1429 4. (a) 0.0608 (b) 0.4695 (c) 0.6578 (d) 0.1584 5. (a) 0.0959 (b) 0.9662 6. (a) 0.00203 (b) 0.00216 (c) 0.2212
300 Mathematics Semester 3 STPM Answers 7. (a) 0.6703 (b) 0.0175 (c) 0.9525 8. (a) 0.1353 (b) 0.9817 9. (a) 1.5 (b) 0.9344 (c) 0.6472 10. (a) 0.0183 (b) 0.0286 11. 0.0527 Exercise 3.12 1. (a) 1 (b) 2 (c) 4 (d) 8 2. (a) 0.8187 (b) 0.0012 (c) 0.0616; 10 3. 3 4. k = 1, 0.3347 5. (a) 0.1108 (b) 2 breakdowns 6. (a) 3 people (b) 0.2149 ; 282 Exercise 3.13 1. (a) 0.1335 (b) 0.99305 (c) 0.1359 (d) 0.1335 (e) 0.99305 (f) 0.1359 (g) 0.6738 (h) 0.0852 (i) 0.4246 (j) 0.4920 2. (a) 0.3026 (b) 0.1041 (c) 0.4236 (d) 0.1000 (e) 0.9524 3. (a) 0.304 (b) – 0.777 (c) –1.417 (d) 2.3 (e) 1.281 (f) 2.326 (g) 1.01 (h) 0.061 4. (a) 0.7734 (b) 0.3830 (c) 0.4079 (d) 0.2660 5. (a) 0.0668 (b) 0.2743 (c) 0.0375 (d) 0.4206 (e) 0.0397 (f) 0.1114 6. (a) 41.35 (b) 40.68 (c) 34.18 (d) 43.29 7. (a) 10.25 (b) 16.43 (c) 14.01 (d) 17.92 Exercise 3.14 1. (a) 0.2660 (b) 0.3307 (c) 0.3377 (d) 0.2236 2. (a) 0.0228 (b) 0.0228 (c) 0.2106 (d) 0.8185 3. (a) 0.1754 (b) 0.0334 4. (a) 0.1360 (b) 0.9738 5. (a) 0.00598 (b) 0.3111 6. (a) 0.8849 (b) 177 7. 75 8. 29 Exercise 3.15 1. 4 2. 8.86 3. 30 4. 52.73, 11.96 5. 100.8, 5.71 6. 51.56, 4.110 7. 23.75 8. 53.87, 16.48 Exercise 3.16 1. (a) 0.0713 (b) 0.4693 (c) 0.1979 (d) 0.2849 (e) 0.1526 (f) 0.0187 2. (a) 0.9984 (b) 0.0352 (c) 0.1213 3. (a) 0.0259 (b) 0.2714 (c) 0.0297 4. 250, 93.75 (a) 0.000816 (b) 0.8357 5. (a) 0.0588 (b) 0.9412 (c) 0.3637 (d) 0.00184 6. (a) 0.0097 (b) 0.00218 (c) 0.6499 7. (a) 0.9967 (b) 0.1345 (c) 0.6344 8. 1 6 , 0.0866, 11 9. (a) 0.00298 (b) 0.9719; 3 10. 0.758 STPM Practice 3 1. (a) x 1 2 3 4 5 6 8 10 12 P(X = x) 10 36 9 36 6 36 5 36 2 36 1 36 1 36 1 36 1 36 (b) 28 9 , 1 015 162 2. (a) 1 4 (b) 1 4 , 1 4 11 3. 1 2 (n + 1), 0.42 4. k = 1 12 , E(X ) = 5 6 , Var(X) = 29 36 5. Var(X ) = 2, E(|X|) = 1 Var(Y ) = 4, E(Y + 3) = 3 6. x 1 2 3 P(X = x) 1 5 3 5 1 5 E(X ) = 2, Var(X) = 2 5 ; E(Y ) = 1, Var(Y) = 2 5 7. x 1 2 3 4 6 8 9 12 16 P(X = x) 1 16 2 16 2 16 3 16 2 16 2 16 1 16 2 16 1 16 (a) 1 4 (b) 25 4 (c) 275 16 8. x –5 –2 1 4 7 10 P(X = x) 32 243 80 243 80 243 40 243 10 243 1 243 Var(X ) = 10 9. (a) 1 8 , 5 24 (b) 2.78 (c) 25 96 10. x 0 1 2 3 4 P(X = x) 1 210 24 210 90 210 80 210 15 210 2.4, 0.8, 1.6, 0.8
