Mathematics 9709/61 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.
When Shona goes to college she either catches the bus with probability 0.8 or she cycles with probability 0.2. If she catches the bus, the probability that she is late is 0.4. If she cycles, the probability that she is late is x. The probability that Shona is not late for college on a randomly chosen day is 0.63. Find the value of x. [3]
Mark scheme
1 0.8 × 0.6 + 0.2 (1 − x ) = 0.63 M1 Equation of form 0.8 × A + 0.2 × B = C, A,B involving 1 – x and
0.6 or 0.4 and C = 0.63 or 0.37
0.2 x = 0.05 M1 Correct unsimplified equation
x = 0.25 A1
Alternative method for question 1
0.8 × 0.4 + 0.2 x = 1 − 0.63 M1 Equation of form 0.8 × A + 0.2 × B = C, A,B involving x and 0.6
or 0.4 and C = 0.63 or 0.37
0.2 x = 0.05 M1 Correct unsimplified equation
x = 0.25 A1
3
Annan has designed a new logo for a sportswear company. A survey of a large number of customers found that 42% of customers rated the logo as good.
(i) A random sample of 10 customers is chosen. Find the probability that fewer than 8 of them rate the logo as good. [3]
(ii) On another occasion, a random sample of n customers of the company is chosen. Find the smallest value of n for which the probability that at least one person rates the logo as good is greater than 0.995. [3]
Mark scheme
2(i) 1 – (10C2 0.428 0.582 + 10C9 0.429 0.581 + 0.4210) M1 Binomial term of form
10
Capa(1 – p)b 0 < p < 1 any p, 0 ⩽ a,b ⩽ 10
A1 Correct unsimplified expression
0.983 A1
3
2(ii) 1 – P(0) > 0.995 0.58n < 0.005 M1 Equation or inequality involving 0.58n or 0.42n and 0.995 or
0.005
log 0.005 M1 Attempt to solve using logs or Trial and Error. May be implied
n> by their answer (rounded or truncated)
log 0.58
n > 9.727
n = 10 A1 CAO
3
The mean and standard deviation of 20 values of x are 60 and 4 respectively.
(i) Find the values of Σ x and Σ x2 . [3]
Another 10 values of x are such that their sum is 550 and the sum of their squares is 40 500.
(ii) Find the mean and standard deviation of all these 30 values of x. [4]
Mark scheme
3(i) ∑ x = 60 × 20 = 1200 B1
∑ x2 M1 Correct variance formula used, condone = 4
− 602 = 42
20
∑ x 2 = 3616 × 20 = 72320 A1 Exact value
3
3(ii) ∑ x = 1200 + 550 = 1750 M1 Summing both values of ∑ x and ∑ x 2
2
∑ x = 72320 + 40500 = 112800
their 1750 B1FT FT their 1750 (not 550 or 1200)/their(20+10), accept
Mean = = 58.3 unsimplified
30
their 112820 their 1750
2 M1 substitute their Σx and Σx2 into correct variance formula
Variance = − ( = 357.89 )
30 30
s.d. = 18.9 A1
4
In a probability distribution the random variable X takes the values −1, 0, 1, 2, 4. The probability distribution table for X is as follows.
x −1 0 1 2 4
P X = x 1 p p 3 4p
4 8
(i) Find the value of p. [2]
(ii) Find E X and Var X . [3]
(iii) Given that X is greater than zero, find the probability that X is equal to 2. [2]
Mark scheme
4(i) 1 3 M1 Unsimplified sum of probabilities equated to 1
+ + p + p + + 4p =1
4 8
1 A1 If method FT from their incorrect (i), expressions for E(X) and
p= Var(X) must be seen unsimplified with all probabilities <1,
16
condone not adding to 1
2
4(ii) 1 1 6 25 M1 May be implied by use in Variance, accept unsimplified
[E(X)] = − + + + 1 =
4 16 8 16
1 1 12 16 25
2 M1 Substitute into correct variance formula, must have
[Var(X)] = + + + − their ‘– their mean2’
4 16 8 4 16
863 A1 OE
or 3.37
256
3
4(iii) 3 M1 Conditional probability formula used consistent with their
P ( X = 2) probabilities
P(X = 2|X >0) = = 8
P(X > 0) 11
16
6 A1
or 0.545
11
2
Ransha measured the lengths, in centimetres, of 160 palm leaves. His results are illustrated in the cumulative frequency graph below.
