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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.
9709_s22_qp_529709_s22_ms_529709/52/M/J/22
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
For n values of the variable x, it is given that Σ x − 200 = 446 and Σ x = 6846. Find the value of n. [3]
Mark scheme
1 Σ x − Σ 200 = Σ ( x − 200 ) B1 Forming a correct 3-term (linear) equation from Σ x, Σ 200 and Σ ( x − 200 ) . Accept 6846 – 200n = 446 OE. Condone 1 sign error. Σ 200 = 200n B1 SOI [ 200n = 6846 − 446 = 6400] n = 32 B1 WWW 3
Question
A fair 6-sided die has the numbers 1, 2, 2, 3, 3, 3 on its faces. The die is rolled twice. The random variable X denotes the sum of the two numbers obtained. (a) Draw up the probability distribution table for X . [3] (b) Find E X and Var X . [3]
Mark scheme
2(a) B1 Table with correct X values and at least one probability. x 2 3 4 5 6 Condone any additional X values if probability stated as 0. p 1 4 10 12 9 B1 3 correct probabilities linked with correct outcomes. 36 36 36 36 36 Accept 3 sf decimals. 0.02778 0.1111 0.2778 0.3333 0.25 B1 2 further correct probabilities linked with correct outcomes. Accept 3 sf decimals. 3 SC B1 for 5 probabilities (0 < p <1) that sum to 1 with less than 3 correct probabilities. 2(b) If method FT from their incorrect (a), expressions for E(X) and Var(X) must be seen at the stage shown in bold (or less simplified) in the scheme with all probabilities < 1. 1 × 2 + 4 × 3 + 10 × 4 + 12 × 5 + 9 × 6 2 + 12 + 40 + 60 + 54 M1 Accept unsimplified expression. May be calculated in variance. E ( X ) = = 36 FT their table with 4 or more probabilities summing to 36 0.999 ⩽ total ⩽ 1 (0 < p <1). 1 × 22 + 4 × 32 + 10 × 42 + 12 × 52 + 9 × 62 2 M1 Appropriate variance formula using their (E(X))2 value. Var ( X ) = − ( their E ( X ) ) = FT their table with 3 or more probabilities (0 < p <1) which 36 need not sum to 1 and the calculation in bold (or less 2 1 × 4 + 4 × 9 + 10 × 16 + 12 × 25 + 9 × 36 14 simplified) seen. − their 36 3 4 + 36 + 160 + 300 + 324 14 2 − their 36 3 168 14 A1 Answers for E(X) and Var(X) must be identified. E( X ) = , , 4.67 E(X) may be identified by correct use in Variance. 36 3 Condone E, V, μ, σ2 etc. 10 1 1440 Var ( X ) = , 1 , 1.11, If M0 earned SC B1 for identified correct final answers. 9 9 1296 3
Question
The back-to-back stem-and-leaf diagram shows the diameters, in cm, of 19 cylindrical pipes produced by each of two companies, A and B. Company A Company B 4 33 1 2 8 9 8 3 2 0 34 1 6 8 9 9 8 7 5 4 1 1 35 1 2 2 3 9 6 5 2 36 5 6 4 3 1 37 0 3 4 38 2 8 Key: 1 35 3 means the pipe diameter from company A is 0.351 cm and from company B is 0.353 cm. (a) Find the median and interquartile range of the pipes produced by company A. [3] It is given that for the pipes produced by company B the lower quartile, median and upper quartile are 0.346 cm, 0.352 cm and 0.370 cm respectively. (b) Draw box-and-whisker plots for companies A and B on the grid below. [3] (c) Make one comparison between the diameters of the pipes produced by companies A and B. [1]
Mark scheme