301 Mathematics Semester 3 STPM Answers 11. (a) 3 16 (b) 17 32 ; Var(X ) = 615 1 024 12. (a) 1 15 625 (b) 624 625 (c) 1 077 3 125 ; 1, 6 144 15 625 13. (a) p5 (6 – 5p) (b) 1 – (1 – p + np)(1 – p)n – 1 (c) 0.0306 14. (a) (i) 0.0729 (ii) 0.0086 (iii) 0.6630 (b) 0.0479 15. (a) 211 243 (b) 211 243 (c) 1 243 ; 1 or 2 lines, 80 243 16. (a) 24 625 (b) 243 3 125 (c) 32 625 17. (a) 4.8, 0.980 (c) 0.737 (d) 0.388 18. 0.647, 0.185 19. (a) 0.135 (b) 0.271 (c) 0.0291 (d) 2.00 (e) 8.19 20. (a) 0.1353 (b) 0.3233 (c) 0.0527 (d) 0.1954 21. (a) 0.419 (b) 0.676 22. (a) X ~ B(1 000, 0.003) (b) 538 23. (a) 0.5664 (b) 0.5374 24. 0.7787 25. (a) (i) 0.192 (ii) 0.39 (iii) 0.0112 (b) (i) 0.6988 (ii) 0.9662 (iii) 0.0273 26. (a) Yes (b) No (c) No 27. (a) —1 4 —1 2 —3 4 1 1 2 3 0 x f(x) (b) (i) 61 64 (ii) 3 8 28. c = 1 10 , k = – 3 10 29. (a) k = 1 2 (b) 71 144 (c) 0.5408 30. k = 1 4 , median = 2, E(X) = 8 3 ; a = 20 31. (a) a = – 3 16 (b) 19 80 (d) 2 32. (a) 3 0 1 f(x) x (b) 3 4 (c) 0, x 0, F(x) = 1 – (1 – x 3 ) 3 , 0 x 3, 1, x 3. 33. (b) E(X) = 2 5 , Var(X) = 19 350 (c) 0, x 0, F(x) = 4x —3 2 – 3x 2 , 0 x 1, 1, x 1. 34. E(X) = 3 4 , Var(X) = 19 80 0, x 0, F(x) = x 16 (12 – x 2 ), 0 x 2, 1, x 2. 0.007 35. (a) 1 3 (c) 4.39 (d) 25 2 36. (a) 2 3 (b) 2x 3 , 0 x 1, f(x) = 1 – x 3 , 1 x 3, 0 , otherwise. E(X ) = 4 3 , Var(X) = 7 18 (c) 1.27; 0.875 37. (a) α = – 1 16 , β = 1 (b) 1 8 x, 0 x 4, f(x) = 0, otherwise. 38. (a) 117 125 (b) 1 125 39. 3.12, 0.169 40. (a) 0.00112 (b) 0.00794 41. (a) 0.0571 (b) 99.11% (c) 0.3962 42. 0.0255 43. 3 44. (a) 2.17; 0.979 45. (a) (i) 0.1488 (ii) 0.5695 (b) (i) 173 (ii) 0.8799 46. (a) 0.8796 (b) 9.69, 0.2339 (c) 0.1801 47. 10C1 (0.05)(0.95)9 (a) 0.195 (b) 0.840
