Cambridge A Level Mathematics 9709 — 2024 May/June Paper 6 · Variant 3

1 The random variable X has the distribution B ( 4000 , 0.001) . (a) Use a suitable approximating distribution to find P ( 2 G X 1 5 ) . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Justify your approximating distribution in this case. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) [λ =] 4 B1 e−4 ( 2 3 4 4 4 4 2! 3! 4!   ) or e−4 (8 + 10.67 + 10.67) or 0.1465 + 0.19537 + 0.19537 M1 Allow one end error. Any λ. Expression must be seen. = 0.537 (3sf) A1 SC B1 B1 for unsupported correct answer. SC B2 for use of Binomial leading to 0.537. Note: use of normal could score B1 only for mean = 4. 3 1(b) n = 4000 > 50 and either np = 4 < 5 or p = 0.001 < 0.1 B1 Explicit values seen. 1

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2 The widths, w cm, of a random sample of 150 leaves of a certain kind were measured. The sample mean of w was found to be 3.12 cm. Using this sample, an approximate 95% confidence interval for the population mean of the widths in centimetres was found to be [3.01, 3.23]. (a) Calculate an estimate of the population standard deviation. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Explain whether it was necessary to use the Central Limit theorem in your answer to part (a). [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 3.12 + z × 150  = 3.23 Any z, but must be a z value. z = 1.96 B1 σ = 0.687 (3sf) [cm] A1 3 2(b) Yes, because population [of widths] not given to be normally distributed B1 Or ‘underlying distribution’ instead of population. Allow ‘yes, because population distribution not known’. Need both statements. 1

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3 The masses in kilograms of large and small bags of cement have the independent distributions N(50, 2.4) and N(26, 1.8) respectively. Find the probability that the total mass of 5 randomly chosen large bags of cement is greater than the total mass of 10 randomly chosen small bags of cement. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 Diff ~ N(5×50 − 10×26, 5×2.4 + 10×1.8) [= N(−10, 30) ] B1 For N and mean = ± (5×50 − 10×26) SOI B1 For var = 5×2.4 + 10×1.8 SOI 0 (' 10') '30'  [= 1.826] M1 Standardising with their values. 1 − Φ(‘1.826’) M1 For area consistent with their values. = 0.0339 or 0.034[0] (3sf) A1 5

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4 In this question you should not use an approximating distribution. At an election in Menham last year, 24% of voters supported the Today Party. A student wishes to test whether support for the Today Party has decreased since last year. He chooses a random sample of 25 voters in Menham and finds that exactly 2 of them say that they support the Today Party. Test at the 5% significance level whether support for the Today Party has decreased. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 H0: p = 0.24 H1: p < 0.24 B1 P(X < 2) = 0.7625 + 25×0.7624×0.24 + 25C2×0.7623×0.242 or 0.0010479 + 0.0082732 + 0.0313513 M1 Expression must be seen. No end errors. = 0.0407 A1 SC B1 for unsupported 0.0407. 0.0407 < 0.05 M1 For valid comparison. [Evidence to reject H0.] There is sufficient evidence to suggest that the support for the Today Party has decreased. A1FT FT their probability. In context, not definite, no contradictions. SC: if H1: p ≠ 0.24 and compare with 0.025; max B0 M1 A1 M1 A0. 5

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5 A random variable X has probability density function f given by ax - x 3 0 G x G 2, f ( )x = ) 0 otherwise, where a is a constant. (a) Show that a = 2 . 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(b) Find the median of X. 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(c) Find the exact value of E(X ). 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Mark scheme: 5(a) 2 3 0 ( )d ax x x   = 1 2 4 2 2 4 0 x x a      = 1 4 4 a  = 1 A1 Correct integration and substitute correct limits. a = 2 A1 AG Convincingly obtained and no errors seen. 3 5(b)   3 0 2 m x x dx    = 1 2 M1 Attempt integrate f(x) with limits 0 to m (or m to 2) and equate to 1 2. 4 2 4 m m  = 1 2 A1 For correct quartic in any form. m4 − 4m2 + 2 = 0 m2 = 4 16 8 2   [= 2 ± 2 ] M1 For solving their three term quartic to find m2. m = 2 2  or 0.765 (3sf) A1 4 Question Answer Marks Guidance 5(c) 2 2 4 0 (2 )d x x x   M1 Attempt to integrate xf(x). Ignore limits. 3 5 2 3 5 2 0 x x      A1 Correct integration and correct limits. [= 4 2 4 2 3 5  ] = 8 15 2 A1 OE For single exact term. 3

