Cambridge A Level Mathematics 9709 — 2016 May/June Paper 6 · Variant 1

1 The height of maize plants in Mpapwa is normally distributed with mean 1.62 m and standard deviation 3 m. The probability that a randomly chosen plant has a height greater than 1.8 m is 0.15. Find the value of 3. [3]

Mark scheme: Question Answer Marks Guidance 1 z = 1.037 B1 Rounding to 1.04 1.8 − 1.62 1.037 = M1 Standardising attempt allow cc no sq rt σ must have a z-value i.e. not 0.8023 or 0.5596. σ = 0.18/1.037 = 0.174 A1 [3]

More questions on The normal distribution

3 The probability that the school bus is on time on any particular day is 0.6. If the bus is on time the probability that Sam the driver gets a cup of coffee is 0.9. If the bus is not on time the probability that Sam gets a cup of coffee is 0.3. (i) Find the probability that Sam gets a cup of coffee. [2] (ii) Given that Sam does not get a cup of coffee, find the probability that the bus is not on time. [3]

Mark scheme: 3 (i) P (cup of coffee) = 0.6×0.9 + 0.4× 0.3 M1 Summing two 2-factor probabilities = 0.66 A1 [2] Correct answer accept 0.660 (ii) P(Not on time no cup of coffee) M1 0.4×0.7 seen as num or denom of a fraction P (noton time ∩ nocup) 0.4 × 0.7 = = M1 Attempt at P(no cup) as 0.1×p1 + 0.7×p2 P (nocup) 1 − 0.66 or as 1 – (i) seen anywhere 0.28 = = 0.824 A1 [3] 0.34

More questions on Probability

6 (a) (i) Find how many numbers there are between 100 and 999 in which all three digits are different. [3] (ii) Find how many of the numbers in part (i) are odd numbers greater than 700. [4] (b) A bunch of flowers consists of a mixture of roses, tulips and daffodils. Tom orders a bunch of 7 flowers from a shop to give to a friend. There must be at least 2 of each type of flower. The shop has 6 roses, 5 tulips and 4 daffodils, all different from each other. Find the number of different bunches of flowers that are possible. [4]

Mark scheme: 6 (a) (i) 9 × 9 × 8 M1 M1 Logical listing attempt = 648 A1 [3] OR 900 – 28 × 9 = 648 (ii) (7….in 1 × 8 × 4 = 32 ways M1 Listing #s starting with 7 or 9 and ending odd 8 …in 1 × 8 × 5 = 40 M1 9… in 1 × 8 × 4 = 32 M1 Total 104 ways A1 [4] (b) R(6 ) T(5) D(4) 2 2 3 = 6C2×5C2×4C3 = 600 M1 Mult 3 combs, 6Cx×5Cy×4Cz 2 3 2 = 6C2×5C3×4C2 = 900 M1 Summing 2 or 3 three-factor outcomes 3 2 2 = 6C3×5C2×4C2 = 1200 can be perms, + instead of × A1 2 options correct unsimplified Total = 2700 A1 [4]

More questions on Permutations and combinations

7 The amounts spent by 160 shoppers at a supermarket are summarised in the following table. Amount spent ($x) 0 < x ≤30 30 < x ≤50 50 < x ≤70 70 < x ≤90 90 < x ≤140 Number of shoppers 16 40 48 26 30 (i) Draw a cumulative frequency graph of this distribution. [4] (ii) Estimate the median and the interquartile range of the amount spent. [3] (iii) Estimate the number of shoppers who spent more than $115. [2] (iv) Calculate an estimate of the mean amount spent. [2]

Mark scheme: 7 (i) cf 16, 56, 104, 130, 160 M1 Attempt at cf table (up to 160) no graph needed accept %cf but give final cf 160 B1 linear scale minimum 0 to 160 and 0 to 120 120 80 M1 Attempt to plot points at (30, 16), (50, 56), (70,104), (90, 130), (140, 160) up 40 to 2 errors can have a polygon 50 100 150 A1 [4] All points correct from their scale and Amount spent $ joined up, with (0,0) as well (ii) median $59 B1 accept 57–60 or ft their graph if used lb, midpts instead of ub or assume linear interpolation. IQR = 82 – 43 M1 Subt a (sensible) LQ from a sensible = $39 A1 [3] UQ (generous) Ans ft need a cf graph and UQ 80–84, LQ (iii) 160 – 149 M1 41–46 = 11 A1 [2] Subtracting from 160 can be implied OR 115 is mid pt of last interval so # of Correct answer accept 9–16 shoppers is 30/2 = 15 (can be implied) (iv) mean = (15×16+ 40×40 +60×48+ 80×26 + M1 Using Σxf/160 with mid-points 115×30)/160 A1 [2] = 10250/160 = $64.1= $64.1

More questions on Representation of data