Cambridge A Level Mathematics 9709 — 2018 May/June Paper 6 · Variant 3

1 The masses in kilograms of 50 children having a medical check-up were recorded correct to the nearest kilogram. The results are shown in the table. Mass (kg) 10 −14 15 −19 20 −24 25 −34 35 −59 Frequency 6 12 14 10 8 (i) Find which class interval contains the lower quartile. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) On the grid, draw a histogram to illustrate the data in the table. [4]

Mark scheme: 1(i) 15–19 (kg) cao B1 kg not necessary; condone 14.5 – 19.5 Total: 1 1(ii) fd = 1.2, 2.4, 2.8, 1, 0.32 3 fd 2 1 0 9.5 19.5 39.5 59.5 Mass (kg) M1 Attempt at fd [f/(attempt at cw)] or scaled freq (may be implied by 4 correct) A1 Correct heights seen on diagram with linear vertical scale from (x, 0) B1 Correct bar widths (1:1:1:2:5) visually no gaps with linear horizontal scale from (9.5,y) and first bar starting at (9.5, y) B1 Histogram, using attempted fds, with labels (mass, kg and fd seen) and at least 3 linearly spaced values on each axis. Horizontal axis must range from at least 9.5 to 59.5 If horizontal axis clearly starts from zero, either a break in the scale must be indicated or the scale must be linear from zero.

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2 The random variable X has the distribution N −3, 32 . The probability that a randomly chosen value of X is positive is 0.25. (i) Find the value of 3. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that, of 8 random values of X, fewer than 2 will be positive. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) z = 0.674 B1 z value ±0.674 0.674 = σ 3 0 − − M1 ±Standardising with 0 and equating to a z-value σ = 4.45 A1 Correct answer www ie not ignoring a minus sign Total: 3 2(ii) P(0, 1) = (0.75)8 + 8C1(0.25)(0.75)7 M1 Any bin of form 8Cx(0.75)x (0.25)8–xany x M1 Correct unsimplified answer, may be implied by numerical values 0.1001+ 0.2670 = 0.367 A1 Correct answer Method 2 1 – P(8,7,6,5,4,3,2) = 1 – (0.25)8 – 8C1(0.75)(0.25)7 – … – 8C2(0.75)6 (0.25)2 = 0.367 M1 Any bin of form 8Cx(0.75)x (0.25)8–xany x M1 Correct unsimplified answer A1 Correct answer Total: 3

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3 The members of a swimming club are classified either as ‘Advanced swimmers’ or ‘Beginners’. The proportion of members who are male is x, and the proportion of males who are Beginners is 0.7. The proportion of females who are Advanced swimmers is 0.55. This information is shown in the tree diagram. Advanced 0.55 swimmers Females Beginners Advanced x swimmers Males 0.70 Beginners For a randomly chosen member, the probability of being an Advanced swimmer is the same as the probability of being a Beginner. (i) Find x. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given that a randomly chosen member is an Advanced swimmer, find the probability that the member is male. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) (1– x) and 0.45 (or 0.3 ) B1 Seen, either on tree diagram or elsewhere Beginners: 0.7 × x + ‘0.45’ × ‘(1 – x)’ = 0.5 Or Advanced: ‘0.3’ × x + 0.55 × ‘(1 – x)’ = 0.5 Or 0.7 × x + ‘0.45’× ‘(1 – x)’ = ‘0.3’ × x + 0.55× ‘(1 – x)’ M1 One of the three correct probability equations x = 0.2 oe A1 Correct answer Total: 3 3(ii) P(M A) = ) ( ) ( A P A M P ∩ = 5.0 3.0 2.0 × M1 ‘i’ × 0.3 as num or denom of a fraction M1 0.5 (or (1 – ‘i’) × 0.55 + ‘i’ × 0.3 unsimplified) seen as denom of a fraction = 0.12 3 25       A1 Correct answer Total: 3

More questions on Probability

4 Farfield Travel and Lacket Travel are two travel companies which arrange tours abroad. The numbers of holidays arranged in a certain week are recorded in the table below, together with the means and standard deviations of the prices. Number of Mean price Standard holidays deviation $ Farfield Travel 30 1500 230 Lacket Travel 21 2400 160 (i) Calculate the mean price of all 51 holidays. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The prices of individual holidays with Farfield Travel are denoted by $xF and the prices of individual holidays with Lacket Travel are denoted by $xL. By first finding Σ x2F and Σ x2L, find the standard deviation of the prices of all 51 holidays. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) M1 Multiply by 30 and 21, summing and dividing total by 51 45000 50400 51 +       = 1870 (1870.59) A1 correct answer (to 3sf) Total: 2 4(ii) 2302 = 2 2 1500 30 − Σ Fx so 2 F x Σ = 69 087 000 M1 One correct substitution into a correct variance formula A1 Correct ΣxF 2 (rounding to 69 000 000 2sf) 1602 = 2 2 2400 21 − Σ Lx so 2 L x Σ = 121 497 600 A1 Correct ΣxL 2 (rounding to 121 000 000 3sf) New var = 69087000 121497600 51 + – 1870.5882 = 237 853 M1 using ‘ΣxF 2’+ ‘’ ΣxL 2 dividing by 51 and subtracting ‘i’ squared. (Correct ‘ΣxF 2’ + ‘’ ΣxL 2 = 190 584 600) New sd = 488 A1 Correct answer accept anything between 486 and 490 Total: 5

