Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
UNIVERSITY of CAMBRIDGE Page 1 of 2 Local Examinations Syndicate FINAL MARK SCHEME 2642 Probability and Statistics 2 M1 A1 2 Correct formula used Answer, art 0.159 Poisson, = 14 0.26(00) 2 N , M1 A1 B1 Poisson distribution, 14 stated or implied 2 Answer, 0.26 or 0.260(0) 1 Normal, parameters correct [not “²/n”] If this notation used, do not allow /8. 148.0 = 18.5 8 2809.68 s2 = x 2 [= 8.96] 8 B1 18.5 only, not i.s.w. M1 Correct method for s2 1 2 2 (i) e (ii) (i) (ii) 4 3 4 ( 4 23 ) 3 =0.159(28) 3! 8 x ˆ 2 3 8 s 2 = 10.24 7 (i) Number pupils Use figures, 3 at a time, ignoring those over 200, or multiply decimal by 200, and select corresponding pupils. (ii) X ~ B(8, 0.6) P(X 6) = 1 – 0.6846 = 0.315(4) (i) P( > 13) = 1 – 2p P( > 13) = p by symmetry 1 – 2p = p and p = 13 AG (ii) 3 1 23 = 0.431 = 6.96(0) 5 (i) 3 10 / n 6 (ii) (iii) (i) (ii) January 2004 1.96 n > 1.96 10/3 = 6.53 n > 42.68 so nmin = 43 0.05 Null hypothesis is correct B(10, 0.4) P( 6) = 0.9452, P( 7) = 0.9877 so need 7 packets A1 3 Answer, 10.24 or 10.2, not i.s.w. B1 B1 2 State that elements of sample frame are numbered Any valid method for using random numbers, need to mention numbers out of range M1 M1 A1 M1 A1 A1 B(8, 0.6) stated or implied Tables (or formula) correctly used, allow 1 – P( 6) 3 Answer in range [0.315, 0.316] One correct statement, e.g. “symmetry” stated Second correct statement 3 Fully correct argument [3/ = p, etc: 0] M1 Standardise with –1, not ; 0.7477 is M0B0 [2: M1B1A0] Correct value seen, in range [0.43, 0.431] Correct answer, in range [6.96, 6.98], no sign errors Standardise with n or n LHS correct, equated to –1, allow cc 0.5 1.96 seen A.r.t. 6.53 or 42.7 seen, allow sign errors Final answer 43, cwo (allow =, , ) Probability 0.05 or 5%, cao Statement that H0 is correct and/or rejected B(10, 0.4) stated or implied One of these probabilities seen Answer 7 [no probabilities seen: M1M0A1] SR: Po(4) or N(4, 2.4): M1M0A0 B(70, 0.4) stated or implied N(np, npq); 28, 16.8 correct Standardise 35 ( 0.5) with np and npq or npq, allow wrong or no cc Both 35.5 and npq, their npq Answer, a.r.t. 0.966 SR: B(35, 0.4): M0M1A0M1A1A0 Po() N(, ): M0M1A0M1A1A0 B1 A1 M1 A1 B1 A1 A1 B1 B1 M1 M1 A1 B(70, 0.4) N(28, 16.8) () 35.5 28 M1 M1A1 M1 16.8 = 1.83 (1.83) = 0.966(4) A1 A1 3 5 1 1 3 6 UNIVERSITY of CAMBRIDGE Page 2 of 2 Local Examinations Syndicate FINAL MARK SCHEME 2642 Probability and Statistics 2 7 January 2004 y (i) B1 M1 A1 Rectangle, ignore outside range Sad parabola, symmetric, not going below axis unless clearly goes through (1, 0), (–1, 0) 3 Roughly correctly related, no labels needed x (ii) Model 1: equally likely to be anything between –1 and 1 Model 2: more likely to be near 0 (iii) k1 = ½ (iv) = 0 stated 1 1 1 = (v) 8 2 k1 x dx ; 1 3 x3 k1 [= 3 1 1 3 ] B1 B1 B1 B1 M1 A1 A1 or 0.577 Results tend to be closer to 0, so SD of Model 2 is less than . H0: = 8 H1: < 8 : P( 2) = 0.0138 Compare with 0.05 : CR is 3, prob 0.0424, Compare 2 Therefore reject H0 Significant evidence that mean number of letters received on Monday is less than 8 (b) P( 3) 0.0424 (ii) (a) Numbers of letters received on the four Saturdays must be independent of one another (b) P(32) N(32, 32) (i) (a) () 25.5 32 [= –1.15] 32 Probability = 1 – 0.8749 = 0.125(1) B2 B1 B1 M1 A1 A1 M1 B1 B1 M1 A1 M1 A1 A1 Correct description of rectangular distribution 2 Correct description of Model 2 Do not allow for description of graphs only 1 Correct value of k1, cwd, allow if seen elsewhere = 0 clearly stated Attempt to integrate x2f(x), limits –1 and 1 Correct indefinite integral 4 Correct answer, aef, e.g. 13 or a.r.t 0.577 [ = 0 not seen: B0M1A1A1] 2 “Less” stated, convincing reason SR: “Less” stated, vague reason: B1 Wrong or no reason: B0 One hypothesis correctly stated, or Other (wrong or no letter: B1 both, but not x) Find P( 2) or critical region from Po(8) [Not P( = 2) or P( 8) etc] Explicitly compare () 0.0138 with 0.05 or 0.0424 () 3 with 2, 0.0424 seen [If 0.0424 seen in (b) but not (a), can recover in ] 5 Conclusion stated in context, on their values Normal approx: B1B1M0 Correct method for critical region 2 0.042(4) or 4.2(4)% [0.0996 from 4: M1B0] 1 “Independent” or “constant rate” stated, in context N(, ) stated or implied, or N(32, …) = 32, allow 2 = 322 or 32 Standardise 25 ( 0.5) with and , allow = , wrong/no cc cc and correct, and 5 Answer, a.r.t. 0.125