Mathematics 9709/63 May/June 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The time taken, in minutes, by a ferry to cross a lake has a normal distribution with mean 85 and standard deviation 6.8.
(i) Find the probability that, on a randomly chosen occasion, the time taken by the ferry to cross the lake is between 79 and 91 minutes. [3]
(ii) Over a long period it is found that 96% of ferry crossings take longer than a certain time t minutes.
Find the value of t. [3]
Mark scheme
1(i) 79 − 85 91 − 85 M1 Using ± standardisation formula for either 79 or 91, no continuity
P(79 < X < 91) = P <Z< correction
6.8 6.8
= P(–0.8824 < Z < 0.8824)
= Φ ( 0.8824 ) − Φ ( −0.8824) M1 Correct area ( Φ − Φ ) with one +ve and one –ve z-value or 2 Φ – 1
or 2( Φ − 0.5)
= 0.8111 – (1 – 0.8111)
= 0.622 A1 Correct answer
3
1(ii) z = –1.751 B1 ± 1.751 seen
t − 85 M1 An equation using ± standardisation formula with a z-value,
–1.751 = condone σ2 or √σ
6.8
t = 73.1 A1 Correct answer
3
Megan sends messages to her friends in one of 3 different ways: text, email or social media. For each message, the probability that she uses text is 0.3 and the probability that she uses email is 0.2. She receives an immediate reply from a text message with probability 0.4, from an email with probability
0.15 and from social media with probability 0.6.
(i) Draw a fully labelled tree diagram to represent this information. [2]
(ii) Given that Megan does not receive an immediate reply to a message, find the probability that the message was an email. [4]
Mark scheme
2(i) B1 Fully correct labelled tree with correct probabilities for ‘Send’
B1 Fully correct labelled branches with correct probabilities for the
‘reply’
2
2(ii) P ( email ∩ NR ) 0.2 × 0.85 M1 P(email) × P(NR) seen as numerator of a fraction, consistent with
P (email NR) = = their tree diagram
P ( NR ) 0.3 × 0.6 + 0.2 × 0.85 + 0.5 × 0.4
0.17 0.17 M1 Summing three appropriate 2-factor probabilities, consistent with
= = their tree diagram, seen anywhere
0.18 + 0.17 + 0.2 0.55
0.55 oe (can be unsimplified) seen as denom of a fraction
17 A1
= 0.309,
55
A1 Correct answer
4
Mr and Mrs Keene and their 5 children all go to watch a football match, together with their friends
Mr and Mrs Uzuma and their 2 children. Find the number of ways in which all 11 people can line up at the entrance in each of the following cases.
(i) Mr Keene stands at one end of the line and Mr Uzuma stands at the other end. [2]
(ii) The 5 Keene children all stand together and the Uzuma children both stand together. [3]
Mark scheme
3(i) 9! × 2 B1 9! seen multiplied by k ⩾ 1, no addition
= 725760 B1 Exact value
2
3(ii) Eg (K1K2K3K4K5) A A A (U1U2) A B1 2! or 5! seen mult by k > 1, no addition (arranging Us or Ks)
= 5! × 2! × 6! B1 6! Seen mult by k > 1, no addition (arranging AAAAKU)
= 172800 B1 Exact value
3
(i) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if there must be at least twice as many men as there are women on the committee. [3]
(ii) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if 2 particular men refuse to be on the committee together. [3]
Mark scheme
4(i) M(8) W(4) B1 One unsimplified product correct
4 2 in 8C4 × 4C2 = 420 ways
5 1 in 8C5 × 4C1 = 224 ways
6 0 in 8C6 × 4C0 = 28 ways
M1 Summing the number of ways for 2 or 3 correct scenarios (can be
unsimplified), no incorrect scenarios
Total 672 ways A1 Correct answer
3
4(ii) Total number of selections = 12C6 = 924 (A) M1 12
Cx – (subtraction seen), accept unsimplified
Selections with males together = 10C4 = 210 (B) A1 Correct unsimplified expression
Total = (A) – (B) = 714 A1 Correct answer
Alternative method for question 4(ii)
10
No males + Only male 1 + Only male 2 M1 Cx + 2 x 10Cy , x ≠ y seen, accept unsimplified
= 10C6 + 10C5 + 10C5
= 210 + 252 + 252 A1 Correct unsimplified expression
= 714 A1 Correct answer
Alternative method for question 4(ii)
Pool without male 1 + Pool without male 2 – Pool without either M1 2 x 11Cx – 10Cx
male
= 11C6 + 11C6 – 10C6 A1 Correct unsimplified expression
= 462 + 462 – 210
= 714 A1 Correct answer
3
On average, 34% of the people who go to a particular theatre are men.
