Mathematics 9709/53 May/June 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The time taken, t minutes, to complete a puzzle was recorded for each of 150 students. These times are summarised in the table.
Time taken (t minutes) t ≤ 25 t ≤ 50 t ≤ 75 t ≤ 100 t ≤ 150 t ≤ 200
Cumulative frequency 16 44 86 104 132 150
(a) Draw a cumulative frequency graph to illustrate the data. [2]
(b) Use your graph to estimate the 20th percentile of the data. [1]
Mark scheme
1(a) Cumulative frequency (cf) graph M1 At least 3 points plotted accurately at class upper
end points (25,16) (50,44) (75,86) (100,104)
(150, 132) (200, 150).
Linear cf scale 0 ⩽ cf ⩽ 150 and linear time scale
0 ⩽ time(mins) ⩽ 200 with at least 3 values
identified on each axis.
A1 All points plotted correctly, curve drawn (within
tolerance) and joined to (0,0).
Axes labelled cumulative frequency (cf), time (t)
and minutes (min), or a suitable title.
2
1(b) Line from cumulative frequency = 30 to meet graph at t is between 37.5 and B1 FT Not from wrong working. Must be an increasing
42 cumulative frequency graph.
1
Twenty children were asked to estimate the height of a particular tree. Their estimates, in metres, were as follows.
4.1 4.2 4.4 4.5 4.6 4.8 5.0 5.2 5.3 5.4
5.5 5.8 6.0 6.2 6.3 6.4 6.6 6.8 6.9 19.4
(a) Find the mean of the estimated heights. [1]
(b) Find the median of the estimated heights. [1]
(c) Give a reason why the median is likely to be more suitable than the mean as a measure of the central tendency for this information. [1]
Mark scheme
2(a) 123.4 B1 1234
20 = 6.17
Accept 6 m 17 cm, .
200
1
2(b) 10th + 11th 5.4 + 5.5 B1 Accept 5 m 45 cm.
= = 5.45 (m)
2 2
1
2(c) The mean is unduly influenced by an extreme value, 19.4. B1 Comment must be within context.
1
The random variable X takes the values −2, 1, 2, 3. It is given that P X = x = kx2 , where k is a constant.
(a) Draw up the probability distribution table for X , giving the probabilities as numerical fractions.
[3]
(b) Find E X and Var X . [3]
Mark scheme
3(a) 1 B1 SOI
k= (4k+ k +4k +9k = 18k = 1)
18
M1 Table with correct x values and at least one
x −2 1 2 3
probability accurate using their k.
Values need not be in order, lines may not be
4 1 4 9
P(X=x) drawn, may be vertical, x and P(X=x) may be
18 18 18 18 omitted.
Condone any additional X values if probability
stated as 0.
A1 Remaining probabilities correct.
3
3(b) 4 × −2 + 1 × 1 + 4 × 2 + 9 × 3 M1 −8k + k + 8k + 27k
E ( X ) = 18
= May be implied by use in Variance. Accept
unsimplified expression.
−8 + 1 + 8 + 27
FT their table if probabilities sum to 1 or 0.999.
18 SC B1 28k.
4 × (−2) 2 + 1 × 12 + 4 × 22 + 9 × 32 2 M1 16k + k + 16k + 81k − (their mean)2
Var ( X ) = − ( their E ( X ) ) = FT their table even if probabilities not summing to
18 1.
2
16 + 1 + 16 + 81 28 114
− ( their E ( X ) ) M1.
2
= − their Note: If table is correct,
18 18 18
SC B1 114k – (their mean)2.
14 5 317 74 A1 Answers for E(X) and Var(X) must be identified.
E(X) = , 1 , 1.56 , Var (X) = , 3 , 3.91
9 9 81 81 3.91 ⩽ Var(X) ⩽ 3.914
3
Ramesh throws an ordinary fair 6-sided die.
(a) Find the probability that he obtains a 4 for the first time on his 8th throw. [1]
(b) Find the probability that it takes no more than 5 throws for Ramesh to obtain a 4. [2]
Ramesh now repeatedly throws two ordinary fair 6-sided dice at the same time. Each time he adds the two numbers that he obtains.
