Mathematics 9709/63 October/November 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
There are 300 students at a music college. All students play exactly one of the guitar, the piano or the flute. The numbers of male and female students that play each of the instruments are given in the following table.
Guitar Piano Flute
Female students 62 35 43
Male students 78 40 42
(i) Find the probability that a randomly chosen student at the college is a male who does not play the piano. [1]
(ii) Determine whether the events ‘a randomly chosen student is male’ and ‘a randomly chosen student does not play the piano’ are independent, justifying your answer. [2]
Mark scheme
1(i) 120 B1 OE
= 0.4
300
1
1(ii) 160 225 8
3 2 M1 P(M) × P(P ' ) seen
P(male) × P(not piano) = × × = Can be unsimplified but the events must be named in a
300 300 15 4 5
product
120 2 A1 Numerical comparison and correct conclusion
As P(male ∩ not piano) also = =
300 5
The events are Independent
Alternative method for question 1(ii)
120 225 M1 P(M|P ' ) or P(P ' |M) unsimplified seen with their probs with
P(male ∩ not piano) = ; P(not piano) = correctly named events
300 300
120 A1 Numerical comparison with P(M) or P(P ' ) and correct
120 8 conclusion
P(M | not piano) = 300 = = = P(male)
225 225 15
300
or
120
120 3
P(not piano | M) = 300 = = = P(not piano)
160 160 4
300
Therefore the events are Independent
2
(i) How many different arrangements are there of the 9 letters in the word CORRIDORS? [2]
(ii) How many different arrangements are there of the 9 letters in the word CORRIDORS in which the first letter is D and the last letter is R or O? [3]
Mark scheme
2(i) 9! B1 9! Divided by at least one of 2! or 3!
= 30240
2!3!
B1 Exact value
2
2(ii) 7! B1 7! Seen alone or as numerator in a term, can be multiplied
D _ _ _ _ _ _ _ R: = 1260 not + or –
2!2!
7!
D _ _ _ _ _ _ _ O: = 840
3!
B1 One term correct, unsimplified
Total = 2100 B1 Final answer
3
A sports team of 7 people is to be chosen from 6 attackers, 5 defenders and 4 midfielders. The team must include at least 3 attackers, at least 2 defenders and at least 1 midfielder.
(i) In how many different ways can the team of 7 people be chosen? [4]
The team of 7 that is chosen travels to a match in two cars. A group of 4 travel in one car and a group of 3 travel in the other car.
(ii) In how many different ways can the team of 7 be divided into a group of 4 and a group of 3? [2]
Mark scheme
3(i) 3A 2D 2M : 6C3 × 5C2 × 4C2 (= 1200) M1 6
Cx × 5Cy × 4Cz, x + y + z = 7
4A 2D 1M : 6C4 × 5C2 × 4C1 (= 600)
3A 3D 1M : 6C3 × 5C3 × 4C1 (= 800)
A1 2 correct products, allow unsimplified
M1 Summing their totals for 3 correct scenarios only
Total = 2600 A1 Correct answer
SC1 6C3 × 5C2 × 4C1 × 9C1 = 7200
4
7
3(ii) 7
C4 × 1 B1 C3 or 7C4 seen anywhere
35 B1
2
The heights of students at the Mainland college are normally distributed with mean 148 cm and standard deviation 8 cm.
(i) The probability that a Mainland student chosen at random has a height less than h cm is 0.67.
Find the value of h. [3]
120 Mainland students are chosen at random.
(ii) Find the number of these students that would be expected to have a height within half a standard deviation of the mean. [4]
Mark scheme
4(i) P(h < 148) = 0.67 B1 z = ±0.44 seen
h − 148 M1 (h − 148)
= 0.44 z-value = ±
8 8
151.52 ≈ 152 A1 CAO
3
4(ii) 144 − 148 152 − 148 M1 Using ± standardisation formula for either 144 or 152,
P(144 < X < 152) = P <Z < µ = 148, σ = 8 and no continuity correction, allow σ2 or √σ
8 8
1 1 M1 Correct final area legitimately obtained from
= P − < Z < = 0.6915 – (1 – 0.6915) = 2 × 0.6915 – 1 phi(their z2) – phi(their z1)
2 2
= 0.383 A1 Final probability answer
0.383 × 120 = 45.96 B1FT Their prob (to 3 or 4 sf) × 120, rounded to a whole number
Accept 45 or 46 only or truncated
4
Last Saturday, 200 drivers entering a car park were asked the time, in minutes, that it had taken them to travel from home to the car park. The results are summarised in the following cumulative frequency table.
Time (t minutes) t ≤ 10 t ≤ 20 t ≤ 30 t ≤ 50 t ≤ 70 t ≤ 90
Cumulative frequency 16 50 106 146 176 200
(i) On the grid, draw a cumulative frequency graph to illustrate the data. [2]
(ii) Use your graph to estimate the median of the data. [1]
(iii) For 80 of the drivers, the time taken was at least T minutes. Use your graph to estimate the value of T . [2]
(iv) Calculate an estimate of the mean time taken by all 200 drivers to travel to the car park. [4]
Mark scheme
5(i) Correct labels and scales B1 Axes labelled ‘cumulative frequency’ (or cf) and ‘time (or t)
[in] min(utes)’, linear scales from 0 to 90 and 0 to 200 with
at least 3 values marked on each axis.