180
160
140
Cumulative frequency
120
100
80
60
40
20
0
0 5 10 15 20 25 30
Length in centimetres
(i) Estimate how many leaves have a length between 14 and 24 centimetres. [1]
(ii) 10% of the leaves have a length of L centimetres or more. Estimate the value of L. [2]
(iii) Estimate the median and the interquartile range of the lengths. [3]
Sharim measured the lengths, in centimetres, of 160 palm leaves of a different type. He drew a box-and-whisker plot for the data, as shown on the grid below.
0 5 10 15 20 25 30
Length in centimetres
(iv) Compare the central tendency and the spread of the two sets of data. [2]
Mark scheme
5(i) 156 – 55 = 99 B1 98 ⩽ answer < 100
1
5(ii) 90% of 160 = 144 M1 144 seen, may be marked on graph
(L =) 22 A1
2
5(iii) Median = 15.6 B1 15.5 < median < 15.8
UQ = 18.8, LQ = 12.7
IQR = 18.8 – 12.7 M1 18.5 < UQ < 19 – 12.5 < LQ < 13
6.1 A1 6.0 ⩽ IQR ⩽ 6.2
3
5(iv) The Median higher for Ransha (1st set of data) B1 Any correct comparison of central tendency, must mention
median
IQR lower for Ransha (1st set of data) B1 Any correct comparison of spread, must refer to IQR
2
(i) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that all four Es are together. [1]
(ii) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that the Ss are not next to each other. [4]
Four letters are selected from the 12 letters of the word STEEPLECHASE.
(iii) Find the number of different selections if the four letters include exactly one S. [4]
Mark scheme
6(i) 9! B1 Exact value
= 181 440
2!
1
6(ii) 12! B1 Accept unevaluated
Total no of ways = = 9 979 200 (A)
2!4!
11! B1 Accept unevaluated
With Ss together = = 1 663 200 (B)
4!
With Ss not together = (B) – (A) M1 12! 8!
Correct or − , m, n integers > 1
m n
or their identified total – their identified Ss together
8 316 000 A1 Exact value
Alternative method for question 6(ii)
_T_E_E_P_L_E_C_H_A_E_ B1 10! × k in numerator k integer ⩾ 1
10! 11 × 10 B1 4! × k in numerator k integer ⩾ 1
×
4! 2!
their 10! 11 M1 OE
× C2 or 11P2
their 4!
8 316 000 A1 Exact value
4
6
6(iii) SEEE:1 M1 Cx seen alone or times K > 1
6
SEE_: 6
C1 = 6 B1 C3 or 6C2 or 6C1 alone
6
SE__: C2 = 15
6
S___: C3 = 20
Add 3 or 4 correct scenarios M1 No extras
Total = 42 A1
4
The shortest time recorded by an athlete in a 400 m race is called their personal best (PB). The PBs of the athletes in a large athletics club are normally distributed with mean 49.2 seconds and standard deviation 2.8 seconds.
(i) Find the probability that a randomly chosen athlete from this club has a PB between 46 and
53 seconds. [4]
(ii) It is found that 92% of athletes from this club have PBs of more than t seconds. Find the value of t. [3]
Three athletes from the club are chosen at random.
(iii) Find the probability that exactly 2 have PBs of less than 46 seconds. [3]
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(i) 46 − 49.2 53 − 49.2 M1 Using ± standardisation formula for either 46 or 53, no
P(46 < X < 53) = P <Z < continuity correction, σ2 or √σ
2.8 2.8
P( −1.143 < Z < 1.357) A1 Both standardisations correct unsimplified
Φ (1.357 ) + Φ (1.143) − 1 M1 Correct final area
= 0.9126 + 0.8735 – 1
0.786 A1 Final answer
4
7(ii) t − 49.2 B1 ±1.406 seen
= −1.406
2.8
M1 An equation using ± standardisation formula with a z-value,
condone σ2 or √σ
45.3 A1
3
7(iii) P(X < 46) = 0.1265 M1 Calculated or ft from (i)
P(2PB < 46) = 3 (1 − 0.1265 ) 0.12652 M1 3(1-p)p2, 0<p<1
0.0419 A1
3
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9709/61 October/November 2019 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/61, October/November 2019 (9709_w19_ms_61), and the question paper (9709_w19_qp_61), to read or download, with each question's marking points written out underneath.
How many marks is 9709/61 out of?+
Mathematics 9709/61 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/61 answers marked?+
Yes. Upload your answers to 9709/61/O/N/19 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.