302 Mathematics Semester 3 STPM Answers 48. (a) 0.160 (b) 0.251 (c) 0.106; 0.00436 or 0.00420 49. (a) 0.194 (b) 0.986 (c) 0.933 50. (a) 0.55 (b) 0.18 51. 19 30 52. k = 9 53. (a) x 0 1 2 3 4 f(x) 0.2 0 0.5 0 0.3 (b) 2.2, 1.96 54. (a) k = 1 4 (b) 0 , x , 0, F(x) = 1 – e x –— 2 ( 1 2 x + 1), 0 x ∞, 1 , x > ∞ (c) 0.0732 55. (a) 22 (b) 0.594 56. (a) 0.1056 (b) 1340 57. (a) 1 30 (b) 3 1 3 , 2.91 (c) 8, 8.73 58. (b) 8 9 , 17 81 59. (a) 0.0368 (b) 0.2584 (c) 0.0276 0.0011 60. (a) k = 1 8 (b) 0 , x , 0 x 8 , 0 < x , 2 F(x) = x2 16 , 2 < x , 4 1 , x > 4 (c) 1 – 4 1 – 2 3 – 4 1 0 2 4 x F(x) (d) 7 16 61. (a) X = x 0 1 2 3 4 P(X = x) 0.35 0.3 0.15 0.1 0.1 (b) 1.3 62. (a) (i) 0.2001 (ii) 0.6482 (b) X is the number of people choosing swimming as their leisure activity. Since np > 5 and nq > 5, use normal approximation i.e. X~N(10, 8). 0.4623 63. (a) 2 3 (c) 3 4 64. (a) 1 4 (b) 29 16 65. (a) µ = 544, s = 40 (b) 0.9366 (c) 2.10 (d) Mean number of mangoes on a branch is not constant. 4 Sampling and Estimation Exercise 4.1 1. (a) 4, 6 (b) X – = 1 3 (X1 + X2 + X3 ) 1 2 3 4 5 6 7 Number of samples 1 3 6 7 6 3 1 P(X – = x –) 1 27 3 27 6 27 7 27 6 27 3 27 1 27 (c) 4, 2 2. (a) (i) µ = 0.5, σ 2 = 0.45 (ii) µ = 0.6, σ 2 = 9.24 (b) (i) E(X – ) = 0.5, Var(X – ) = 0.225 (ii) E(X – ) = 0.6, Var(X – ) = 4.62 3. (a) µ = 5, σ 2 = 7.5 (b) X – = 1 2 (X1 + X2 ) 2.5 4 4.5 5.5 6 7.5 Number of samples 2 2 2 2 2 2 P(X – = x –) 1 6 1 6 1 6 1 6 1 6 1 6 (c) E(X – ) = 5, Var(X – ) = 2.5 (e) s2 (N – n) n(N – 1) = s2 n 4. (a) µ = 0, σ2 = 7 6 (b) X – = 1 2 (X1 + X2 ) –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 P(X – = x –) 1 144 6 144 17 144 30 144 36 144 30 144 17 144 6 144 1 144 (c) 0, 7 18 5. sA = 8.11, sB = 5.59 A bigger sample size will have a smaller standard deviation. Exercise 4.2 1. 0.034 2. (a) 3.6, 3.24 (b) 0.8542 3. (a) 4.5, 0.75 (b) 0.8682 4. (a) 0.2902 (b) 0.8071 5. 0.8185 6. 40 7. 385 8. (a) 0.00637 (b) 0.00295 (c) 0.5521 (d) 0.4479 9. (a) 0.79 (b) 0.9565 (c) 0.7085 (d) 0.0179 10. 0.0912, 0.0559 The larger sample size, the smaller the standard error.