(c) For 10 randomly chosen throws of the two dice, find the probability that Ramesh obtains a total of less than 4 on at least three throws. [4]
Mark scheme
4(a) 5 7 1 78125 B1 0.0465 ⩽ p < 0.04652
× = 0.0465,
6 6 1679616
1
4(b) 5
5 2 3
1 5 1 5 1 5 1 5 1
4 M1 1 – pn, 0 < p < 1, n = 4, 5, 6 or sum of 4, 5 or 6
P(X < 6) = 1 − or + + + + terms p × (1 − p ) for n = 0,1, 2,3, 4 ( 5 ) .
n
6 6 6 6 6 6 6 6 6 6
4651 A1
0.598,
7776
2
4(c) 3 1 B1 SOI
[Probability of total less than 4 is] or
36 12
[1 – P(0, 1, 2)] M1 One term 10Cx p x (1 − p )
10 − x
, for 0 < x < 10,
0 10 1 9 2 8
1 11 1 11 1 11 0 < p < 1.
= 1 − ( 10C0 + 10C1 + 10C2 )
12 12 12 12 12 12
1 – (0.418904 + 0.380822 + 0.155791) A1 FT Correct expression. Accept unsimplified.
0.0445 A1 0.04448 ⩽ p ⩽ 0.0445
4
Farmer Jones grows apples. The weights, in grams, of the apples grown this year are normally distributed with mean 170 and standard deviation 25. Apples that weigh between 142 grams and
205 grams are sold to a supermarket.
(a) Find the probability that a randomly chosen apple grown by Farmer Jones this year is sold to the supermarket. [4]
Farmer Jones sells the apples to the supermarket at $0.24 each. He sells apples that weigh more than
205 grams to a local shop at $0.30 each. He does not sell apples that weigh less than 142 grams.
The total number of apples grown by Farmer Jones this year is 20 000.
(b) Calculate an estimate for his total income from this year’s apples. [3]
Farmer Tan also grows apples. The weights, in grams, of the apples grown this year follow the distribution N 182, 202 . 72% of these apples have a weight more than w grams.
(c) Find the value of w. [3]
Mark scheme
5(a) 142 − 170 205 − 170 M1 Use of ± standardisation formula once substituting
[P(142 < X < 205)] = P <z< 170, 25 and either 142 or 205 appropriately..
25 25
Condone 252 and continuity correction ±0.5.
P( −1.12 < z < 1.4) A1 Both correct. Accept unsimplified.
Φ (1.4 ) − (1 − Φ (1.12 ) ) = 0.9192 + 0.8686 – 1 M1 Calculating the appropriate area from stated phis of
z-values.
0.788 A1 AWRT, not from wrong working
4
5(b) P(X > 205) = 1 – 0.9192 = 0.0808 B1 FT Correct or FT from part 5(a).
( 0.0808 × 0.30 + their 0.788 × 0.24 ) × 20000 M1 Correct or
their 0.0808 × 0.30 × k + their 0.788 × 0.24 × k,
k positive integer.
[$]4266.24 A1 4265 < income ⩽ 4270, not from wrong working
3
5(c) w − 182 B1 0.5828 ⩽ z ⩽ 0.583 or -0.583 ⩽ z ⩽ –0.5828 seen.
[P( Z > ) = 0.72 ]
20
w − 182 M1 182 and 20 substituted in ± standardisation
= −0.583 formula, no continuity correction, not σ2, √ σ,
20
equated to a z-value.
w = 170 A1 170 ⩽ w < 170.35
3
Sajid is practising for a long jump competition. He counts any jump that is longer than 6 m as a success.
On any day, the probability that he has a success with his first jump is 0.2. For any subsequent jump, the probability of a success is 0.3 if the previous jump was a success and 0.1 otherwise. Sajid makes three jumps.
(a) Draw a tree diagram to illustrate this information, showing all the probabilities. [2]
(b) Find the probability that Sajid has exactly one success given that he has at least one success. [5]
On another day, Sajid makes six jumps.
(c) Find the probability that only his first three jumps are successes or only his last three jumps are successes. [3]
Mark scheme
6(a) 1st 2nd 3rd B1 First and second jumps correct with probabilities
and outcomes identified.
0.3 S
0.3 S B1 Third jump correct with probabilities and outcomes
0.7 F identified.
S 0.1 S
0.2 0.7
F F
0.9
0.3 S
0.8 0.1 S
0.7
F F
0.9 0.1 S
F
0.9 F
2
6(b) SFF 0.2 × 0.7 × 0.9 = 0.126 M1 Two or three correct 3 factor probabilities added,
FSF 0.8 × 0.1× 0.7 = 0.056 correct or FT from part 6(a). Accept unsimplified.