(i) A random sample of 14 people who go to the theatre is chosen. Find the probability that at most
2 people are men. [3]
(ii) Use an approximation to find the probability that, in a random sample of 600 people who go to the theatre, fewer than 190 are men. [5]
Mark scheme
5(i) P(0, 1, 2) = (0.66)14 + 14C1(0.34)(0.66)13 + 14C2(0.34)2(0.66)12 M1 Binomial term of form 14Cxpx(1 – p)14 – x 0 < p < 1 any p, x ≠ 14,0
= 0.0029758 + 0.02146239 + 0.071866 A1 Correct unsimplified answer
= 0.0963 A1 Correct answer
3
5(ii) Mean =600 × 0.34 = 204, Var = 600 × 0.34 × 0.66 = 134.64 B1 Correct unsimplified np and npq (or sd = 11.603 or Variance =
3366/25)
M1 Substituting their µ and σ, (no σ2 or √σ) into the Standardisation
Formula with a numerical value for ‘189.5’. Condone ±
189.5 − 204 standardisation formula
P(< 190) = P z < = P(z < –1.2496)
134.64
M1 Using continuity correction 189.5 or 190.5 within a Standardisation
formula
= 1 – Φ (1.2496) M1 Appropriate area Φ from standardisation formula P(z<….) in final
solution, (<0.5 if z is –ve, >0.5 if z is +ve)
= 1 – 0.8944 = 0.106 A1 Correct final answer
5
A fair five-sided spinner has sides numbered 1, 1, 1, 2, 3. A fair three-sided spinner has sides numbered
1, 2, 3. Both spinners are spun once and the score is the product of the numbers on the sides the spinners land on.
(i) Draw up the probability distribution table for the score. [4]
(ii) Find the mean and the variance of the score. [3]
(iii) Find the probability that the score is greater than the mean score. [2]
Mark scheme
6(i) B1 Probability distribution table with correct scores, allow extra score
score 1 2 3 4 6 9 values if probability of zero stated
prob 3 4 4 1 2 1 B1 2 probabilities (with correct score) correct
15 15 15 15 15 15
B1 3 or more correct probabilities with correct scores
B1 FT Σp = 1, at least 4 probabilities
4
6(ii) (3 + 8 + 12 + 4 + 12 + 9) 48 B1
mean = = (3.2)
15 15
(3 + 16 + 36 + 16 + 72 + 81) M1 FT Substitute their attempts at scores in correct var formula,
− ( their 3.2 )
2
Var = must have “– mean2 ” (condone probabilities not summing to 1)
15
224 352 A1
= – 3.22 = 4.69
15 75
3
6(iii) Score of 4, 6, 9 M1 Identifying relevant scores from their mean and their table
4 A1 Correct answer
Prob (0.267)
15
SC B1 for 4/15 with no working
2
The times in minutes taken by 13 pupils at each of two schools in a cross-country race are recorded in the table below.
Thaters School 38 43 48 52 54 56 57 58 58 61 62 66 75
Whitefay Park School 45 47 53 56 56 61 64 66 69 73 75 78 83
(i) Draw a back-to-back stem-and-leaf diagram to illustrate these times with Thaters School on the left. [4]
(ii) Find the interquartile range of the times for pupils at Thaters School. [2]
The times taken by pupils at Whitefay Park School are denoted by x minutes.
(iii) Find the value of Σ x − 602 . [2]
(iv) It is given that Σ x − 60 = 46. Use this result, together with your answer to part (iii), to find the variance of x. [2]
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) Thaters School Whitefay Park School B1 Correct stem can be upside down, ignore extra values,
8 3
8 3 4 5 7 B1 Correct Thaters School labelled on left, leaves in order from right to
8 8 7 6 4 2 5 3 6 6 left and lined up vertically, no commas
6 2 1 6 1 4 6 9
5 7 3 5 8 B1 Correct Whitefay Park School labelled on same diagram on right
8 3 hand side in order from left to right and lined up vertically, no
commas
Key 8 | 4 | 5 represents 48 minutes for Thaters School and 45 B1 FT Correct key for their diagram, need both teams identified and
minutes for Whitefay Park School. ‘minutes’ stated at least once here or in leaf headings or title.
SC If 2 separate diagrams drawn, SCB1 if both keys meet these
criteria
4
7(ii) LQ = 50 B1 Both quartiles correct
UQ = 61.5
IQ range = 61.5 – 50 = 11.5 B1 FT 61 ⩽ UQ ⩽ 62 – 48 ⩽ LQ ⩽ 52
2
7(iii) Σ(x – 60)2 = (–15)2 + (–13)2 + (–7)2 + (–4)2 + (–4)2 + 12 + 42 + 62 + M1 Summing squares with at least 5 correct unsimplified terms
92 + 132 + 232 + 152 + 182
= 1856 A1 Exact value
2
7(iv) Var = mean of coded squares – (coded mean)2 M1
∑ ( x − 60 ) ∑ ( x − 60 )
2 2
= −
13 13 Using two coded values in correct formula (variance or sd)
their 1856 46
2 A1 Correct answer
Var = −
13 13 SC if correct variance obtained by another method give SCB1
= 130
2
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Mathematics 9709/63 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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