Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 6 · Variant 1

2 The people living in two towns, Mumbok and Bagville, are classified by age. The numbers in thousands living in each town are shown in the table below. Mumbok Bagville Under 18 years 15 35 18 to 60 years 55 95 Over 60 years 20 30 One of the towns is chosen. The probability of choosing Mumbok is 0.6 and the probability of choosing Bagville is 0.4. Then a person is chosen at random from that town. Given that the person chosen is between 18 and 60 years old, find the probability that the town chosen was Mumbok. [5]

Mark scheme: 2 either 55/90 (11/18) B1 oe or 95/160 (19/32) seen P(M and 18 – 60) = 0.6 × 55/90 M1 0.6 mult by 55/90 seen as num / denom = 0.367 (11 / 30) of a fraction P(18 – 60) = 0.6 × 55/90 + 0.4 × 95/160 M1 Summing 2 two-factor products seen (= 29/48 or 0.604) anywhere P ( M ∩ 18 − 60 ) P(M │ 18 – 60) = A1 Correct unsimplified answer seen as P (18 − 60 ) num/denom of a fraction = 88/145 (0.607) A1 5 Correct answer

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4 The following are the house prices in thousands of dollars, arranged in ascending order, for 51 houses from a certain area. 253 270 310 354 386 428 433 468 472 477 485 520 520 524 526 531 535 536 538 541 543 546 548 549 551 554 572 583 590 605 614 638 649 652 666 670 682 684 690 710 725 726 731 734 745 760 800 854 863 957 986 (i) Draw a box-and-whisker plot to represent the data. [4] An expensive house is defined as a house which has a price that is more than 1.5 times the interquartile range above the upper quartile. (ii) For the above data, give the prices of the expensive houses. [2] (iii) Give one disadvantage of using a box-and-whisker plot rather than a stem-and-leaf diagram to represent this set of data. [1]

Mark scheme: 4 (i) B1 Linear scale or 5 values shown and labels or in heading, need thousands of dollars, B1 Correct median B1 Correct quartiles 200 300 400 500 600 700 800 900 1000 House price, 000’s dollars B1 4 Correct end points of whiskers not through box (ii) 1.5 × 170 = 255 M1 Mult their IQ range by 1.5 Expensive houses above 690 + 170 × 1.5 = 945 i.e. 957 and 986 thousands of dollars A1 2 Correct answers from correct wkg need thousands of dollars (iii) doesn’t show all the data items B1 1 Need to see ‘individual items’ oe GCE AS/A LEVEL – October/November 2013 9709 61

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5 Lengths of a certain type of carrot have a normal distribution with mean 14.2 cm and standard deviation 3.6 cm. (i) 8% of carrots are shorter than c cm. Find the value of c. [3] (ii) Rebekah picks 7 carrots at random. Find the probability that at least 2 of them have lengths between 15 and 16 cm. [6]

Mark scheme: 5 (i) z = –1.406 B1 Rounding to ± .1 41 seen c − 142. = −.1406 M1 Standardising allow sq rt no cc 6.3 c = 9.14 A1 3 Correct answer  15 − 142.   16 − 142.  (ii) P   < z <   M1 2 attempts at standardising no cc no sq rt  6.3   6.3  = Φ(0.5) – Φ(0.222) M1 Subt two Φs (indep mark) = 0.6915 – 0.5879 = 0.1036 A1 Needn’t be entirely accurate, rounding to 0.10 P(at least 2) = 1 – P(0, 1) M1 Binomial term with 7Crpr(1–p)7–r seen r ≠ 0 = 1 – (0.8964)7 – (0.8964)6(0.1036)7C1 any p < 1 = 1 – 0.8413 M1 1 – P(0), 1 – P(1), 1 – P(0, 1) seen their p = 0.159 A1 6 Correct answer accept 3sf rounding to 0.16

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6 A shop has 7 different mountain bicycles, 5 different racing bicycles and 8 different ordinary bicycles on display. A cycling club selects 6 of these 20 bicycles to buy. (i) How many different selections can be made if there must be no more than 3 mountain bicycles and no more than 2 of each of the other types of bicycle? [4] The cycling club buys 3 mountain bicycles, 1 racing bicycle and 2 ordinary bicycles and parks them in a cycle rack, which has a row of 10 empty spaces. (ii) How many different arrangements are there in the cycle rack if the mountain bicycles are all together with no spaces between them, the ordinary bicycles are both together with no spaces between them and the spaces are all together? [3] (iii) How many different arrangements are there in the cycle rack if the ordinary bicycles are at each end of the bicycles and there are no spaces between any of the bicycles? [3]

Mark scheme: 6 (i) M R O 3 1 2 = 7C3 × 5C1 × 8C2 = 4900 M1 Summing more than one 3term option involving combs (can be added) 3 2 1 = 7C3 × 5C2 × 8C1 = 2800 M1 Mult 3 combs only (indep) 2 2 2 = 7C2 × 5C2 × 8C2 = 5880 A1 1 option correct unsimplified Total = 13580 A1 4 Correct answer (ii) 4 groups in 4! ways M1 4! seen mult by something 3 mountain in 3! ways 2 ordinary in 2! ways M1 Mult by 3! for racing or 2! for ordinary 4! × 3! × 2 = 288 A1 3 Correct answer (iii) e.g. s O x x x x O s s s M1 2! or 4! seen mult Ordinary in 2! Rest of bikes in 4! M1 Mult by 5 (ssssb) Bikes and spaces 5 groups in 5 ways 2! × 4! × 5 = 240 A1 3 Correct answer GCE AS/A LEVEL – October/November 2013 9709 61

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