303 Mathematics Semester 3 STPM Answers Exercise 4.3 1. (a) 19.12, 0.25 (b) 23.5, 4.43 (c) 9.71, 621.12 (d) 3.17, 1.5769 2. 0.4 Exercise 4.4 1. (a) (9.26, 11.52) 2. (a) x – = 178.2 cm (b) σ = 5 cm (c) (177.037 cm, 179.363 cm) 3. (a) (8.161, 8.639) (b) (8.117, 8.683) 4. (a) (0.1096, 0.1904) (b) (0.102, 0.198) 5. 2155 6. n = 56 7. n = 134 8. 257 9. 384 10. 68 11. 23 12. (a) (13.43, 16.17) (b) 118 (c) 86.6 13. (a) 0.383 (b) 0.886 (c) (1.34, 2.52) 14. (a) 0.24 (b) 0.0191 (c) 500 15. 0.2516 (a) 0.6305 (b) 0.5467 (c) 0.0375 (d) 0.0478 16. (a) 48 (b) 92.8 17. (a) (2.92, 2.96)A (b) (2.78, 2.80)B Orchard A because the two interval do not overlap and 2.936 ∈ (2.92, 2.96)A 18. (a) Simple random sample (b) (25.8, 31.8) (c) Sample is not representative, maybe weak candidates (d) 37 19. (a) (15.87, 18.33), width is proportional to magnitude of confidence level. (b) 0.60 (c) 17.1 20. (a) 4.2, 3.89 (b) 3.9, 3.15 (c) second sample, larger size (d) (3.45, 4.69) STPM Practice 4 1. (a) 25.3, 3.63 (b) (24.85, 25.75) 2. (a) (0.245, 0.455) (b) (0.352, 0.548) (c) No, because (0.245, 0.455) intersects with (0.352, 0.548), i.e. pA pB . pB pA 0.245 0.455 0.352 0.548 3. (a) 114.4, 88.27 (b) (108.577, 120.223) 4. (a) 2.7, 1.384 (b) 1.915 (c) (1.49, 3.91) (d) fair = estimated (s2 ) = s2 (e) confidence interval = 95% confidence mean population within the interval (1.49, 3.91) 5. (a) (45.595, 49.381) (b) 0.0321 6. (b) 69.34% 7. (a) 68% (b) (61.5%, 74.5%) (c) 929 8. (a) 8.309, 1.319 (b) 2164 9. (a) 91.32, 7.42 (b) (90.48, 92.16) 10. (a) 1062.5, 120 (b) (1011, 1114) 11. (a) 265 (b) (i) sample size is bigger (ii) sample size is smaller 12. (a) 0.8971 (b) 0.3274 (c) (56.5, 63.3)B , No difference in mean age because mean age of 61 in country A is in the confidence interval. 13. (a) (14.98, 16.62) (b) 49 (c) 89 14. (a) N(0.8, 0.0018) (b) (i) 0.5620 (ii) 0.2398 (iii) 0.547 (c) 15% 15. µ = 5.088, s = 1.891 (a) 0.466 (b) 7.51, n = 50 16. (15.92, 18.88) 17. (a) (0.748, 0.852) (b) 96% confident 74.8% to 85.2% of people supported party A. Allow for calculations of errors of values. 18. (a) 0.168 (b) (0.134, 0.202) (c) 215 19. x – = 1300 (a) (RM1225, RM13 785) (b) 86 (c) 178 20. (a) p ^ ~ N(0.46, 0.002484) (b) 0.0355 21. (a) 0.6232 (b) 0.5682 22. (a) p ^ ~ N(0.75, 0.001875) (b) (i) 0.124 (ii) 0.383 23. (a) 139, p ^ = 0.85 (b) (i) n is bigger (ii) n is smaller 24. (a) x – ~ N(188, 36 n ) (b) 36 25. (a) 0.264 (b) 22.1 26. (a) 0.769 (b) (0.165, 0.315) (c) (122 000, 187 000) 27. (a) 236, 7.58 (b) 0.5714 28. 0.0397 29. 90 30. (a) X ~ N(4.5, 0.075) (b) 0.034 (c) (3.9295, 4.8305)
304 Mathematics Semester 3 STPM Answers 5 Hypothesis Testing Exercise 5.1 1. H0 : The defendant is innocent, H1 : The defendant is guilty. 2. H0 : The drug is neither safe and nor effective, H1 : The drug is safe and effective. 3. H0 : p = 0.2, H1 : p . 0.2. 4. H0 : μ = 8 kg, H1 : μ ≠ 8 kg. 5. H0 : μ = 1000 mg, H1 : μ , 1000 mg. 6. α; β. 7. (a) Type II error. (b) Type I error. 8. A new teaching technique and the conventional classroom procedure are in fact equally effective. The decision made is that the new teaching technique is either inferior or superior to the conventional procedure, thus Type I error is committed. The new teaching technique is in fact either inferior or superior to the conventional procedure. The decision made is that the new teaching technique and the conventional classroom procedure are equally effective, thus Type II error is committed. 9. (a) Decrease. (b) Increase. 10. (a) H0 : μ = 5 hours, H1 : μ , 5 hours. (b) Type I error, (c) Type II error, (d) a one-tailed test. 11. (a) X , 25 or X . 75, (b) 25 , X , 75, (c) X = 25 and X = 75, (d) 0.05. 12. (a) X , 31, (b) X . 31. (c) X = 31, (d) 0.01. 