Mathematics 9709/62 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.
Twelve tourists were asked to estimate the height, in metres, of a new building. Their estimates were as follows.
50 45 62 30 40 55 110 38 52 60 55 40
(i) Find the median and the interquartile range for the data. [3]
(ii) Give a disadvantage of using the mean as a measure of the central tendency in this case. [1]
Mark scheme
1(i) Median = 51 B1
UQ = 57.5, LQ = 40
IQR = UQ – LQ M1 55 ⩽ UQ ⩽ 62 – 38 ⩽ LQ ⩽ 45
17.5 A1 NFWW
3
1(ii) Result will be disproportionately affected by 110 B1 Affected by an extreme/large value
There is a large outlier
…contains outliers such as 110…
Not ‘mean affected by extreme values’
1
Benju cycles to work each morning and he has two possible routes. He chooses the hilly route with probability 0.4 and the busy route with probability 0.6. If he chooses the hilly route, the probability that he will be late for work is x and if he chooses the busy route the probability that he will be late for work is 2x. The probability that Benju is late for work on any day is 0.36.
(i) Show that x = 0.225. [2]
(ii) Given that Benju is not late for work, find the probability that he chooses the hilly route. [3]
Mark scheme
2(i) 0.4x + 0.6 × 2x = 0.36 or 0.4(1 − x) + 0.6(1 − 2x) = 0.64 M1 0.4a + (1 – 0.4)b = 0.36 or 0.64, a,b terms involving x
1.6x = 0.36 A1 Fully justified by algebra
x = 0.225 AG
2
2(ii) P( H L') = M1 Correct numerical numerator of a fraction. Allow unsimplified.
0.4 (1 − x ) 0.4 × (1 − 0.225 ) 0.4 × 0.775
= =
1 − 0.36 0.64 0.4 × 0.775 + 0.6 × 0.55
M1 Denominator 0.36 or 0.64. Allow unsimplified.
31 A1
or 0.484
64
3
The speeds, in km h−1 , of 90 cars as they passed a certain marker on a road were recorded, correct to the nearest km h−1 . The results are summarised in the following table.
Speed (km h−1 ) 10 − 29 30 − 39 40 − 49 50 − 59 60 − 89
Frequency 10 24 30 14 12
(i) On the grid, draw a histogram to illustrate the data in the table. [4]
(ii) Calculate an estimate for the mean speed of these 90 cars as they pass the marker. [2]
Mark scheme
3(i) 0.5 2.4 3 1.4 0.4 M1 At least 3 frequency densities calculated (frequency ÷ class
width)
10 10 10
e.g. , or may be read from graph using their scale,
20 19 19.5
3SF or exact
All heights correct on graph. A1
Bar ends of 9.5, 29.5, 39.5, 59.5, 89.5 B1
Axes labelled: Frequency density (fd) and speed/km h-1 (or B1
appropriate title). Linear scales 9.5 ⩽ horizontal axis ⩽ 89.5, 0
⩽ vertical axis ⩽ 3, 5 bars with no gaps
4
3(ii) 19.5 × 10 + 34.5 × 24 + 44.5 × 30 + 54.5 × 14 + 74.5 × 12 M1 Uses at least 4 midpoint attempts (e.g. 19.5 ± 0.5).
their 90 Allow unsimplified expression.
195 + 828 + 1335 + 763 + 894
=
90
4015 803
= or
90 18
11 A1 Final answer not an improper fraction
44 or 44.6 (km h −1 ) NFWW
18
2
In Quarendon, 66% of households are satisfied with the speed of their wifi connection.
(i) Find the probability that, out of 10 households chosen at random in Quarendon, at least 8 are satisfied with the speed of their wifi connection. [3]
(ii) A random sample of 150 households in Quarendon is chosen. Use a suitable approximation to find the probability that more than 84 are satisfied with the speed of their wifi connection. [5]
Mark scheme
4(i) P(8, 9, 10) = 10C8 0.668 0.342 +10C9 0.669 0.341 + 0.6610 M1 Correct binomial term, 10Ca 0.66a(1–0.66)b
a+b = 10, 0 < a,b < 10
A1 Correct unsimplified expression
0.284 B1 CAO
3
4(ii) np = 0.66 × 150 = 99 B1 Accept evaluated or unsimplified µ, σ2 numerical expressions,
npq = 0.66 × (1 − 0.66 ) × 150 = 33.66 condone σ = 33.66 = 5.8017 or 5.802
CAO
84.5 − 99 M1 x − their 99
P(X > 84) = P Z > ± Standardise, , condone σ2, x a value
33.66 their 33.66
M1 84.5 or 83.5 used in their standardisation formula
(= P ( Z > −2.499 ) ) M1 Correct final area
0.994 A1 Final answer (accept 0.9938)
SC if no standardisation formula seen,
B2 P(Z > -2.499) = 0.994
5
A fair red spinner has four sides, numbered 1, 2, 3, 3. A fair blue spinner has three sides, numbered
−1, 0, 2. When a spinner is spun, the score is the number on the side on which it lands. The spinners are spun at the same time. The random variable X denotes the score on the red spinner minus the score on the blue spinner.
