Mathematics 9709/61 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 times, t seconds, taken to swim 100 m were recorded for a group of 9 swimmers and were found to be as follows.
95 126 117 135 120 125 114 119 136
(i) Find the values of Σ t − 120 and Σ t − 1202 . [2]
(ii) Using your values found in part (i), calculate the variance of t. [2]
Mark scheme
1(i) Σ(t – 120) = –25 + 6 − 3 + 15 + 0 + 5 – 6 – 1+ 16 = 7 M1 Attempt to sum both (t – 120) and (t – 120)2 Correct ans using
Σt – 9 × 120 and Σ (t – 120)2 M1A1
Σ(t – 120)2 = 252 + 62 + 32 + 152 + 02 + 52 + 62 + 12 + 162 A1 Both correct, www, SC correct ans no working B1B1
= 1213
2
1(ii) ∑ ( t − 120 )
2
∑ ( t − 120 ) their 1213 their 7 2
2 M1 Using two coded values in correct formula including finding Σt from 7 etc
Var = − = −
9 9 9 9
= 134(.2) A1 Correct answer
SC if correct variance obtained by another method from raw data
give SCB1
2
Jameel has 5 plums and 3 apricots in a box. Rosa has x plums and 6 apricots in a box. One fruit is chosen at random from Jameel’s box and one fruit is chosen at random from Rosa’s box. The probability that both fruits chosen are plums is 41 . Write down an equation in x and hence find x. [3]
Mark scheme
2 5 x M1 Their 2 probabilities for P(plum) multiplied and equated to 1/4
Jameel: P(plum) = , Rosa: P(plum) =
8 x+6
A1 Correct equation oe
5 x 1
× =
8 x+6 4
(x =) 4 A1 SC correct answer with no appropriate equations i.e. common sense
B1
3
A fair six-sided die is thrown twice and the scores are noted. Event X is defined as ‘The total of the two scores is 4’. Event Y is defined as ‘The first score is 2 or 5’. Are events X and Y independent?
Justify your answer. [4]
Mark scheme
3 3 1 B1
P(X) = oe
36 12
12 1 B1
P(Y) = oe
36 3
1 M1 Independent method to find P(X∩Y) without multiplication, either
P(X∩Y) = stated or by listing or circling numbers on a probability space
36
diagram. OR condititional prob with a single fraction numerator
P(X) × P(Y) = P(X∩Y), independent A1 Numerical comparison and conclusion, www
4
The Mathematics and English A-level marks of 1400 pupils all taking the same examinations are shown in the cumulative frequency graphs below. Both examinations are marked out of 100.
1500
1400
1300
1200
1100 English
Mathematics
1000
Cumulative frequency
900
800
700
600
500
400
300
200
100
0
0 10 20 30 40 50 60 70 80 90 100
Marks
Use suitable data from these graphs to compare the central tendency and spread of the marks in
Mathematics and English. [6]
Mark scheme
4 Median Maths = 40 M1 Indication of finding medians, such as mark on graph or reference
marks to 700 pupils, condone poor terminology such as ‘mean’
Median English = 55 A1 Both values correct, condone 54<English<56 but 54, 56 get A0
Median of English is larger than median of Maths B1 Correct statement, median must be referenced within answer.
No credit if statement references ‘means’
Range Maths is 100 or IQ range Maths = 80 – 12 = 68 M1 Evidence of finding either both ranges or both IQ ranges i.e. see a
minus
Range English is 60 or IQ range English = 62 – 42 = 20 A1 Both ranges or IQR correct
Maths marks have more spread then English marks B1 Correct conclusion. Accept standard deviation but must see some
figures
6
In a certain country the probability that a child owns a bicycle is 0.65.
(i) A random sample of 15 children from this country is chosen. Find the probability that more than
12 own a bicycle. [3]
(ii) A random sample of 250 children from this country is chosen. Use a suitable approximation to find the probability that fewer than 179 own a bicycle. [4]
Mark scheme
5(i) (P > 12) = P(13, 14, 15) M1 Binomial term of form 15Cxpx(1 – p)15 – x 0 < p < 1 any p, x ≠ 15,0
= 15C13(0.65)13(0.35)2 + 15C14(0.65)14(0.35)1 + (0.65)15 A1 Correct unsimplified answer
= 0.0617 A1 SC if use np and npq with justification give (12.5 – 9.75)/√3.41 M1
1–F(1.489) A1 0.0681 A0
3
5(ii) mean = 250 × 0.65 = 162.5 B1 Correct unsimplified np and npq
variance = 250 × 0.65 × 0.35 = 56.875
178.5 − 162.5 M1 Substituting their µ and σ (condone σ2) into the Standardisation
P(< 179) = P(z < ) = P(z < 2.122) Formula with a numerical value for ‘178.5’. Continuity correct not
56.875
required for this M1. Condone ± standardisation formula
Using continuity correction 178.5 or 179.5 M1
= 0.983 A1 Correct final answer
4
At a funfair, Amy pays $1 for two attempts to make a bell ring by shooting at it with a water pistol.
³ If she makes the bell ring on her first attempt, she receives $3 and stops playing. This means that overall she has gained $2.