13. One-tailed test because the rejection region lies in the right tail of the distribution curve. = 70 Nonrejection region Critical value Rejection region – x 14. Two-tailed test because the rejection region falls in both tails of the distribution curve. – x = 15 – 2 Nonrejection region Critical value Critical value Rejection region Rejection region – 2 15. α = 0.2497 16. z = 1.928 Exercise 5.2 1. –2.05 0 0.02 2. x – = 824.1, 857.9 3. Do not reject H0 4. Reject H0 5. For n = 20, do not reject H0 ; for n = 200, reject H0 6. Reject the company’s claim 7. The manufacturer’s claim is not valid 8. Do not reject H0 9. Do not reject H0 10. 80.36 , μ , 97.64, z —a 2 = –1.645, 1.645 11. The machine slipped out of normal operation 12. A mean cholesterol level still exceeds the national average after taking new drug 13. Reject the claim 14. Agree with this claim 15. 348.3 ml and 351.7 ml. I am not convinced Exercise 5.3 1. P(X < 3) = 0.1071, P(X > 14) = 0.0171 2. P(X < 13) = 0.0867 , 0.10, reject H0 3. Do not reject the claim 4. There is evidence of gender bias in trainee selection 5. There is no evidence to support the claim 6. Reject sociologist’s claim 7. It is a valid estimation 8. z = 2.066 9. The exact probability is 0.274. The corresponding normal approximation value is 0.207. 10. Reject the construction firm’s claim 11. These figures confirm that the sex ratio is 50 : 50 12. Reject the club’s claim 13. Do not reject the estimation 14. The claim is not valid 15. The rate has not increased 16. It differs from the percentage of the population having type O blood. The conclusion is reverse if the significance level changes from 0.1 to 0.05.
305 Mathematics Semester 3 STPM Answers STPM Practice 5 1. (a) Do not reject H0 (b) P(Type I error) = 0.05 2. Reject H0 3. Reject H0 4. Do not reject H0 5. (a) H0 : p = 0.5 against H1 : p > 0.5 (b) Reject the person’s claim 6. (a) H0 : μ = 9.8 against H1 : p > 9.8 (b) The cleanup project at a lake is effective 7. There has been no change in the proportion 8. H0 : p = 0.6 against H1 : p ≠ 0.6; do not reject the claim 9. Do not reject H0 10. There has been a change in the average height of male students 11. The data support the null hypothesis 12. The daily smokers nowadays smoke less 13. There is sufficient evidence to reject the shop’s claim 14. The machine is not out of control 15. The new patent medicine is not superior 16. The fast food restaurant management have grounds to complain 17. (a) µ ^ = 36, s ^ = 324 (b) 32.06 , μ , 39.94 (c) Do not reject H0 (d) In part (b), the confidence interval does contain the value μ0 = 34, so we would not reject H0 18. Reject the claim 19. The die is unbiased 20. The authority’s worry is valid 21. (a) The mean time to collect a bill is less than 120 seconds (b) Customers’ bill payment is a random sample 22. Reject H0 23. (a) X . 6 (b) X = 7, α = 0.057 (c) If X = 7, reject H0 24. (a) 5.416 , μ , 6.984 (b) Assume that reaction time is approximately normal (c) The null hypothesis would be rejected since the hypothesised value of 7.0 seconds is not included in the confidence level found in (a). 25. Do not reject H0 . 26. Reject H0 27. Reject H0 28. (a) Do not reject H0 (b) 13.82% 29. 25.171 , X , 25.229 6 Chi-squared Tests Exercise 6.1 1. 