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6 The numbers of green sweets in 200 randomly chosen packets of Frutos are summarised in the table. Number of green sweets 0 1 2 3 2 3 Number of packets 32 50 97 21 0 (a) Calculate an unbiased estimate for the population mean of the number of green sweets in a packet of Frutos, and show that an unbiased estimate of the population variance is 0.783 correct to 3 significant figures. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The manufacturers of Frutos claim that the mean number of green sweets in a packet is 1.65 . Anji believes that the true value of the mean, n, is less than 1.65 . She uses the results from the 200 randomly chosen packets to test the manufacturers’ claim. (b) State suitable null and alternative hypotheses for the test. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Show that the result of Anji’s test is significant at the 5% level but not at the 1% level. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) It is given that Anji made a Type I error. Explain how this shows that the significance level that Anji used in her test was not 1%. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) 200 or 1.535 Σx2f = 627, 2 Est( )  = 2 200 '627' 199 200 ( '1.535' )  or   2 '307' 1 199 200 '627' M1 Use of a correct formula with their values. = 0.783 A1 AG Correctly obtained with no errors seen. 3 6(b) H0: µ = 1.65 H1: µ < 1.65 B1 Accept ‘population mean’ but not just ‘mean’. 1 Question Answer Marks Guidance 6(c) '1.535' 1.65 0.783 200   M1 Standardising with their mean. = −1.838 or −1.84 A1* Φ(0.05) and Φ(0.01) attempted M1 Or P(z < −`1.838`) attempted. SC: Condone Φ(0.025) = 2.807 and Φ(0.005) = 3.291 following two-tailed test in (b). −1.645 > −1.838 > −2.326 [Hence significant at 5% but not 1% level] DA1 AG = 0.033 and 0.05 > 0.033 > 0.01 SC: use of 1.54 or 1.53 for the mean leading to -1.645 > –1.758 > – 2.326 or –1.645 > -1.918 > –2.326 or 0.95 < 0.9606 or 0.9724 < 0.99 scores M1 M1 A1. Accept use of critical value method 1.535 < 1.547 or accept 1.65 > 1.638. 4 6(d) At the 1% level H0 is not rejected Or a Type I error can only occur if H0 is rejected. B1 OE 1

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7 The independent random variables X and Y have the distributions Po(1.9) and Po(2.2) respectively. (a) Find P ( X + Y 1 4 ) . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the probability that X = 2 given that X + Y 1 4 . 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(c) A sample of 60 randomly chosen pairs of values of X and Y is taken, and the value of X + Y is calculated for each pair. The sample mean of these 60 values is found. Find the probability that the sample mean of X + Y is less than 4.0 . 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Mark scheme: 7(a) λ = 1.9 + 2.2 [= 4.1] B1 e−4.1(1 + 4.1 + 2 4.1 2! + 3 4.1 3! ) or e−4.1(1 + 4.1 + 8.405 + 11.487) or 0.01657 + 0.06795 + 0.13929 + 0.19037 M1 Allow any λ. Allow one end error. Must see expression. = 0.414 (3sf) A1 SC: unsupported answer 0.414 scores B1 B1. 3 7(b) P(X + Y < 4 and X = 2) = P(2, 0 or 2, 1) M1 Stated or implied. = e−1.9× 2 1.9 2 (e−2.2 + e−2.2×2.2) [= 0.0957] M1 P(X = 2 | X + Y < 4) = '0.0957' '0.414' M1 Attempt P 4 and 2 ( ) P( 4) . X X Y X Y      Prob for denominator can be found in (a). 0.231 (3sf) A1 4 Question Answer Marks Guidance 7(c) E(X + Y) = 4.1 Var(X + Y) = 4.1 or Po(246) B1 SOI Normal and var = 4.1 60 Or normal and var = 246 M1 4.0 4.1 4.1 60   or totals method 240 246 246  or use of continuity correction M1 No mixed methods. Or continuity correction: 1 120 4.0 4.1 (4.1 60)    or 239.5 246 246 .  Condone incorrect continuity correction for M1. = −0.383 (3sf) A1 = −0.414 Φ(‘−0.383’) = 1 − Φ(‘0.383’) M1 Φ(‘−0.414’) = 1 − Φ(‘0.414’) = 0.351 (3sf) A1 = 0.340 or 0.339 6

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