7 correctly plotted points above upper boundaries joined in a curve or line B1 (0, 0); (10, 16); (20, 50); (30, 106); (50, 146); (70,176);
segments (90,200)
2
5(ii) 29 B1 28 ⩽ median ⩽ 30
1
5(iii) 120 seen M1 For seeing 120 in a calculation or marked on the graph
37 A1FT 36 ⩽ Ans ⩽ 39 or FT from their graph
SC1 unsupported answer in range
2
5(iv) Frequencies 16 34 56 40 30 24 B1 Seen. Allow unsimplified
5 × 16 + 15 × 34 + 25 × 56 + 40 × 40 + 60 × 30 + 80 × 24 M1 At least 4 correct midpoints (5, 15, 25, 40, 60, 80) used in a
Est. Mean = calculation
200
7310 M1 Summing products of their 6 mid-points (not lower or upper
200 bound or class width) × their frequencies / 200 (or their ∑f),
unsimplified
36.55 A1 Accept 36.6
4
A box contains 3 red balls and 5 white balls. One ball is chosen at random from the box and is not returned to the box. A second ball is now chosen at random from the box.
(i) Find the probability that both balls chosen are red. [1]
(ii) Show that the probability that the balls chosen are of different colours is 15
28
. [2]
(iii) Given that the second ball chosen is red, find the probability that the first ball chosen is red. [2]
The random variable X denotes the number of red balls chosen.
(iv) Draw up the probability distribution table for X . [2]
(v) Find Var X . [3]
Mark scheme
6(i) 3 2 3 B1 OE
P(RR) = × =
8 7 28
1
6(ii) P(RW) + P(WR) M1 Method shown, numerical calculations identified, may
3 5 5 3 include replacements
× + ×
8 7 8 7
15 A1 AG, Fully correct calculations
=
28
Alternative method for question 6(ii)
1 – (P(RR) + P(WW) M1 Method shown, numerical calculations identified, may
3 5 4 include replacements
1− + ×
28 8 7
15 A1 AG, Fully correct calculations
=
28
2
6(iii) 3 2 3 M1 Conditional probability formula used consistent with their
× probabilities or correct
their (i) 8 7
P(first red|second red) = = = 28
5 3 3 2 5 3 21
their (i) + × × + ×
8 7 8 7 8 7 56
2 A1 OE
=
7
2
6(iv) B1 Probability distribution table with correct values of x and at
x 0 1 2
least one correct probability placed. Extra x values allowed
p with probability of zero stated.
10 15 3
28 28 28
B1FT Fully correct
FT P(2) = their (i), P(1) = their (ii), Σ(p) = 1.
2
6(v) 30 12 42 3 B1 May be implied by use in variance formula
E(X) = + = =
56 56 46 4
30 24 3
2 M1 Substitute into correct variance formula, must have
Var(X) = + − their ‘ – their mean2’
56 56 4
Must be for 2 or more non-zero x-values
45 A1 Correct final answer
or 0.402
112
3
7(i)(a) P(0, 1, 2) = 6C0 0.30 0.76 + 6C1 0.31 0.75 + 6C2 0.32 0.74 M1 Binomial term of form 6Cxpx(1 – p)6– x 0 < p < 1
any p, x ≠ 6,0
0.1176 ... + 0.3025 ... + 0.3241 ... A1 Correct unsimplified answer
0.744 A1 Correct final answer
3
7(i)(b) P(support neither choir) = 1 – (0.3 + 0.45) = 0.25 M1 0.25n seen alone, 1 < n ⩽ 6
P(6 support neither choir) = 0.256 A1 Correct final answer
1
= 0.000244 or
4096
2
A competition is taking place between two choirs, the Notes and the Classics. There is a large audience for the competition.
³ 30% of the audience are Notes supporters.
³ 45% of the audience are Classics supporters.
³ The rest of the audience are not supporters of either of these choirs.
³ No one in the audience supports both of these choirs.
(i) A random sample of 6 people is chosen from the audience.
(a) Find the probability that no more than 2 of the 6 people are Notes supporters. [3]
(b) Find the probability that none of the 6 people support either of these choirs. [2]
(ii) A random sample of 240 people is chosen from the audience. Use a suitable approximation to find the probability that fewer than 50 do not support either of the choirs. [5]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(ii) Mean = 240 × 0.25 = 60 B1FT Correct unsimplified 240p and 240pq
Variance = 240 × 0.25 × 0.75 = 45 where p =their P(support neither choir) or 0.25
49.5 − 60 M1 Substituting their µ and σ (condone σ2) into the
P(X < 50) = P Z < = P(Z < –1.565) ±Standardisation Formula with a numerical value for ‘49.5’.
45
M1 Using continuity correction 49.5 or 50.5 within a
standardisation expression
1 – 0.9412 M1 Appropriate area Φ from standardisation formula P(z< …)
in final solution, (< 0.5 if z is –ve, > 0.5 if z is +ve)
0.0588 A1 Correct final answer
5
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Mathematics 9709/63 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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