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Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w22_qp_539709_w22_ms_539709/53/O/N/22
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Question
50 values of the variable x are summarised by Σ x − 20 = 35 and Σ x2 = 25 036. Find the variance of these 50 values. [3]
Mark scheme
1 ∑ x − 50 × 20 = 35 ; ∑ x = 1035 B1 Correct value for ∑ x or x . 35 1035 or x = + 20 = [= 20.7] 50 50 25036 Σ x 2 25036 1035 2 M1 25036 Σx 2 − = − − their 50 50 50 50 50 50 72.23 A1 Exact answer only SC B1 for 72.23 with no substitution in formula. 3
Question
In a large college, 32% of the students have blue eyes. A random sample of 80 students is chosen. Use an approximation to find the probability that fewer than 20 of these students have blue eyes. [5]
Mark scheme
2 Mean = 80 × 0.32 = 25.6, B1 25.6 and 17.4[08] seen, allow unsimplified. var = 80 × 0.32 × 0.68 = 17.408 4.172… implies correct variance. 19.5 − 25.6 M1 Substituting their 25.6 and 17.408 into ±standardisation P(X < 20) = P( Z < ) = P( Z < −1.462) formula (any number for 19∙5), not σ2, √ σ. 17.408 M1 Using continuity correction 19∙5 or 20∙5 in their standardisation formula. = [1 − Φ (1.462)] = 1 – 0.9282 M1 Appropriate area Φ, from final process, must be probability. (Expect final ans < 0∙5 ). Note: the correct final answer may imply M1 from use of calculator. 0.0718 A1 0.0718 ⩽ p ⩽0.0719 5
Question
The times, t minutes, taken to complete a walking challenge by 250 members of a club are summarised in the table. Time taken (t minutes) t ≤ 20 t ≤ 30 t ≤ 35 t ≤ 40 t ≤ 50 t ≤ 60 Cumulative frequency 32 66 112 178 228 250 (a) Draw a cumulative frequency graph to illustrate the data. [2] (b) Use your graph to estimate the 60th percentile of the data. [1] It is given that an estimate for the mean time taken to complete the challenge by these 250 members is 34.4 minutes. (c) Calculate an estimate for the standard deviation of the times taken to complete the challenge by these 250 members. [4]
Mark scheme
3(a) Cumulative frequency graph M1 At least 3 points plotted accurately at class upper end points: (20,32), (30, 66), (35, 112), (40, 178), (50, 228), (60, 250). Linear cf scale 0 ⩽ cf ⩽ 250 and linear time scale 0 ⩽ time ⩽ 60 with at least 3 values identified on each. A1 All points plotted correct, curve drawn (within tolerance) and joined to (0,0). Axes labelled cumulative frequency (cf), time (t) and minutes (min or m) – or a suitable title. Axes can be the other way round. 2 3(b) Line drawn from 150 on cf axis to meet graph at about B1 FT Must be an increasing cf graph with correct upper bounds. t =38 minutes Use of graph must be seen. Expect an answer in range 37 ⩽ t ⩽ 39 for a correct graph 1 3(c) [Frequencies] [32] 34 46 66 50 22 B1 May be unsimplified and/or in variance calculation. [Midpoints] 10 25 32.5 37.5 45 55 M1 At least 5 correct midpoints seen , may be unsimplified. [Variance] = M1 Correct unsimplified Variance formula with their midpoints 32 × 102 + 34 × 252 + 46 × 32.52 + 66 × 37.52 + 50 × 452 + 22 × 552 and their frequencies for var or sd. − 34.42 250 ( − mean2 included) 333650 [= − 34.42 = 151.24] 250 [Sd =] 12.3 A1 Awrt WWW SC B1 for 12.3 if second M1 not awarded. 4
Question
Three fair 4-sided spinners each have sides labelled 1, 2, 3, 4. The spinners are spun at the same time and the number on the side on which each spinner lands is recorded. The random variable X denotes the highest number recorded. 7. (a) Show that P X = 2 = 64 [3] (b) Complete the probability distribution table for X . [2] x 1 2 3 4 7 19 P X = x 64 64 On another occasion, one of the fair 4-sided spinners is spun repeatedly until a 3 is obtained. The random variable Y is the number of spins required to obtain a 3. (c) Find P Y = 6. [1] (d) Find P Y > 4. [2]