3(a) Median = 0.355 B1 Identified condone Q2. [IQR =] 0.366 – 0.348 M1 0.365 ⩽ UQ ⩽ 0.369 – 0.343 ⩽ LQ ⩽ 0.349. Subtraction may be implied by answer. 0.018 A1 If 0/3 scored SC B1 for figs Median = 355 IQR = 18. 3 3(b) Box-and-whisker plot on provided grid B1 All 5 key values for B plotted accurately in standard format using their scale. Labelled B. Check accuracy in the middle of vertical line. B1 FT All 5 key values for A, FT from part 3(a), plotted in standard format accurately using their scale. Labelled A. Check accuracy in the middle of vertical line. B1 Whiskers not through box for both, not drawn at corners of boxes, single linear scale with at least 3 values stated, covering at least 0.34 to 0.38 and labelled diameter (d etc) and cm. Accept as a title. 3 If both plots attempted and plot(s) not labelled, SC B1 for at least 1 fully correct set of values plotted. 3(c) A comparison in context B1 Single comment comparing spread or central tendency in context. Must reference either diameter or pipes. Not a simple numerical comparison of statistical values such as median, range, IQR or min/max. 1
Question
The weights, in kg, of bags of rice produced by Anders have the distribution N 2.02, 0.032 . (a) Find the probability that a randomly chosen bag of rice produced by Anders weighs between 1.98 and 2.03 kg. [3] The weights of bags of rice produced by Binders are normally distributed with mean 2.55 kg and standard deviation 3 kg. In a random sample of 5000 of these bags, 134 weighed more than 2.6 kg. (b) Find the value of 3. [4]
Mark scheme
4(a) 1.98 − 2.02 2.03 − 2.02 M1 Use of ±standardisation formula once with 2.02, 0.03 and either [P(1.98 < X < 2.03) = ]P( <z< ) 1.98 or 2.03 substituted appropriately. 0.03 0.03 [= P( −1.333 < z < 0.333) ] Condone 0.032 and continuity correction ±0.005, not √0.03. [= Φ ( 0.333) − (1 − Φ (1.333) ) ] M1 Calculating the appropriate probability area from their z-values. (or 0.6304 – 0.09121 or (0.9087 – 0.5) + (0.6304 – 0.5) etc) = 0.6304 + 0.9087 – 1 0.539 A1 0.539 ⩽ z < 0.5395 Only dependent upon 2nd M mark. If M0 scored SC B1 for 0.539 ⩽ z < 0.5395. 3 4(b) 134 B1 4866 2433 [P(X > 2.6) = = 0.0268 ] 0.9732 or or seen. 5000 5000 2500 [P(X<2.6) = 1 – 0.0268 =] 0.9732 2.6 − 2.55 M1 Use of ±standardisation formula with 2.6 and 2.55 substituted, = 1.93 σ no σ 2 , σ or continuity correction. M1 Their standardisation formula with values substituted equated to z-value which rounds to ±1.93. σ = 0.0259 A1 5 AWRT 0.0259 or . 193 If M0 earned, SC B1 for correct final answer. 4
Question
In a large college, 28% of the students do not play any musical instrument, 52% play exactly one musical instrument and the remainder play two or more musical instruments. A random sample of 12 students from the college is chosen. (a) Find the probability that more than 9 of these students play at least one musical instrument. [3] A random sample of 90 students from the college is now chosen. (b) Use an approximation to find the probability that fewer than 40 of these students play exactly one musical instrument. [5]
Mark scheme
5(a) [P(10, 11, 12) =] M1 One term 12Cx p x (1 − p ) 12 − x , for 0 < x < 12, 0 < p < 1. 12 C10 0.7210 0.282 + 12C11 0.7211 0.281 + 12C12 0.7212 0.280 = 0.193725 + 0.0905726 + 0.0194084 A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.304 B1 Final answer 0.3036 < p ⩽ 0.304. Alternative method for question 5(a) [1 – P(0,1,2,3,4,5,6,7,8,9) =] M1 One term 12Cx p x (1 − p ) 12 − x , for 0 < x < 12, 0 < p < 1. 1 – (12C0 0.720 0.2812 + 12C1 0.721 0.2811 + 12C2 0.72 2 0.2810 + 12 C3 0.723 0.289 + 12C4 0.72 4 0.288 + 12C5 0.725 0.287 + A1 Correct expression, accept unsimplified, no terms omitted, 12 C6 0.726 0.286 + 12C7 0.727 0.285 + 12C8 0.728 0.284 + leading to final answer. 12 C9 0.729 0.283 ) 0.304 B1 Final answer 0.3036 < p ⩽ 0.304. 3 5(b) Mean = [ 0.52 × 90 = ]46.8, var =[ 0.52 × 0.48 × 90] = 22.464 B1 46.8 and 22.464 or 22.46 seen, allow unsimplified, (4.739 < σ ⩽ 4.740 imply correct variance). 39.5 − 46.8 M1 Substituting their mean and their variance into ±standardisation [P(X < 40) =] P z < formula (any number for 39.5), not σ2, √ σ. 22.464 M1 Using continuity correction 39.5 or 40.5 in their standardisation formula. = [P( Z < −1.540)] = 1 − 0. 9382 M1 Appropriate area Φ, from final process, must be probability. 0.0618 A1 0.06175 ⩽ p ⩽ 0.0618 5