16.919 2. (a) 0.99 (b) 0.10 (c) 0.05 3. (a) 2.558 (b) 31.41 (c) 7.261, 25.00 4. (a) 11.52 (b) 4.575 (c) 7.015 5. (a) µ = 8 and s = 4 (b) (iv) Exercise 6.2 1. (a) 16.81 (b) 14.68 2. Do not reject H0 3. The three coins are biased 4. The deck was honest 5. The package does contain the mixed beans in the ratio 5 : 3 : 1 : 1 6. Reject the hypothesis that the colours of the jelly beans occur with equal frequency 7. The data does follow Poisson distribution 8. The distribution of marks is normal 9. The normal distribution offer a good fit for the data distribution 10. Evidence suggests that the data is good fit to the Poisson distribution 11. Evidence suggests rejecting the observed frequencies are normally distributed Exercise 6.3 1. (a) ν = 15 (b) 30.58 (c) The chi-squared test values of 39.2 falls into the critical region 2. The row and column variables are independent 3. Gender is not independent of handedness 4. There is association between gender and preference of food 5. There is no association between cigarette smoking and the risk of diabetes. 6. The supplier is not associated with the lens quality 7. The preference for the different formulation does not change with age 8. The outcome is not independent of treatment 9. The distribution of blood type for Malay is not different across the three states 10. These data provide sufficient evidence to infer that the machine breakdown is not independent of the shift STPM Practice 6 1. (a) 0.95 (b) 11.591 2. The number of shirts sold is not uniformly distributed 3. The three dice are bias 4. Proportions of cars going straight, turning left and turning right differ significantly from the officer’s assertion 5. The two cough remedies are equally effective 6. The number of goals per match has a Poisson distribution 7. The normal distribution provides a good fit for the distribution of battery lives. 8. The distribution of marks is not normal 9. The recorded data is poorly fitted by the binomial distribution 10. It is adequate to use the normal distribution as a model for these data 11. The leaves9 length can be approximately modelled by the normal distribution
306 Mathematics Semester 3 STPM Answers 12. There is association between the exposure to the virus and the development of the disease 13. There is association between gender and employment categories 14. There is association between smoking habits and socioeconomic status 15. The method of payment is dependent of age group 16. There is no association the level of exposure to the pollutant and the number of brain abnormality in the laboratory mice 17. The time spent on sleeping is not independent of the age of the adult 18. There is association between sterility and genotype 19. There is evidence of an association between gender and courses chosen 20. It is concluded that a person’s skin reacting to ultra-violet light is related to the person’s eye colour 21. There is no relationship between the gender and the programmes watched 22. Reject H0 and there is evidence that the distribution of grades is not uniform 23. Reject H0 and there is relationship between the smoking habit and hypertension 1. (a) Group the data in a frequency table. Number of patients 21-30 31-40 41-50 51-60 61-70 71-80 81-90 Frequency 2 2 4 5 4 2 1 Split the values into the tens and units, for example 25 = 20 + 5. Put the tens values, dropping the trailing zero, on the left and the units values in ascending order on the right in the diagram below. Write the key which explains what the values in the diagram represents. Stem Leaf 2 3 4 5 6 7 8 9 0 5 3 4 8 8 9 9 2 5 5 5 6 3 3 5 5 4 5 0 Key: 4| 8 means 48 Arrange the data in ascending order 20 25 33 34 48 48 49 49 52 55 55 55 56 63 63 65 65 74 75 90 1 2 (20) = 10, so the median, Q2 is 1 2 (x10 + x11) = 1 2 (55 + 55) = 55 \ The median is 55 1 4 (20) = 5, so the lower quartile, Q1 is 1 2 (x5 + x6 ) = 1 2 (48 + 48) = 48 3 4 (20) = 15, so the upper quartile, Q3 is 1 2 (x15 + x16) = 1 2 (63 + 65) = 64 \ Interquartile range = 64 – 48 = 16 (b) The lower bound outlier value L = Q1 – 1.5(Q3 – Q1 ) = 48 – 1.5(16) = 24. The upper bound outlier value U = Q3 + 1.5(Q3 – Q1 ) = 64 + 1.5(16) = 88. Hence, 20 and 90 are outliers. (c) On a graph paper, draw a box with the left and right sides collinear with Q1 , Q3 and a divider collinear with Q2 . Draw a whisker from the smallest value which is not an outlier to the left side of the box and then a whisker from the right side of the box to the largest value which is not an outlier. 