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5 A game is played with 3 coins, A, B and C. Coins A and B are biased so that the probability of obtaining a head is 0.4 for coin A and 0.75 for coin B. Coin C is not biased. The 3 coins are thrown once. (i) Draw up the probability distribution table for the number of heads obtained. 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(ii) Hence calculate the mean and variance of the number of heads obtained. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) P(1) = 0.4 × 0.25 × 0.5 + 0.6 × 0.75 × 0.5 + 0.6 × 0.25 × 0.5 = 0.35 P(2) = 0.4 × 0.75 × 0.5 + 0.4 × 0.25 × 0.5 + 0.6 × 0.75 × 0.5 = 0.425 P(3) = 0.4 × 0.75 × 0.5 = 0.15 B1 P(1), P(2) and P(3) M1 Multiply 3 probabilities together from 0.4 or 0.6, 0.25 or 0.75, 0.5 with or without a table No of heads 0 1 2 3 Prob 0.075 3 40       0.35 7 20       0.425 17 40       0.15 3 20       M1 Summing 3 probabilities for P(1) or P(2) with or without a table B1 One correct probability seen. A1 All correct in a table Total: 5 5(ii) E(X) = 0.35 + 2 × 0.425 + 3 × 0.15 = 1.65 33 oe 20       M1 Correct unsimplified expression for the mean using their table, ∑p = 1; can be implied by correct answer 5(ii) Var(X) = 0.35 + 4 × 0.425 + 9 × 0.15 – 1.652 M1 Correct unsimplified expression for the variance using their table and their mean2 subtracted, ∑p = 1 = 0.678 (0.6775) 271 oe 400       A1 Correct answer Total: 3

More questions on Probability

6 The diameters of apples in an orchard have a normal distribution with mean 5.7 cm and standard deviation 0.8 cm. Apples with diameters between 4.1 cm and 5 cm can be used as toffee apples. (i) Find the probability that an apple selected at random can be used as a toffee apple. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) 250 apples are chosen at random. Use a suitable approximation to find the probability that fewer than 50 can be used as toffee apples. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(ii) B1ft Correct unsimplified mean and var – ft their prob for (i) providing (0 < p < 1) Implied by 34.944 5.911 σ = = P(< 50) = P       − < 944 . 34 42 5. 49 z = P(z < 1.2687) M1 ± Standardising using 50, their mean and sd; must have sq rt. M1 49.5 or 50.5 seen as a cc = Φ(1.2687) M1 Correct area Φ(> 0.5 for + z and < 0.5 for –z)in their final answer = 0.898 A1 Correct final answer Total: 5

More questions on Discrete random variables

7 Find the number of ways the 9 letters of the word SEVENTEEN can be arranged in each of the following cases. (i) One of the letter Es is in the centre with 4 letters on either side. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) No E is next to another E. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 5 letters are chosen from the 9 letters of the word SEVENTEEN. (iii) Find the number of possible selections which contain exactly 2 Es and exactly 2 Ns. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iv) Find the number of possible selections which contain at least 2 Es. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) ****E**** Other letters arranged in 8! 2!3! = 3360 ways M1 A1 Correct final answer www OR 8× 7× 6×5× 4× 4×3× 2×1 4!2! = 3360 ways M1 Correct numerator (161 280) A1 Correct final answer www Total: 2 7(ii) * * * * * Arrangements other letters × ways Es inserted = 5! 2! × 6 4 C 6 4 5! 2! 4! P   ×     M1 k mult by 6 4 C or 6 4P oe (ways to insert Es ignoring repeats), k can = 1 or k mult by 5! 2! M1 Correct unsimplified expression or 5! 2! × 6 4P = 900 ways A1 Correct answer OR Total no of ways – no of ways with Es touching 9!/(4! × 2!) – … or 7 560 – … 6! 2! + 6 2 5 2 P ! × ! + 6 2P 5! 2! 2! × + 6 3P 5! 2! 2! × = 360 + 1800 + 900 + 3600 = 6660 M1 7560 unsimplified – k M1 Attempting to find four ways of Es touching (4 Es, 3Es and a single, 2 lots of 2 Es, 2 Es and 2 singles) 7 560 – 6 660 = 900 A1 Correct answer Question Answer Marks Guidance 7(ii) OR Adding the number of ways with the first E in the 1st (E1), 2nd (E2) or 3rd (E3) position. 5! 2! (E1 + E2 + E3) where E1 = 10, E2 = 4, E3 = 1 5! 2! (E1 + E2 + E3) M1 For any values for E1, E2 and E3 M1 For any two correct values of E1, E2 and E3 600 + 240 + 60 = 900 A1 Correct answer Total: 3 7(iii) EENN* in 3 ways B1 Numerical value must be stated Total: 1 Question Answer Marks Guidance 7(iv) EE *** with no N: 1 way EEN** 3C2 or listing 3 ways EENN* 3 ways from (iii) M1 Identifying the three different scenarios of EE, EEE or EEEE A1 Total no of ways with two Es (7 or 3 + 3 + 1) EEE** with no N: 3 ways EEEN* 3 ways EEENN 1 way A1 Total no. of ways with 3 Es (7) EEEE* no N 3 ways EEEEN 1 way Total 18 ways A1 Correct answer stated Method List containing ways with 2Es, 3Es and 4Es List containing at least 8 correct different ways List of all 18 correct ways Total 18 M1 At least 1 option listed for each of EE^^^, EEE^^, EEEE^ A1 Ignore repeated options A1 Ignore repeated/incorrect options A1 Correct answer stated Total: 4

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