Mathematics 9709/52 February/March 2023: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability and Statistics 1, February/March 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

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Paper: Paper 5 Probability and Statistics 1
Total marks: 50
Time: 1 hour 15 minutes
Code on the paper: 9709/52/F/M/23

9709_m23_ms_52.pdf

9709/52 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 18 marks

Question

Each year the total number of hours, x, of sunshine in Kintoo is recorded during the month of June. The results for the last 60 years are summarised in the table. x 30 ≤ x < 60 60 ≤ x < 90 90 ≤ x < 110 110 ≤ x < 140 140 ≤ x < 180 180 ≤ x ≤ 240 Number 4 8 14 25 7 2 of years (a) Draw a cumulative frequency graph to illustrate the data. [3] (b) Use your graph to estimate the 70th percentile of the data. [2] (c) Calculate an estimate for the mean number of hours of sunshine in Kintoo during June over the last 60 years. [3]

Mark scheme

1(a) B1 All cumulative frequencies stated. Upper value 60 90 110 140 180 240 May be under data table, condone omission of 4. May be read accurately from graph, must include 4. cf 4 12 26 51 58 60 M1 At least 5 points plotted at class upper end points, daylight rule tolerance. Linear cf scale 0 ⩽ cf ⩽ 60, linear time scale 30 ⩽ time ⩽ 240 with at least 3 values identified on each axis. A1 All points plotted correctly. Curve drawn (within tolerance), no ruled segments, and joined to (30, 0). Axes labelled ‘cumulative frequency’ and ‘hours [of sunshine]’ (OE including appropriate title). 3 1(b) [60 × 0.7 = ] 42 M1 42 may be implied by clear use on graph. 126 A1 FT Must be clear evidence on graph of use of 42, e.g. an appropriate mark on either axis, appropriate mark on curve. FT from increasing cf graph only read at 42 only. 2 1(c) Midpoints: 45, 75, 100, 125, 160, 210 B1 At least 5 correct mid-points seen, check by data table or used in formula. 4 × 45 + 8 × 75 + 14 ×100 + 25 ×125 + 7 ×160 + 2 × 210 M1 Correct mean formula using their 6 midpoints (must be within [Mean =] class, not upper bound, lower bound), condone 1 data error 60 If correct midpoints seen accept 6845 180 + 600 + 1400 + 3125 + 1120 + 420 = 60 . 60 1 A1 Accept 114.1, 114.08[3…] = 114, 114 1 12 If A1 not awarded, SC B1 for 114, 114 , 114.1 or 12 114.08[3…]. 3

Question 27 marks

Question

Alisha has four coins. One of these coins is biased so that the probability of obtaining a head is 0.6. The other three coins are fair. Alisha throws the four coins at the same time. The random variable X denotes the number of heads obtained. (a) Show that the probability of obtaining exactly one head is 0.225. [3] (b) Complete the following probability distribution table for X . [2] x 0 1 2 3 4 P X = x 0.05 0.225 0.075 (c) Given that E X = 2.1, find the value of Var X . [2]

Mark scheme

2(a) 0.6(0.5)3 + 0.4 ( 0.5) × 3 B1 Either 0.6(0.5)3 + a 3 ( ) or b + 0.4 ( 0.5) × 3 or 3 C1 , 0 < a, b < 1 seen. 3 M1 0.6(0.5)3 + 0.4 ( 0.5) × d seen, d = 1, 3. 3 Condone 0.075 + 0.05 × d, d = 1, 3. = 0.225 A1 AG full supporting working required. Scenarios identified and linked to calculations. 3 2(b) B1 3 x 0 1 2 3 4 Either [P(2) =] 0.375, 8 P(X = x) 0.05 0.225 0.375 0.275 0.075 11 or [P(3) =] 0.275, seen. 40 Condone not in table if identified. B1 FT Both values in table. FT P(2) + P(3) = 0.650 . 2 2(c) Var(X) M1 Appropriate variance formula from their probability = [1 ×]0.225 + 2 × their 0.375 + 3 × their 0.275 + 4 × 0.075 − 2.1 2 2 2 2 2 distribution table with at least 4 terms, 0 < their P(x) < 1. Condone 4.41 for 2.12 . Condone mean clearly recalculated inaccurately. Or 0.225 + 4 × their 0.375 + 9 × their 0.275 + 16 × 0.075 − 2.12 Condone 2.12 for 4.41. 5.4 − 2.12 = 0.99 [0] A1 If M0 awarded SC B1 for 0.99[0] WWW. 2

Question 36 marks

Question

80% of the residents of Kinwawa are in favour of a leisure centre being built in the town. 20 residents of Kinwawa are chosen at random and asked, in turn, whether they are in favour of the leisure centre. (a) Find the probability that more than 17 of these residents are in favour of the leisure centre. [3] (b) Find the probability that the 5th person asked is the first person who is not in favour of the leisure centre. [1] (c) Find the probability that the 7th person asked is the second person who is not in favour of the leisure centre. [2]