20 30 40 50 60 70 80 90 (d) The shape of the frequency distribution is near symmetrical because Q2 – Q1 ≈ Q3 – Q2 2. P(A) = 0.6, P(B) = 0.5 and P(A < B) = 0.8. (a) P(A > B) = P(A) + P(B) – P(A < B) = 0.6 + 0.5 – 0.8 = 0.3 P(A) × P(B) = 0.6 × 0.5 = 0.3 \ P(A > B) = P(A) × P(B) ⇒ Events A and B are independent. (b) P(A < C) = 0.25, P(B < C) = 0.85 and P(A > C) = 3 P(B > C). P(A < C) = P(A) + P(C) – P(A > C) = 0.25 P(A) + P(C) – [3P(B > C)] = 0.25 …… 1 P(B < C) = P(B) + P(C) – P(B > C) = 0.85 …… 2 3 × 2 ⇒ 3P(B) + 3P(C) – 3P(B > C) = 2.55 …… 3 3 – 1: 3P(B) – P(A) + 2P(C) = 2.3 2P(C) = 2.3 – 3P(B) + P(A) = 2.3 – 3 × 0.5 + 0.6 = 1.4 \ P(C) = 0.7 3. Let X be the number of larvae in 100 ml of water. Then X ~ P0 (8).
307 Mathematics Semester 3 STPM Answers (a) Probability that there are most 3 larvae = P(X < 3) = e–81 1 + 8 + 82 2 + 83 6 2 = 0.0424 (b) Now P(X = r + 1) P(X = r) = 1 e–88r + 1 (r + 1)! 2 1 e–88r r! 2 = 1 e–88r + 1 r!(r + 1) 2 1 e–88r r! 2 = 8 r + 1 If P(X = r) < P(X = r + 1) ⇒ P(X = r + 1) P(X = r) > 1 8 r + 1 > 1 ⇒ r + 1 < 8 ⇒ r < 7 When r = 6, P(X = 6) , P(X = 7) = e–81 87 7! 2 = 0.1396 When r = 7, P(X = 7) < P(X = 8) = e–81 88 8! 2 = 0.1396 \ The most likely number of larvae in 100 ml of water is 7 or 8. 4. Let X be the masses of batteries manufactured. Assume that X has a normal distribution. X ~ N(1.48, 0.552 ). (a) Probability that a randomly selected battery has a mass of between 1.3 kg and 1.6 kg = P(1.3 , X , 1.6) = P1 1.3 – 1.48 0.55 , Z , 1.6 – 1.48 0.55 2 = P(–0.327 , Z , 0.218) = 0.2145 (b) By the central limit theorem, the mean mass of the batteries, X – ~ N1 1.48, 0.552 25 2 The probability that the mean mass of 25 randomly selected batteries has a mass between 1.3 kg and 1.6 kg = P(1.3 , X – , 1.6) = P 1 1.3 – 1.48 0.55 25 , Z , 1.6 – 1.48 0.55 25 2 = P(–1.636 , Z , 1.091) = 0.8114 5. Null hypothesis: H0 : p = 0.53 Alternative hypothesis: H1 : p , 0.53 Let X be the number of blue balls chosen. Assuming that the null hypothesis is true, then X ~ B(10, 0.53). P(X < 3) = 10C0 (0.53)0 (0.47)10 + 10C1 (0.53)1 (0.47)9 + 10C2 (0.53)2 (0.47)8 + 10C3 (0.53)3 (0.47)7 = 0.127 Since P(X < 3) = 0.127 . 0.10 ⇒ Accept H0 i.e. p = 0.53 6. Null hypothesis H0 : There is no association between attending kindergarten and pre-arithmetic achievement Alternative hypothesis H1 : There is an association between attending kindergarten and pre-arithmetic achievement Observed data: Below level At level Above level Total No kindergarten 10 7 7 24 Kindergarten 7 18 11 36 Total 17 25 18 60 Expected data: The expected number of no kindergarten attending students who perform below level is calculated as follows: Expected frequency = (Row total)(Column total) Total sample size = 24 × 17 60 = 6.8 The complete table for the expected frequencies is: Below level At level Above level Total No kindergarten 6.8 10 7.2 24 Kindergarten 10.2 15 10.8 36 Total 17 25 18 60 The table has 2 rows and 3 columns. Thus, the number of degree of freedom, v = (2 – 1)(3 – 1) = 2. Perform a test at the 5% significance level. From chi-squared table, the critical value χ2 for 2 degrees of freedom at 5% level is given by 5.991. H0 is rejected if χ2 . 5.991. Calculate χ2 : Observed (o) Expected (e) (o – e)2 e 10 6.8 1.5059 7 10 0.9 7 7.2 0.0056 7 10.2 1.0039 18 15 0.6 11 10.8 0.0037 ∑ (o – e)2 e = 4.019 As χ2 = ∑ (o – e)2 e = 4.019 , 5.991, so do not reject H0 . There is no association between attending kindergarten and pre-arithmetic achievement.
