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Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w24_qp_629709_w24_ms_629709/62/O/N/24
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
A random variable X has the distribution Be4500 000, o. 1000 000 Use a Poisson distribution to calculate an estimate of P (X H 4) . [3]
Mark scheme
1 λ = 4.5 B1 1 − e−4.5(1 + 4.5 + 2 + 3! ) =1– e−4.5(1 + 4.5 +10.125 +15.1875) 4.52 4.53 M1 Expression must be seen or implied by correct figures. Any λ. Allow one end error. = 1– (0.011109 + 0.049999+0.11248 + 0.16872) Accept fully correct Σ notation. 0.658 (3 sf) A1 SC unsupported 0.658 scores B1 B1. 3
Question
The lengths of a random sample of 50 roads in a certain region were measured. Using the results, a 95% confidence interval for the mean length, in metres, of all roads in this region was found to be [245, 263]. (a) Find the mean length of the 50 roads in the sample. [1] (b) Calculate an estimate of the standard deviation of the lengths of roads in this region. [2] (c) It is now given that the lengths of roads in this region are normally distributed. State, with a reason, whether this fact would make any difference to your calculation in part (b). [1]
Mark scheme
2(a) 254 [m] B1 1 2(b) σ M1 ft their ‘254’ accept 1.96 or 1.645 for M1. 263 = ‘254’ + 1.96× σ oe or 2 × 1.96 × 50 = 18 50 9 50 A1 [σ = 1.96 = ]. s.d. = 32.5 [m] (3 sf) 2 2(c) No B1 Both needed. Because the sample mean is approximately normally distributed [for Or because of the Central Limit theorem. large n] Or because n is large [accept ⩾30 condone ⩾50]. 1
Question
A factory owner models the number of employees who use the factory canteen on any day by the distribution B(25, p). In the past the value of p was 0.8 . A new menu is introduced in the canteen and the owner wants to test whether the value of p has increased. On a randomly chosen day he notes that the number of employees who use the canteen is 23. (a) Use the binomial distribution to carry out the test at the 10% significance level. [5] (b) Given that there are 30 employees at the factory comment on the suitability of the owner’s model. [1]
Mark scheme
3(a) H0: p = 0.8 H1: p > 0.8 B1 [Assuming H0, P(X ⩾ 23) =] 25C23×0.22×0.823 + 25C24×0.2×0.824 + 0.825 M1 No end errors. Expression must be seen or supported by =0.070835 + 0.0236118 + 0.0037779 enough figures to be convinced B(25,0.8) used. Accept correct Σ notation. = 0.0982 A1 SC B1 for 0.0982 unsupported. 0.0982 < 0.1 M1 Valid comparison their 0.0982 must be a tail probability. [There is evidence to reject H0] ftA1 No contradictions. In context, non-definite. There is sufficient evidence to suggest that p has increased Condone ‘there is sufficient evidence that the ‘claim’ is correct’ and condone ‘there is sufficient evidence that the number of employees (using the canteen) has increased’ Note: CR method will include P(X ⩾ 23) so M1 A1 as above, and P(X ⩾22)=0.234>0.1 with at least one probability comparison with 0.1 needed to find CR of 23,24,25 (so 23 in CR) M1 A1ft as above. 5 3(b) Not suitable as model does not allow for more than 25 employees to use B1 Need both (i.e. suitable or not suitable plus reason). the canteen/Not suitable as uses a sample instead of all employees/Not suitable doesn’t include all employees /Not suitable as 30 is only just bigger than 25 should have used 30 OR Suitable as owner knows that not all employees use the canteen, or similar 1
Question
A population is normally distributed with mean 35 and standard deviation 8.1 . A random sample of size 140 is chosen from this population and the sample mean is denoted by X . (a) Find P (X 2 36) . [3] (b) It is given that P (X 1 a) = 0.986 . Find the value of a. [3]
Mark scheme
4(a) 36−35 M1 Ignore inclusion of cc for M1. Must have √140. 8.1÷ 140 [= 1.461] 1 − Φ(‘1.461’) M1 For area consistent with their values. = 0.0720 (3 sf) A1 Allow 0.072. 3 4(b) [Φ−1(0.986)] = 2.197 to 2.198 B1 Seen. Note: 2.2 and nothing better seen scores B0 ± 8.1a÷−35140 = ± ‘2.198’ M1 Must be a z value. a = 36.5 (3 sf) A1 CWO Note: use of 2.2 scores A1 so 2/3. But e.g. 2.196 gives 36.5 but scores B0 M1 A0 so 1/3 3
Question
A machine puts sweets into bags at random. The numbers of lemon and orange sweets in a bag have the independent distributions Po(3.7) and Po(2.6) respectively. A bag of sweets is chosen at random. (a) Find the probability that the number of lemon sweets in the bag is more than 2 but not more than 5. [2] (b) Find the probability that the total number of lemon and orange sweets in the bag is less than 4. [3] 10 bags of sweets are chosen at random. (c) Use approximating distributions to find the probability that the total number of lemon sweets in the 10 bags is less than the total number of orange sweets in the 10 bags. [6]
Mark scheme