Mark scheme
4(a) Method 1: Scenarios identified [no of ways for score of 2 are] 222, 211, 212, 221, 122, 112, 121 B1 7 correct scenarios identified, no incorrect. [Total options = 64] 7 7 M1 a [So P(X = 2) =] = , a = their number of correct identified 4× 4× 4 64 4× 4× 4 scenarios > 4 A1 Approach identified, WWW. Method 2: P(2 on all spinners) + P(2 on two spinners and 1 on one spinner) + P(2 on one spinner and 1 on two spinners) 3 1 3 1 1 1 3 1 1 1 B1 3 1 3 1 1 1 + C2 × × + C1 × × 4 4 4 4 4 4 4 ( ) + C2 or C1 × × + d , 0 < d< 1 4 3 4 4 4 M1 1 3 1 3 1 3 + e + f 1 < e <5 and 1 < f < 5 4 4 4 7 A1 Approach identified, WWW. [So P(X = 2) =] = 64 Method 3: P(1 or 2 on each spinner) – P(1 on all spinners) 3 1 1 3 B1 1 3 − − b seen, 0 < b < 1 2 4 2 M1 1 3 −c , 0 < c < ½ 3 2 7 A1 Approach identified, WWW. [So P(X = 2) =] 64 3 4(b) 1 B1 P(X = 1) or P(X = 4) correct. Condone answers not in P( X = 1) = probability distribution table if clearly identified. 64 1 7 19 37 B1 FT All 4 probabilities summing to 1. P(X = 4) = 1 − − − = 64 64 64 64 2 4(c) 3 5 1 243 B1 Accept 0.059326… to 4 or more SF. P(Y = 6) = × = 0.0593, 4 4 4096 1 4(d) 3 4 M1 3 g 4 , g = 4, 5 or p where 0 < p < 1 4 4 81 A1 Accept 0.316406…to 4 or more SF. = ,0.316 256 Alternative method for Question 4(d) 1 3 1 3 2 1 3 3 1 M1 Correct or P(Y >4) = 1 – P( Y⩽4) = 1 − + × + × + × 1 3 1 3 2 1 3 3 1 3 4 4 4 4 4 4 4 4 1− + × + × + × + 4 4 4 4 4 4 4 4 175 = 1 − 256 ( ) or 1 − p + qp + q 2 p + q3 p where 0 < p < 1 and q = 1–p 81 A1 Accept 0.316406…to 4 or more SF. = ,0.316 256 2
Question
Company A produces bags of sugar. An inspector finds that on average 10% of the bags are underweight. 10 of the bags are chosen at random. (a) Find the probability that fewer than 3 of these bags are underweight. [3] The weights of the bags of sugar produced by company B are normally distributed with mean 1.04 kg and standard deviation 0.06 kg. (b) Find the probability that a randomly chosen bag produced by company B weighs more than 1.11 kg. [3] 81% of the bags of sugar produced by company B weigh less than w kg. (c) Find the value of w. [3]
Mark scheme
5(a) [P(0, 1, 2) =] 10C0 0.10 0.910 + 10C1 0.11 0.99 + 10C2 0.12 0.98 M1 One term 10Cx p x (1 − p )10− x , 0 < p < 1, x ≠ 0 = 0.348678+0.38742+0.19371 A1 Correct expression, accept unsimplified. 0.930 B1 0.9298 ⩽ p ⩽ 0.9303 Alternative method for Question 5(a) [1 – P(3, 4, 5, 6, 7, 8, 9, 10) = 1 – (10C3 0.97 0.13 +10C4 0.96 0.14 +10C5 M1 One term 10Cx p x (1 − p )10− x , 0.95 0.15 +10C6 0.94 0.16 +10C7 0.93 0.17 +10C8 0.92 0.18 +10C9 0.91 0.19 0 < p < 1, x ≠ 0 +10C10 0.90 0.110 ) A1 Correct expression, accept unsimplified. 0.930 B1 0.9298 ⩽ p ⩽ 0.9303 3 5(b) 1.11 − 1.04 M1 1.11, 1.04 and 0.06 substituted into ±Standardisation [P(X > 1.11) = ]P( Z > ) = P( Z > 1.167) formula, no continuity correction not 0.062 or √0.06 0.06 = 1 – 0.8784 M1 1 – their 0.8784 as final answer, must be probability. (Expect final ans < 0∙5 ). 0.122 A1 0.1216 ⩽ p ⩽ 0.122 SC M0 M1 B1 for 0.122 with no standardisation formula. 3 5(c) w − 1.04 B1 0.8775 < z ⩽ 0.878 or −0.878 ⩽ z < −0.8775 seen. [P(X < w) = P( Z < ) = 0.81] 0.06 w − 1.04 M1 1.04 and 0.06 substituted in ±standardisation formula, no = 0.878 continuity correction, not σ2, √ σ, equated to a z-value. 0.06 w = 1.09 A1 1.09 ⩽ w ⩽ 1.093 3