FFS 0.8 × 0.9 × 0.1= 0.072
127 A1 Accept unsimplified.
[Total = probability of 1 success =] 0.254
500
44 B1 FT Accept unsimplified.
[Probability of at least 1 success = 1− 0.8 × 0.9 × 0.9 =]0.352
125
their 0.254 M1 Accept unsimplified.
P(exactly 1 success | at least 1 success)=
their 0.352
127 A1 0.7215 < p ⩽ 0.722
0.722,
176
5
6(c) 0.8 × 0.9 × 0.9 × 0.1× 0.3 × 0.3 = 0.005832 [FFFSSS] M1 a × b × c × d × e × f FT from their tree diagram.
0.2 × 0.3 × 0.3 × 0.7 × 0.9 × 0.9 = 0.010206 [SSSFFF] Either a, b and c all = 0.8 or 0.9 (at least one of
each) and d, e and f all = 0.1 or 0.3 (at least one of
each).
Or a, b, c = 0.2 or 0.3 (at least one of each) and d,
e, f = 0.7 or 0.9 (at least one of each).
A1 Either correct. Accept unsimplified.
[Total =] 0.0160[38] A1
3
12
A group of 15 friends visit an adventure park. The group consists of four families.
• Mr and Mrs Kenny and their four children
• Mr and Mrs Lizo and their three children
• Mrs Martin and her child
• Mr and Mrs Nantes
The group travel to the park in three cars, one containing 6 people, one containing 5 people and one containing 4 people. The cars are driven by Mr Lizo, Mrs Martin and Mr Nantes respectively.
(a) In how many different ways can the remaining 12 members of the group be divided between the three cars? [3]
The group enter the park by walking through a gate one at a time.
(b) In how many different orders can the 15 friends go through the gate if Mr Lizo goes first and each family stays together? [3]
In the park, the group enter a competition which requires a team of 4 adults and 3 children.
(c) In how many ways can the team be chosen from the group of 15 so that the 3 children are all from different families? [2]
(d) In how many ways can the team be chosen so that at least one of Mr Kenny or Mr Lizo is included? [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(a) C5 × 7C4 [× 3C3] M1 12
Cr × q, r = 3, 4, 5 q a positive integer > 1,
no + or − .__
12
M1 Cs × 12–sCt [× 12-s–tCu]
s = 3, 4, 5; t = 3, 4, 5 ≠ s; u = 3, 4, 5 ≠ s ,t
Alternative method for question 7(a)
12! M1 12! ÷ by a product of three factorials.
5!× 3!× 4!
M1 n!
5!× 3!× 4!
[792 × 35 =] 27 720 A1 CAO
3
7(b) 4! (Lizo) × 6! (Kenny) × 2! (Martin) × 2! (Nantes) M1 Product involving at least 3 of 4!, 6!, 2!, 2!
× 3! (orders of K, M and N) M1 w × 3! , w integer > 1.
414 720 A1 WWW CAO
3
7
7(c) C4 (adults) × 4C1 × 3C1 M1 7
C4 × b, b integer > 1 no + or – .
420 A1
2
5
7(d) K not L C3 × 8C3 = 560 M1 8
C3(or 8P3) × c for one of the products
5
L not K C3 × 8C3 = 560 or 5C3 (or 5P3)× c, positive integer >1 for first 2
5
L and K C2 × 8C3= 560 products only.
M1 Add 2 or 3 correct scenarios only values, no
additional incorrect scenarios, no repeated
scenarios. Accept unsimplified.
[Total or Difference=] 1680 A1
Alternative method for question 7(d)
8
Total no of ways – neither L nor K M1 C3 × c, c a positive integer >1.
Total = 7C4 × 8C3 = 1960
Neither K nor L = 5C4 × 8C3 =280 M1 Subtracting the number of ways with neither from
their total number of ways.
[Total or Difference=] 1680 A1
7(d) Alternative method for question 7(d)
8
Subtracting K and L from sum of K and L M1 C3 × c, c a positive integer >1.
6
K C3 × 8C3 = 1120
6
L C3 × 8C3 = 1120 M1 Subtracting number of ways with both from sum of
L and K 5C2 × 8C3= 560 number of ways with K and number of ways with
1120 + 1120 – 560 = 1680 L.
[Total or Difference=] 1680 A1
3
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Mathematics 9709/53 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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