2 The amounts of money, x dollars, that 24 people had in their pockets are summarised by Σ(x 36) = 60 and Σ(x 36) 2 = Find Σ x and Σ x 2.

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2 2 1 Ashok has 3 green pens and 7 red pens. His friend Rod takes 3 of these pens at random, without replacement. Draw up a probability distribution table for the number of green pens Rod takes. [4] 2 The amounts of money, x dollars, that 24 people had in their pockets are summarised by Σ(x 36) = 60 and Σ(x 36) 2 = Find Σ x and Σ x 2. [5] 3 Lengths of rolls of parcel tape have a normal distribution with mean 75 m, and 15% of the rolls have lengths less than 73 m. (i) Find the standard deviation of the lengths. [3] Alison buys 8 rolls of parcel tape. (ii) Find the probability that fewer than 3 of these rolls have lengths more than 77 m. [3] 4 Prices in dollars of 11 caravans in a showroom are as follows (i) Represent these prices by a stem-and-leaf diagram. [3] (ii) Write down the lower quartile of the prices of the caravans in the showroom. [1] (iii) 3 different caravans in the showroom are chosen at random and their prices are noted. Find the probability that 2 of these prices are more than the median and 1 is less than the lower quartile. [3] 5 A company set up a display consisting of 20 fireworks. For each firework, the probability that it fails to work is 0.05, independently of other fireworks. (i) Find the probability that more than 1 firework fails to work. [3] The 20 fireworks cost the company $24 each. 450 people pay the company $10 each to watch the display. If more than 1 firework fails to work they get their money back. (ii) Calculate the expected profit for the company. [4] 6 Ana meets her friends once every day. For each day the probability that she is early is 0.05 and the probability that she is late is Otherwise she is on time. (i) Find the probability that she is on time on fewer than 20 of the next 96 days. [5] (ii) If she is early there is a probability of 0.7 that she will eat a banana. If she is late she does not eat a banana. If she is on time there is a probability of 0.4 that she will eat a banana. Given that for one particular meeting with friends she does not eat a banana, find the probability that she is on time. [4] UCLES /61/O/N/12

3 3 7 (a) In a sweet shop 5 identical packets of toffees, 4 identical packets of fruit gums and 9 identical packets of chocolates are arranged in a line on a shelf. Find the number of different arrangements of the packets that are possible if the packets of chocolates are kept together. [2] (b) Jessica buys 8 different packets of biscuits. She then chooses 4 of these packets. (i) How many different choices are possible if the order in which Jessica chooses the 4 packets is taken into account? [2] The 8 packets include 1 packet of chocolate biscuits and 1 packet of custard creams. (ii) How many different choices are possible if the order in which Jessica chooses the 4 packets is taken into account and the packet of chocolate biscuits and the packet of custard creams are both chosen? [3] (c) 9 different fruit pies are to be divided between 3 people so that each person gets an odd number of pies. Find the number of ways this can be done. [5] UCLES /61/O/N/12

4 4 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. 9709/61/O/N/12

5 Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November P(0) = 7/10 6/9 5/8 = 210/720 P(1) = 3/10 7/9 6/8 3C1 = 378/720 P(2) = 3/10 2/9 7/8 3C2 = 126/720 P(3) = 3/10 2/9 1/8 = 6/720 (1/120) 2 Σx Σ36 = 60 Σx = = 804 OR x = 36 60/ 24 = Σx = = 804 Σx Σx + Σ36 2 = Σx 2 = (27000) OR /24 ( 2.5) 2 = sd 2 = 3.24 Σx 2 /24 (33.5) 2 = 3.24 Σx 2 = (27000) 3 (i) z = = 4 (i) σ = σ (ii) P(> 77) = 0.15 P(< 3) = P(0, 1, 2) = (0.85) C 1 (0.15)(0.85) C 2 (0.15) 2 (0.85) 6 = [4] [2] Finding P(0, 1, 2, 3) 1 or 2 correct 3 correct All correct Expanding brackets ie mult by 24 and subt 60 Dividing by 24 and subt from 36 Expanding brackets with 36Σx and Σ36 2 min Σx Σx + Σ36 2 = seen /24 (their coded mean) 2 seen Σx 2 /24 ( x ) 2 = their var if +ve seen o.e. ± correct z value accept ± Equation with 73, 75, σ and a z value Rounding to correct answer Prob rounding to 0.15 and C x p x (1 p) 8 x seen any p, 0<p< Correct stem Correct leaves Key: 14 3 represents dollars [3] Key need dollars (ii) LQ = [1] (iii) 5/11 4/10 2/9 3C2 = 4/33 (0.121) 5 C2 2C1 OR 11C3 5 (i) P(> 1) = 1 (0.95) 20 (0.95) 19 (0.05) 1 20C 1 = [3] Mult 3 diff fractions or (5C2 or 2C1) seen in num Mult by 3C2 o.e. or correct denom Binomial term 20Cx(0.05) x (0.95) 20 x Correct unsimplified expression Cambridge International Examinations 2012

6 Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November (ii) Profit 19 or 20 work = = 4020 Profit < 19 work = ( ) Expected profit = = $2830 ($2832) Or ( ) = (i) p = 0.2 µ = = 19.2 σ 2 = = P(< 20) = P( z < ) = P( z < ) = (ii) P(OT B) = 0.12 = = (8/59) [4] [5] [4] 4020 seen Multiplying 4020 by their (i) or their (1 (i)) Multiplying 480 by [1 their (i)] and subtracting Rounding to correct answer and seen standardising must have sq rt continuity correction either 19.5 or 20.5 correct area (> 0.5) correct value their 0.2 (0.6 or 0.4) as numerator of a fraction attempt at P(B) or P(NB) anywhere involving sum of 2 or 3 products correct unsimplified num or denom of a fraction correct answer 7 (a) 10! = !4! [2] 10! or 10 P 10 seen in num or alone or dividing by 5! 4! only Correct final answer (b) (i) 8P 4 or 8 C 4 4! = 1680 (ii) 6C2 4! = 360 OR 6P4 or = 360 (c) A B C = 9C1 8C1 7C7 (oe) 3 C 1 = = 9C1 8C3 5C5(oe) 3! = = 9C3 6C3 3C3 (oe) = 1680 Total = 4920 ways [2] [5] 8P 4 or 8C 4 oe seen allow extra multiplication 6C2 or 6P2 seen multiplied Mult by 4! Award full marks Summing at least two options of 1, 1, 7 or 1, 3, 5 or 3, 3, 3 Mult an option by 3C1 or 3! or 3C3 Any one of the 2 nd term being xcy seen mult, fitting with the first ( x could be 2, 4, 5, 6 or 8) and correspnding y Any of unsimplified 72, 504 or 1680 seen Cambridge International Examinations 2012

MATHEMATICS FRIDAY 23 MAY 2008 ADVANCED SUBSIDIARY GCE 4732/01. Probability & Statistics 1. Morning Time: 1 hour 30 minutes

MATHEMATICS FRIDAY 23 MAY 2008 ADVANCED SUBSIDIARY GCE 4732/01. Probability & Statistics 1. Morning Time: 1 hour 30 minutes ADVANCED SUBSIDIARY GCE 4732/01 MATHEMATICS Probability & Statistics 1 FRIDAY 23 MAY 2008 Additional materials (enclosed): Additional materials (required): Answer Booklet (8 pages) List of Formulae (MF1)