Mark scheme

3(a) Method 1 for Question 3(a) [P(X > 17) = P(18, 19, 20) =] ( )( ) M1 One term 20Cx p x 1 − p 20− x , 0 < p < 1, 0 < x < 20 . ( 0.2) + 20C19 ( 0.8) ( 0.2) C18 ( 0.8) 20 18 2 19 1 + 20C20 ( 0.8 ) A1 Correct expression, accept unsimplified, no terms omitted 20 leading to final answer. = 0.13691 + 0.05765 + 0.01153 0.206 B1 Mark the final answer at the most accurate value 0.206 ⩽ p ⩽ 0.2061 . Method 2 for Question 3(a) [P(X > 17) = 1 – P(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) = ] ( )( ) M1 One term 20Cx p x 1 − p 20− x , 0 < p < 1, 0 < x < 20 . 1 – (20C0 ( 0.8) ( 0.2 ) + 20C1 ( 0.8) ( 0.2 ) 0 20 1 19 + 20C2 ( 0.8) ( 0.2 ) ++ 20C16 ( 0.8) ( 0.2 ) A1 Correct expression, accept unsimplified, no terms omitted 2 18 16 4 leading to final answer. If answer correct, condone omission of + 20C17 ( 0.8) ( 0.2) ) 17 3 any 15 of the 16 middle terms. −14 + 8.389 ×10−13 = 1 – ( 1.048 ×10 +3.188 ×1011 ++ 0.2182 + 0.2054 ) 0.206 B1 Mark the final answer at the most accurate value 0.206 ⩽ p ⩽ 0.2061 . Condone omission of brackets. 3 3(b) B1 ( 0.8)4 ( 0.2 ) = 0.08192, 256 Accept 8192 OE. 3125 100000 1 3(c) ( 0.8) ( 0.2) × 6 M1 ( 0.8) ( 0.2) × k or ( 0.8) ( 0.2 ) × k × 0.2 , 5 2 5 2 5 2 ⩽ k ⩽ 7. 8144 A1 786432 = 0.0786, 0.0786 ⩽ p < 0.07865, . 78125 10000000 If A0 awarded, SC B1 for correct answer WWW. 2

Question 43 marks

Question

The probability that it will rain on any given day is x. If it is raining, the probability that Aran wears a hat is 0.8 and if it is not raining, the probability that he wears a hat is 0.3. Whether it is raining or not, if Aran wears a hat, the probability that he wears a scarf is 0.4. If he does not wear a hat, the probability that he wears a scarf is 0.1. The probability that on a randomly chosen day it is not raining and Aran is not wearing a hat or a scarf is 0.36. Find the value of x. [3]

Mark scheme

4 (1 – x) × 0.7 × 0.9 = 0.36 M1 (1 − x ) × a × b = 0.36, a = 0.7 or 0.3, b = 0.9 or 0.1 B1 (1-x)×0.7×0.9=0.36, (1-x)×0.63=0.36, 0.36 0.63 – 0.63x = 0.36 or 1 − x = seen. 0.63 Condone recovery from omission of brackets. 3 A1 Accept 0.428571 to at least 3 sf. x= Condone 0.4285 rounding to 0.429 . 7 3 If M0 awarded, SC B1 for x = or 0.428571 to at least 3 sf. 7 3

Question 53 marks

Question

Marco has four boxes labelled K , L, M and N . He places them in a straight line in the order K , L, M , N with K on the left. Marco also has four coloured marbles: one is red, one is green, one is white and one is yellow. He places a single marble in each box, at random. Events A and B are defined as follows. A: The white marble is in either box L or box M . B: The red marble is to the left of both the green marble and the yellow marble. Determine whether or not events A and B are independent. [3]

Mark scheme

5 1 8 1 B1 Both stated, accept unsimplified. P(A) = , P(B) = = , 2 24 3 1 M1 Evidence that independence properties not used. P(A B ) = 6 1 1 1 A1 Evaluated and conclusion stated. P(A) × P(B) = × = P(A) × P(B) and P(A B ) seen. 2 3 6 so events are independent 3

Question 611 marks

Question

In a cycling event the times taken to complete a course are modelled by a normal distribution with mean 62.3 minutes and standard deviation 8.4 minutes. (a) Find the probability that a randomly chosen cyclist has a time less than 74 minutes. [2] (b) Find the probability that 4 randomly chosen cyclists all have times between 50 and 74 minutes. [4] In a different cycling event, the times can also be modelled by a normal distribution. 23% of the cyclists have times less than 36 minutes and 10% of the cyclists have times greater than 54 minutes. (c) Find estimates for the mean and standard deviation of this distribution. [5]