Question
(a) Find the number of different arrangements of the 9 letters in the word CROCODILE. [1] (b) Find the number of different arrangements of the 9 letters in the word CROCODILE in which there is a C at each end and the two Os are not together. [3] (c) Four letters are selected from the 9 letters in the word CROCODILE. Find the number of selections in which the number of Cs is not the same as the number of Os. [3] (d) Find the number of ways in which the 9 letters in the word CROCODILE can be divided into three groups, each containing three letters, if the two Cs must be in different groups. [3]
Mark scheme
6(a) 9! B1 2!2! = 90 720 1 6(b) Method 1 Arrangements Cs at ends – Arrangements Cs at ends and Os together 7! M1 w! [Os not together =] − 6! [= 2520 – 720] – y, w = 6, 7 y an integer. 2! 2! w! Condone 2 × − y . 2! M1 a – 6! or a – 720, a an integer resulting in a positive answer. 1800 A1 Method 2 identified scenarios R ^ ^ ^ R 6×5 M1 5! × b, b integer >1. [Os not together =] 5! × = 2! M1 6×5 6 6 P c× or C 2 or 2 or15 , c integer > 1. 2! 2! 1800 A1 3 5 6(c) CCO _ C1 = 5 B1 Correct outcome/value for 1 identified scenario. 5 CC _ _ C2 = 10 Accept unsimplified. 5 OOC _ C1 = 5 WWW 5 OO _ _ C2 = 10 5 C___ C3 = 10 M1 Add 5 or 6 values of appropriate scenarios only, no additional 5 O___ C3 = 10 incorrect scenarios, no repeated scenarios. Accept unsimplified. Condone use of permutations. [Total =] 50 A1 3 5 6(d) Both Os in group with a C C2 = 10 B1 A correct scenario calculated accurately. 5 3 Both Os in group without a C C2 × C2 = 30 Accept unsimplified. 5 One O in a C group, one not C1 × 4C2 = 30 One O with each C (5C1 × 4C1) ÷2! = 10 M1 Add 3 or 4 correct scenario values, no incorrect scenarios, accept repeated scenarios. Accept unsimplified. [Total =] 80 A1 Alternative method for question 6(d) CCO O^^ ^^^ = 5C2 = 10 B1 A correct scenario calculated accurately. CC^ O^^ O^^ = 5C1 ×4C2 = 30 Accept unsimplified. CC^ OO^ ^^^ = 5C1 × 4C1 = 20 9 C6 × 6 C3 M1 Total subtract 2 or 3 correct scenario values, no incorrect Total ways of making three groups = 140 scenarios. 2× 2×3 Accept unsimplified. 140 – (their 10+ their 30+ their 20) 80 A1 3
Question
Hanna buys 12 hollow chocolate eggs that each contain a sweet. The eggs look identical but Hanna knows that 3 contain a red sweet, 4 contain an orange sweet and 5 contain a yellow sweet. Each of Hanna’s three children in turn randomly chooses and eats one of the eggs, keeping the sweet it contained. (a) Find the probability that all 3 eggs chosen contain the same colour sweet. [4] (b) Find the probability that all 3 eggs chosen contain a yellow sweet, given that all three children have the same colour sweet. [2] (c) Find the probability that at least one of Hanna’s three children chooses an egg that contains an orange sweet. [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) 5 4 3 60 1 M1 Either 12 × 11 × 10 in denominator YYY: × × = , 12 11 10 1320 22 or a × (a –1) × ( a – 2), a = 5, 4, 3 in numerator seen in at least one expression. 4 3 2 24 1 OOO: × × = , A1 a a −1 a − 2 12 11 10 1320 55 One expression × × , a = 5, 4, 3 (consistent in 12 11 10 3 2 1 6 1 expression). RRR: × × = , Correct order of values in the numerator is essential. 