5(a) M1 Expression must be seen or implied by correct figures. e−3.7( 3! + 4! + 5! ) = e−3.7 (8.44217 + 7.80900 + 5.77866) = 3.73 3.74 3.75 Any λ. Allow one end error. 0.20872 + 0.19307 +0.14287 Accept fully correct Σ notation. = 0.545 (3 sf) A1 SC 0.545 unsupported scores B1. 2 5(b) [λ] = 6.3 B1 e−6.3(1 + 6.3 + 2! + 3! ) = e−6.3(1 + 6.3 +19.845 + 41.6745) = 6.32 6.33 M1 Expression must be seen or implied by correct figures. Any λ. Allow one end error. 0.0018363 + 0.011569 + 0.0364415 + 0.076527 Accept fully correct Σ notation. = 0.126 (3 sf) A1 SC 0.126 unsupported scores B1 B1 3 5(c) L~N(37, 37), O~N(26, 26) B1 SOI. (O − L) ~ N(−11, 63) B1 For N(±11, …..) SOI. M1 For var = 37+26 SOI. 0 − ' −11' 0+0.5−'−11' M1 Standardising with their values (wrong cc scores M1). '63' [=1.386] or '63' [=1.449] 1 − Φ(‘1.386’) or 1 − Φ(‘1.449’) M1 For area consistent with their working. = 0.0828 or 0.0829 (3 sf) or = 0.0737 or 0.0736 (3 sf) A1 SC1 10 used twice N(37,37), N(26,26) and use of 10O- 10L>0 Apply MR rules max B1 B1 M1 M1 M1 A0 (MR) SC2 P(11) giving N(11,11) scores B0 B1 M0 M1 M1 A0 6
Question
The time, X hours, taken by a large number of people to complete a challenge is modelled by the probability density function given by 1 a G x G b, f (x) = * x 2 0 otherwise, where a and b are constants. (a) State what the constants a and b represent in this context. [1] b (b) Show that a = . [3] b+1 It is given that E (X ) = ln 3. (c) Show that b = 2 and find the value of a. [4] (d) Find the median of X. [3]
Mark scheme
6(a) Min and max times [to complete challenge] B1 In context (e.g. min and max x scores B0). 1 6(b) b M1 Attempt to integrate f(x) and =1, ignore limits. a 1 x2 dx = 1 b A1 For correct equation using correct limits into correct [ − x = 1 ] − b + a = 1 1 1 1 integration and = 1. a −a + b = ab or b = a(b + 1) A1 Convincingly obtained. No errors seen. b OE a = b +1 AG. 3 6(c) b M1 Attempt to integrate xf(x). Limits a and b or b/(b+1) and b E(X) = a 1 x dx (condone a and 2 for M1) See SC for use of limits 2/3 and 2 = ln b − ln a or ln b – ln (b/(b+1) A1 Correct integration and limits substituted. Condone ln 2 – ln a. [= ln b − (ln b − ln (b + 1)) ] = ln b – ln(b/(b+1) = ln 3 A1 For correct equation in b only (i.e. using part (b)). b = ln (b + 1) = ln 3 or b+1 =3 or b2 + b = 3b or =3 b b +1 b = 2 (AG) a = 23 A1 Both obtained correctly (Note: if b=2 not shown but used can score M1 A1, A1/A0 depending on where b=2 is introduced, A0) SC verification: using b=2 and a=2/3 then integrating xf(x) from 2/3 to 2 scores M1 A1 for integration and limits substituted, then A1 for showing =ln 3 Final A0 (as verified not shown) max ¾. 4 6(d) m 2 M1 Attempt to integrate f(x) equated to 0.5 and correct limits ' 23 ' 1 x2 dx = 0.5 or m 1 x2 dx = 0.5 stated. m 2 A1FT Correct integration FT their a . − 1x 2 = 0.5 or − 1x = 0.5 ' ' 3 m [ − m1 + 32 = 0.5] or [ − 12 + m1 = 0.5] A1 m=1 3
Question
The heights of one-year-old trees of a certain variety are known to have mean 2.3 m. A scientist believes that, on average, trees of this age and variety in her region are slightly taller than in other places. She plans to carry out a hypothesis test, at the 2% significance level, in order to test her belief. (a) State the probability that she will make a Type I error. [1] She takes a random sample of 100 such trees in her region and measures their heights, h m. Her results are summarised below. n = 100 / h = 238 / h 2 = 580 (b) Carry out the test at the 2% significance level. [7] (c) The scientist carries out the test correctly, but another scientist claims that she has made a Type II error. Comment on this claim. [1] Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
7(a) 0.02 or 2% B1 <0.02 B0 1 7(b) H0: μ = 2.3 H1: μ > 2.3 B1 Accept ‘population mean’ for µ (not just mean) If not seen here, can be awarded if correctly seen in part (a) s2 = 100 ( 580 − (2.38)2 ) ) or 1/99 (580 – 2382/100) 99 100 M1 Correct substitution in s2 or s 2 formula. = 0.137 = 113/825 or s = 0.370 (3 sf) and x = 238/100 [= 2.38] A1 x and s2 (or s) correct. (SC biased estimate 0.1356 and x = 2.38 scores B1). 2.38 − 2.3 M1 ' 0.137 ' [=2.161 or 2.162] 100 = 2.16 (3 sf) OR 0.0153/0.0154 if area comparison used A1 ‘2.16’ > 2.054 (or 2.055) OR ‘0.0153 or 0.0154’<0.02 M1 Valid comparison. [There is evidence to reject Ho.] A1FT No contradictions. In context, non-definite. There is sufficient evidence to suggest that the [mean] height [in Accept CV method x = 2.376<2.38 or x = 2.304>2.3 M1 A1 scientist’s region] is greater than 2.3 [m] OR there is sufficient evidence for x and M1 A1ft for comparison and conclusion to suggest that the scientist’s claim is justified. Two tail test can score B0 M1 A1 M1 A1 M1 (comparison with 0.01oe) A0ft max 5/7 7 7(c) Not possible since Ho was rejected. B1FT Need both. Accept No as H0 was rejected. Follow through their conclusion in (b) . Condone a definite statement. 1
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