Mark scheme

6(a) 74 − 62.3 M1 Use of ± standardisation formula with 74, 62.3 and 8.4 [P(X < 74) =] P Z < = P ( Z < 1.393) 8.4 2 substituted appropriately, not 8.4 , not √8.4, no continuity correction. = 0.918 A1 0.918 ⩽ p ⩽ 0.9185 . 2 6(b) 50 − 62.3 74 − 62.3 M1 Use of ± standardisation formula with both 74 (may be seen in [P ( 50 < X < 74) = P] <Z < 6(a) if their value seen) & 50, 62.3 and 8.4 substituted 8.4 8.4 appropriately. [P (−1.464 < Z < 1.393)] Condone use of 8.42 , 8.4 and continuity correction ±0.5 (73.5 or 74.5 and 49.5 or 50.5). [Φ (1.464) + Φ (1.393) −1 ] M1 Calculating the appropriate probability area from stated Φ of z-values (leading to their final answer > 0.5) but not 0.9285 + 0.9182 – 1 symmetrical values. = 0.847 A1 0.8465 ⩽ p < 0.8475 . SC B1 for 0.8465 ⩽ p < 0.8475 if M0A0 awarded. ( 0.8467 ) = 0.514 B1 FT Accept 0.513 ⩽ p ⩽ 0.514 . 4 FT (their 4-figure p)4, 0 < p < 1. 4 6(c) 36 − μ B1 −0.740 < z1 < −0.738 or 0.738 < z1 < 0.740 . z1 = = −0.739 σ B1 z2 = ±1.282 (critical value). 54 − μ z2 = = 1.282 σ M1 Use of the ±standardisation formula once with μ, σ and a z- value (not 0.23, 0.77, 0.90, 0.10, ±0.261, ±0.282…). Condone continuity correction ±0.5, not σ 2 , σ . Solve, obtaining values for μ and σ M1 Solve using the elimination method, substitution method or μ = 42.6, σ = 8.91 other appropriate approach to obtain values for both μ and σ. A1 42.58 ⩽ µ ⩽ 42.6, 8.90 ⩽ σ ⩽ 8.91 . 5

Question 712 marks

Question

(a) Find the number of different arrangements of the 9 letters in the word DELIVERED in which the three Es are together and the two Ds are not next to each other. [4] (b) Find the probability that a randomly chosen arrangement of the 9 letters in the word DELIVERED has exactly 4 letters between the two Ds. [5] Five letters are selected from the 9 letters in the word DELIVERED. (c) Find the number of different selections if the 5 letters include at least one D and at least one E. [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) Method 1: Arrangements with 3 Es together – arrangements with 3 Es together and 2 Ds together 7! B1 7! − 6! – e, e a positive integer (including 0). 2! 2! M1 f – 6!, f > 6! M1 7! 6! − , a,c = 1, 2 and b,d = 1, 3. a !b ! c !d ! 1800 A1 Method 2: Identified scenarios ^ EEE ^ ^ ^ 6×5 B1 5! × j, j a positive integer (j = 1 may be implied). 5!× 2 M1 k! 6×5 k! 6 k ! 6 P2 7×6 × , × C2 , × or k !× , m! 2 m! m! 2 n k a positive integer (k = 1 may be implied), m = 1, 2 n = 1, 2, 3. M1 m × ( m − 1) k !× k a positive integer > 1, m = 10, 9, 8, 7, 6 and n n = 1, 2. 1800 A1 4 7(b) First 2 marks: Method 1 – Number of arrangements with 2 Ds in one position with 4 letters in between – repeats allowed 7! × 4 × 2 M1 7! × s, s = positive integer > 1. M1 t! × 4 × 2, t = 8, 7, 6. Condone t! × 8. First 2 marks: Method 2 – Picking 2Ds, arranging 4 letters from remaining letters between and then arranging terms 7 P4 × 4!× 2! M1 7 P4 × a!× b! , 1 ⩽ a ⩽ 6 and b = 1, 2, 3. M1 7 Pc × 4!× 2! , c = 3, 4, 5. First 2 marks: Method 3 – Identified scenarios involving Es between Ds D ^ ^ ^ ^ D E E E = 4C4 × 4! × 4! ×2! = 1152 M1 1 identified scenario value correct. D E ^ ^ ^ D E E ^ = 4C3 × 4! × 4! × 3 ×2! = 13824 D E E ^ ^ D E ^ ^ = 4C2 × 4! × 4! × 3 ×2! = 20736 M1 4 appropriate scenarios added, no incorrect. D E E E ^ D ^ ^ ^ = 4C1 × 4! × 4! ×2! = 4608 7(b) Final 3 marks for Methods 1, 2 and 3 40320 A1 If A0 scored, SC B1 for 40320 WWW. [Total number of arrangements =] [9! =] 362880 B1 Accept unsimplified. May be seen as denominator of probability. 40320 1 B1FT their 40320 Probability = = , accept unsimplified. 362880 9 their 362880 B1FT if their 40320 and their 362880 supported by work in this part. Condone their 362880 supported by calculation in 7(a). 5 7(c) Scenarios B1 1 correct unsimplified outcome/value for one identified 4 DE___ C3 4 scenario excluding DDEEE. 4 DEE__ C2 6 Note: 4C1 cannot be used for 4C3 . 4 DEEE_ C1 4 4 M1 Add values of 6 appropriate scenarios, no additional, incorrect DDE__ C2 6 4 or repeated scenarios. Accept unsimplified. DDEE_ C1 4 DDEEE [4C0] 1 [Total =] 25 A1 3

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