³ If she makes the bell ring on her second attempt, she receives $1.50 and stops playing. This means that overall she has gained $0.50.
³ If she does not make the bell ring in the two attempts, she has lost her original $1.
The probability that Amy makes the bell ring on any attempt is 0.2, independently of other attempts.
(i) Show that the probability that Amy loses her original $1 is 0.64. [2]
(ii) Complete the probability distribution table for the amount that Amy gains. [4]
Amy’s gain ($)
Probability 0.64
(iii) Calculate Amy’s expected gain. [1]
Mark scheme
6(i) P(loses $1) = P( F and F) = 0.8 × 0.8 M1 0.8 x 0.8 or (1 – 0.2)(1-0.2) or P(F) × P(F) or P(F)+P(F) seen or
implied
= 0.64 AG A1 Must see probabilities multiplied together with final answer and a
clear probability statement or implied by labelled tree diagram
2
6(ii) B1 –1 linked with 0.64 in table
Amount
–1 0.50 2
gained ($) B1 0.5 seen in table
Prob 0.16 0.2 B1 0.16 seen in table linked to their 0.5
B1 FT P(2.00 gained) = 0.36 – P(0.50 gained) or correct, and all
amount gained linked correctly in table
4
6(iii) E(winnings) = –1 × 0.64 + 0.5 × 0.16 + 2 × 0.2 B1 FT Accept ($)0.16 or 16 cents loss. FT unsimplified E(winnings)
= –($)0.16, –16 cents from their table provided Σp = 1
1
The weight of adult female giraffes has a normal distribution with mean 830 kg and standard deviation
120 kg.
(i) There are 430 adult female giraffes in a particular game reserve. Find the number of these adult female giraffes which can be expected to weigh less than 700 kg. [4]
(ii) Given that 90% of adult female giraffes weigh between 830 − w kg and 830 + w kg, find the value of w. [3]
The weight of adult male giraffes has a normal distribution with mean 1190 kg and standard deviation
3 kg.
(iii) Given that 83.4% of adult male giraffes weigh more than 950 kg, find the value of 3. [3]
Mark scheme
7(i) 700 − 830 M1 Using ± standardisation formula, no continuity correction, not σ2 or
P (< 700) = P z < = P(z < –1.083) √σ
120
= 1 – 0.8606 M1 Appropriate area Φ from standardisation formula P(z<….) in final
probability solution, (<0.5 if z is –ve, >0.5 if z is +ve)
= 0.1394 A1 Correct final probability rounding to 0.139
Expected number of female adults = 430 × their 0.1394 B1 FT their 3 or 4 SF probability, rounded or truncated to integer
= 59.9 So 59 or 60
4
7(ii) P(giraffe < 830+w) = 95% so z = 1.645 B1 ±1.645 seen (critical value)
(830 + w ) − 830 = w = 1.645 M1 An equation using the standardisation formula with a z-value
(not 1 – z), condone σ2 or √σ not 0.8519, 0.8289
120 120
w = 197 A1 Correct answer
3
7(iii) P(male > 950) = 0.834, so z = –0.97 B1 ± 0.97 seen
950 − 1190 M1 Using ± standardisation formula, condone continuity correction, σ2
= –0.97 or √σ, condone equating with non z-value not 0.834, 0.166
σ
σ = 247 A1 Condone –σ = –247. www.
3
Freddie has 6 toy cars and 3 toy buses, all different. He chooses 4 toys to take on holiday with him.
(i) In how many different ways can Freddie choose 4 toys? [1]
(ii) How many of these choices will include both his favourite car and his favourite bus? [2]
Freddie arranges these 9 toys in a line.
(iii) Find the number of possible arrangements if the buses are all next to each other. [3]
(iv) Find the number of possible arrangements if there is a car at each end of the line and no buses are next to each other. [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
8(i) (9C4 =) 126 B1
1
7 7
8(ii) C2 B1 Cx or yC2 (implied by correct answer) or 7Px or 7Py, seen alone
= 21 B1 correct answer
2
8(iii) _ C1 (B1 B2 B3 ) C2 _ C3 _ C4 _ C5 _ C6 B1 3! or 6! seen alone or multiplied by k > 1 need not be an integer
3! × 6! × 7 B1 3! and 6! seen multiplied by k > 1, integer, no division
= 30240 B1 Exact value
Alternative method for question 8(iii)
C1 (B1 B2 B3 ) C2 C3 C4 C5 C6 B1 3! or 7! seen alone or multiplied by k > 1 need not be an integer
3! × 7! B1 3! and 7! seen multiplied by k > or = 1, no division
= 30240 B1 Exact value
3
8(iv) C1 _ C2 _ C3 _ C4 _ C5 _C6 B1 6! or 4! X 6P2 seen alone or multiplied by k > 1, no division
(arrangements of cars)
6! × 5P3 or 6! × 5 × 4 × 3 or 6! x 3! x10 B1 Multiply by 5P3 oe i.e. putting Bs in between 4 of the Cs
OR multiply by 3! x n where n = 7, 8, 9, 10 (number of options)
= 43200 B1 Correct answer
3
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Mathematics 9709/61 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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