(i) Draw up the probability distribution table for X . [4]
(ii) Find Var X . [3]
Mark scheme
5(i) B1 Table with correct values of x, at least 1 probability, all
probabilities ⩽ 1
x –1 0 1 2 3 4
p 1 1 3 2 3 2
12 12 12 12 12 12
B1 2 probabilities correct, may not be in table
B1 2 more probabilities correct, may not be in table
B1 All correct, values in table
SC1 No more than 1 correct probability and at least 5
probabilities summing to 1 in table
4
5(ii) −1 + 0 + 3 + 4 + 9 + 8 23 M1 May be implied by use in variance. Allow unsimplified
[E(X)] = = expression
12 12
1 + 0 + 3 + 8 + 27 + 32 ( = 71) 23 2 M1 Appropriate variance formula using their E(X)2
[Var(X)] = −
12 12
323 35 A1 CAO
2.24 or or 2
144 144
3
The heights, in metres, of fir trees in a large forest have a normal distribution with mean 40 and standard deviation 8.
(i) Find the probability that a fir tree chosen at random in this forest has a height less than 45 metres.
[2]
(ii) Find the probability that a fir tree chosen at random in this forest has a height within 5 metres of the mean. [2]
In another forest, the heights of another type of fir tree are modelled by a normal distribution. A scientist measures the heights of 500 randomly chosen trees of this type. He finds that 48 trees are less than 10 m high and 76 trees are more than 24 m high.
(iii) Find the mean and standard deviation of the heights of trees of this type. [5]
Mark scheme
6(i) 45 − 40 M1 ± Standardise, no continuity correction, σ2 or σ , formula must
P(X < 45) = P Z <
8 be seen
= P(Z < 0.625)
0.734(0) A1 CAO
2
6(ii) 1 − 2 (1 − ( i ) ) = 2 ( i ) − 1 = 2((i) – 0.5) M1 Use result of part (i) or recalculated to find area
OE
0.468 A1ft 0 < FT from (i) < 1 or correct.
2
6(iii) P(X < 10) = 48/500 = 0.096 B1 z = ± 1.305
z = –1.305
P(X > 24) = 76/500 = 0.152 B1 z = ± 1.028
z = 1.028
10 − µ = −1.305σ M1 Form 1 equation using 10 or 24 with , , –value. Allow
24 − µ = 1.028σ continuity correction, not , √
14 = 2.333σ M1 OE
Solve two equations in σ and µ to form equation in one
variable
σ = 6.[ 00] , µ = 17.8[3] A1 CAO, WWW
5
(i) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that all three Os are together and both Ts are together. [1]
(ii) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that the Ts are not together. [4]
(iii) Find the probability that a randomly chosen arrangement of the 9 letters of the word TOADSTOOL has a T at the beginning and a T at the end. [2]
(iv) Five letters are selected from the 9 letters of the word TOADSTOOL. Find the number of different selections if the five letters include at least 2 Os and at least 1 T. [4]
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) 6! = 720 B1 Evaluated
1
7(ii) 9! B1 Accept unevaluated
Total no of arrangements: = 30240
2!3!
8! B1 Accept unevaluated
No with Ts together = = 6720
3!
With Ts not together: 30 240 – 6720 M1 9! 8!
correct or − , m, n integers > 1
m n
or their identified total – their identified Ts together
23 520 A1 CAO
Alternative method for question 7(ii)
7! 8 × 7 B1 7! × (k > 0) in numerator, cannot be implied by 7P2, etc.
×
3! 2
B1 3! × (k > 0) in denominator
M1 their 7!
× 8C2 or 8P2
their 3!
23 520 A1 CAO
4
7(iii) 7! M1 their identified number of arrangements with T at ends
Number of arrangements =
3! their identified total number of arrangements
7! 7!
their
3! = 840
Probability =
9! 30240 or m m, n integers > 1
their 9!
3!2! n
1 A1 Final answer
or 0.0278
36
2
4
7(iv) OOT_ _ 4
C2 = 6 M1 Cx seen alone or 4Cx x k ≥1, k an integer, 0< x <4
4
OOTT_ C1 = 4 4
OOOT_ 4
C1 = 4 A1 C2 x k, k = 1 oe or 4C1 x m, m = 1 oe alone
OOOTT =1
M1 Add 3 or 4 identified correct scenarios only, accept unsimplified
(Total) = 15 A1 CAO, WWW
Only dependent on 2nd M mark
4
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Mathematics 9709/62 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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