308 Mathematics Semester 3 STPM Answers 7. 1 t . 3 F(t) = ct2 (18 – t 2 ) 0 , t < 3 where c is a constant. 0 t = 0 (a) F(3) = 1 ⇒ c(32 )(18 – 32 ) = 1 \ c = 1 81 . (b) The probability that the life span is more than 2 days = 1 – F(2) = 1 – (22 )(18 – 22 ) 81 = 0.309 (c) Let m be the median ⇒ F(m) = 1 2 (m2 )(18 – m2 ) 81 = 1 2 m4 – 18m2 + 40.5 = 0 m2 = –(–18) ± (–18)2 – 4(1)(40.5) 2(1) Since 0 , t < 3 ⇒ m2 = 2.636 ⇒ m = 1.62 \ The median is 1.62 days. (d) For 0 < t < 3 ; d dx (t 2 (18 – t 2 )) 81 = 36t – 4t 3 81 The probability density function is 4t(9 – t 2 ) 81 0 , t , 3 f(t) = 0 elsewhere Graph of f(t) f(t) t 0 3 8. (a) X is the time taken, in minutes, from receiving an emergency call to the arrival of an ambulance at the location requested for the service. Let E(X) = μ and Var(X) = σ2 . n = 80, ∑x = 1570 and ∑x2 = 31810 The unbiased estimate of the population mean μ is µ ^ , where µ ^ = x = ∑x n = 1570 80 = 19.625 The unbiased estimate of the population variance σ2 is s ^ , where s ^ 2 = ∑(x – x)2 n – 1 = ∑x2 – nx 2 n – 1 = 31 810 – 80 × 19.6252 79 = 12.642 (b) By the central limit theorem, since n is large, X is approximately normal distributed so, X ~ N1 µ, s ^ 2 n 2 with n = 80 x = 19.625 and s ^ = √12.642 = 3.556, Standardising, z = x – µ s ^ √n For a 95% confidence interval, we have 1 – a = 0.95 or a 2 = 0.025. From the standard normal distribution table, z a 2 = 1.96. Therefore, the 95% confidence interval for the population mean μ becomes 1 x ± 1.96 s ^ √n2 or (18.846 minutes, 20.404 minutes). (c) As the mean target waiting time of 19 minutes is located within the 95% confidence level, it can be concluded that there is no significant evidence that the mean target time for the arrival of ambulance is not 19 minutes. (d) Perform the hypothesis test Null hypothesis H0 : μ = 19 Alternative hypothesis H1 : μ , 19 X ~ N1 µ, s ^ 2 n 2 with n = 80 and s ^ = 3.556. If H0 is true, μ = 19, so X ~ N1 19, 3.5562 80 2 Use a one-tailed test, at the 5% significance level. The critical z-value for a one-tailed 5% test is –1.645. Reject H0 if z , –1.645, where Z = x – 19 3.556 √80 From the sample, x = 19.625. Thus Z = 19.625 – 19 3.556 √80 = 1.572 Since z . –1.645, do not reject H0 . There is no significance evidence that the mean target waiting time for the arrival of ambulance is less than 19 minutes.