Question
(a) Find the number of different arrangements of the 9 letters in the word ACTIVATED. [2] (b) Find the number of different arrangements of the 9 letters in the word ACTIVATED in which there are at least 5 letters between the two As. [3] Five letters are selected at random from the 9 letters in the word ACTIVATED. (c) Find the probability that the selection does not contain more Ts than As. [5]
Mark scheme
6(a) 9! M1 h! , h = 7, 8, 9; j = 1, 2 2!2! 2!× j ! 90720 A1 2 6(b) Arrangements with 5 letters between As + Arrangements with 6 letters between As + Arrangements with 7 letters between As 7! M1 7! With gap of 5: × 3 [= 7560] × k , k positive integer 1< k < 7 2! 2! 7! With gap of 6: × 2 [= 5040] M1 Add their no of ways for 3 identified correct scenarios, no 2! additional incorrect scenarios, accept unsimplified. 7! With gap of 7: ×1 [= 2520] 2! 7! A1 [Total no = × 6 =] 15120 2! 3 6(c) Method 1: Summing number of ways AT _ _ _ 2×2× 5C3 40 B1 Correct no of ways for 4 correctly identified scenarios, A____ 2×5C4 10 accept unsimplified. 5 AATT _ C1 5 AAT _ _ 2×5C2 20 M1 Add no of ways for 5 or 6 identified correct scenarios, no 5 AA _ _ _ C3 10 additional incorrect scenarios, no repeated scenarios, accept 5 _____ C5 1 unsimplified. [Total no of ways not containing more Ts than As = ] A1 All correct and added = 40+10+5+20+10+1 [=86] 86 M1 their 86 Probability = 9 accept numerator unevaluated C5 9C 5 or their identified total 86 43 A1 , , 0.683 126 63 Method 2: Subtracting no of ways with more Ts from total T____ 2×5C4 10 B1 Correct no of ways for 2 correctly identified scenarios, no TTA _ _ 2×5C2 20 additional incorrect scenarios, no repeated scenarios, accept 5 TT _ _ _ C3 10 unsimplified, condone use of permutations M1 Add no of ways for 2 or 3 correct scenarios and subtract from their total no of ways All correct and subtracted Total no of ways with more Ts than As =40 A1 9 C5 − 40 = 86 86 M1 their 86 Probability = 9 accept numerator unevaluated C5 9C 5 or their identified total 6(c) 43 A1 , 0.683 63 5
Question
Sam and Tom are playing a game which involves a bag containing 5 white discs and 3 red discs. They take turns to remove one disc from the bag at random. Discs that are removed are not replaced into the bag. The game ends as soon as one player has removed two red discs from the bag. That player wins the game. Sam removes the first disc. (a) Find the probability that Tom removes a red disc on his first turn. [2] (b) Find the probability that Tom wins the game on his second turn. [4] (c) Find the probability that Sam removes a red disc on his first turn given that Tom wins the game on his second turn. [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(a) 3 2 5 3 M1 3 2 5 3 [P(SR TR) + P(SW TR) =] × + × × + k or l + × 0 < k,l < 1 8 7 8 7 8 7 8 7 21 3 A1 3 = , , 0.375 SC B1 for with no explanation. 56 8 8 2 7(b) [RRWR, WRRR, WRWR] M1 m n o q × × × 1 ⩽ m,n,o,q ⩽ 5, m ≠ n ≠ o ≠ q 3 2 5 1 5 3 2 1 5 3 4 2 8 7 6 5 × × × + × × × + × × × 8 7 6 5 8 7 6 5 8 7 6 5 1 1 1 A1 Probability for one scenario correct, accept unsimplified. [= + + ] 56 56 14 M1 Adding probabilities for 3 correct scenarios and no incorrect. 180 3 A1 Or 0.1071428… to 4SF or better. = , , 0.107 SC B1 for 3/28 with inadequate explanation. 1680 28 4 7(c) 30 1 M1 3 2 5 1 their P ( RRWR ) or × × × [P(S first disc R |T2 ) =] 1680 = 56 8 7 6 5 3 3 3 their 7 ( b ) − must be a prob or 28 28 28 1 A1 , 0.167 6 2
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