12 11 10 1320 220 M1 5 4 3 4 3 2 3 2 1 × × + × × + × × , either d = 11, e = 10 or 12 d e 12 d e 12 d e d = 12, e = 12. 1 1 1 Condone + + OE 22 55 220 90 3 A1 0.06818. Dependent only upon the second M mark. [Total =] , , 0.0682 1320 44 7(a) Alternative method for question 7(a) 5 C3 10 1 M1 Either 12 C3 in denominator or a C3 in numerator seen in at least YYY: 12 = , C3 220 22 one expression. 4 C3 4 1 OOO: 12 = , A1 a C C3 220 55 One expression 12 3 a = 5, 4, 3 3 C3 C3 1 RRR: 12 = C3 220 M1 5 C3 4 C3 3 C3 12 + 12 + 12 C3 C3 C3 1 1 1 Condone + + OE 22 55 220 90 3 A1 0.06818. Dependent only upon the second M mark. [Total =] , , 0.0682 1320 44 4 7(b) 60 90 M1 60 1 [P(YYY | all same colour) =] ÷ their P ( YYY ) or or 1320 1320 1320 22 90 3 their 7 ( a ) or or 1320 44 2 A1 OE , 0.667 3 2 7(c) In each method, the M mark requires the scenarios to be identifiable. This may be implied by a list of scenarios and then the calculations which will be assumed to be in the same order. A correct value/expression will be condoned as identifying the connected scenario. Method 1 8 7 6 8 C 14 B1 8 7 6 8 C 336 56 [1 – no orange = ]1− × × or 1 − 12 3 = 1 − × × or 12 3 seen, condone or only, not 12 11 10 C3 55 12 11 10 C3 1320 220 OE. M1 f g h 1− × × 12 d e Either d = 11, e = 10 or d = 12, e = 12 8 C or 1 - 12 3 . C3 14 41 Condone 1 − OE (not ). 55 55 41 A1 0.745 ⩽ p ⩽ 0.74545 55 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. 7(c) Method 2 4 3 2 4 5 4 B1 P(1 O)or P(2 O) correct, accept unsimplified. 12 × 11 × 10 + 12 × 11 × 10 + 672 P(1 O)= ×3 = M1 3 correct scenarios added, with at least one 3-term product of 4 5 3 1320 2× × × f g h 12 11 10 form × × seen, either d = 11, e = 10 or d = 12, e = 12. 12 d e 4 3 8 288 P(2O) = × × ×3 = 12 11 10 1320 24 P(3O) = 1320 984 41 A1 0.745 ⩽ p ⩽ 0.74545 [Total =] = , 0.745 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. 1320 55 Method 3 O Y R = 4 C1 × 5 C1 × 3 C1 = 60 B1 Number of ways either 1 or 2 orange sweets obtained correctly (112 or 48). Accept unsimplified O R R = 4 C1 × 3 C 2 = 12 Note 4C1 × 8C2 = 112 or 4C2 × 8C1 = 48 are correct alternatives. 4 5 O Y Y = C1 × C2 = 40 4 5 M1 3 correct scenarios (1, 2 or 3 orange sweets) added on O O Y = C2 × C1 = 30 numerator, denominator 12 C3 O O R = 4 C2 × 3 C1 = 18 4 O O O = C3 =4 Total = 164 164 Prob = 12 C3 984 41 A1 0.745 ⩽ p ⩽ 0.74545 = , 0.745 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. 1320 55 7(c) Method 4 3 2 4 1 B1 17 17 P(R R O) = × × = P(R ^ ^) = or P(Y ^ ^) = . Accept unsimplified. 12 11 10 55 110 66 3 4 1 P(R O ) = × = M1 3 correct scenarios added, with at least one 3-term product of 12 11 11 3 5 4 1 f g h P(R Y O) = × × = form × × seen, either d = 11, e = 10 or d = 12, e = 12. 12 11 10 22 12 d e 4 1 P(O )= = 12 3 5 3 4 1 P(Y R O) = × × = 12 11 10 22 5 4 5 P(Y O ) = × = 12 11 33 5 4 4 2 P(Y Y O) = × × = 12 11 10 33 984 41 A1 0.745 ⩽ p ⩽ 0.74545 = , 0.745 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. 1320 55 7(c) Method 5 4 1 B1 8 28 P(O )= = P(^ O ) = or P(^ ^ O) = . Accept unsimplified. 12 3 33 165 8 4 8 P(^ O ) = × = M1 3 correct scenarios added, with at least one 3-term product of 12 11 33 8 7 4 28 f g h P(^ ^ O) = × × = form × × seen, either d = 11, e = 10 or d = 12, e = 12 12 11 10 165 12 d e with correct numerator. 984 41 A1 0.745 ⩽ p ⩽ 0.74545 = , 0.745 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. 1320 55 3
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