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Cambridge AS & A Level Mathematics (9709), Paper 5 Probability and Statistics, February/March 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_m20_qp_529709_m20_ms_529709/52/F/M/20
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
The 40 members of a club include Ranuf and Saed. All 40 members will travel to a concert. 35 members will travel in a coach and the other 5 will travel in a car. Ranuf will be in the coach and Saed will be in the car. In how many ways can the members who will travel in the coach be chosen? [3]
Mark scheme
1 Cr or nC34 M1 Either expression seen OE, no other terms, condone x1 38 C34 A1 Correct unsimplified OE 73815 A1 If M0, SCB1 38C34 x k, k an integer 3
Question
An ordinary fair die is thrown repeatedly until a 1 or a 6 is obtained. (a) Find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or a 6. [3] On another occasion, this die is thrown 3 times. The random variable X is the number of times that a 1 or a 6 is obtained. (b) Draw up the probability distribution table for X . [3] (c) Find E X . [2]
Mark scheme
2(a) 2 3 1 2 1 2 1 2 4 M1 One correct term with 0 < p < 1 + + 3 3 3 3 3 3 4 8 16 2432 A1 Correct expression, accept unsimplified = + + = 27 81 243 7776 76 A1 = or 0.313 243 3 2(b) B1 Probability distribution table with correct values of x, no x 0 1 2 3 additional values unless with probability of 0 stated, at least one non-zero probability included 8 12 6 1 P(x) 27 27 27 27 2 3 B1 1 correct probability seen (may not be in table) P(0) = or 3 or 4 non-zero probabilities summing to 1 3 2 1 2 P(1) = × 3 B1 All probabilities correct 3 3 2 2 1 P(2) = × 3 3 3 3 1 P(3) = 3 3 2(c) 8 12 6 1 M1 Correct method from their probability distribution table with at E(X) = 0 × + 1 × + 2× + 3× least 3 terms, 0 ⩽ their P(x) ⩽ 1, accept unsimplified 27 27 27 27 0 12 12 3 = + + + 27 27 27 27 =1 A1 2
Question
The weights of apples of a certain variety are normally distributed with mean 82 grams. 22% of these apples have a weight greater than 87 grams. (a) Find the standard deviation of the weights of these apples. [3] (b) Find the probability that the weight of a randomly chosen apple of this variety differs from the mean weight by less than 4 grams. [4]
Mark scheme
3(a) 87 − 82 M1 Using ± standardisation formula, not σ 2 , not σ , P(X > 87) = P Z > = 0.22 σ no continuity correction 5 B1 AWRT ±0.772 seen PZ < = 0.78 B0 for ±0.228 σ 5 = 0.772 σ σ = 6.48 A1 3 3(b) 4 4 M1 Using ±4 used within a standardisation formula (SOI), P − < Z < = P ( −0.6176 < Z < 0.6176 ) σ σ allow σ 2, σ and continuity correction M1 Standardisation formula applied to both their ±4 Φ = 0.7317 M1 Correct area 2Φ − 1 oe linked to final solution Prob ==2Φ − 1 2 ( 0.7317 ) − 1 = 0.463 A1 4
Question
Richard has 3 blue candles, 2 red candles and 6 green candles. The candles are identical apart from their colours. He arranges the 11 candles in a line. (a) Find the number of different arrangements of the 11 candles if there is a red candle at each end. [2] (b) Find the number of different arrangements of the 11 candles if all the blue candles are together and the red candles are not together. [4]
Mark scheme
4(a) R^^^^^^^^^R M1 9! Alone on numerator, 9! 3! × k or 6! × k on denominator 3!6! = 84 A1 2 4(b) ^ (B B B) ^ ^ ^ ^ ^ M1 7! × k or 7k seen, k an integer > 0 6! 7! 8 × 7 M1 n ( n –1 m × ) or m × nC2 or m × n P2 , n=7, 8 or 9, m an integer > 0 × 6! 2 M1 n = 8 used in above expression = 196 A1 Alternative for question 4(b) [Arrangements, blues together – Arrangements with blues M1 9! Seen alone or as numerator with subtraction together and reds together =] 9! 8! − 2!6! 6! = [252 – 56] M1 8! Seen alone or as numerator in a second term and no other terms M1 All terms divided by 6! x k, k an integer = 196 A1 4
Question
In Greenton, 70% of the adults own a car. A random sample of 8 adults from Greenton is chosen. (a) Find the probability that the number of adults in this sample who own a car is less than 6. [3] A random sample of 120 adults from Greenton is now chosen. (b) Use an approximation to find the probability that more than 75 of them own a car. [5]
Mark scheme
5(a) 1 – P(6, 7, 8) M1 One term 8Cx p x (1 − p ) 8− x , 0 < p < 1, x ≠ 0 = 1 – (8C6 0.76 0.32 + 8C7 0.77 0.31 + 0.78 ) = 1 – 0.55177 A1 Correct unsimplified expression, or better = 0.448 A1 Alternative method for question 5(a) P(0, 1, 2, 3, 4, 5) M1 One term 8Cx p x (1 − p ) 8− x , 0 < p < 1, x ≠ 0 = 0.38 + 8C1 0.710.37 +8C2 0.720.36 + 8C3 0.730.35 + 8 C40.740.34 + 8C5 0.750.33 A1 Correct unsimplified expression, or better = 0.448 A1 3 5(b) Mean = 120 × 0.7 = 84 B1 Correct mean and variance, allow unsimplified Var = 120 × 0.7 × 0.3 = 25.2 75.5 − 84 M1 Substituting their µ and σ into the ±standardising formula (any P( more than 75) = P z > number), not σ2, not √σ 25.2 M1 Using continuity correction 75.5 or 74.5 P( z > −1.693) M1 Appropriate area Φ , from final process, must be a probability = 0.955 A1 Allow 0.9545 < p ⩽ 0.955 5
Question
Box A contains 7 red balls and 1 blue ball. Box B contains 9 red balls and 5 blue balls. A ball is chosen at random from box A and placed in box B. A ball is then chosen at random from box B. The tree diagram below shows the possibilities for the colours of the balls chosen. (a) Complete the tree diagram to show the probabilities. [3] Box A Box B Red Red Blue Red Blue Blue (b) Find the probability that the two balls chosen are not the same colour. [2] (c) Find the probability that the ball chosen from box A is blue given that the ball chosen from box B is blue. [4]
Mark scheme
6(a) Box A Box B B1 Both correct probs, box A Red B1 2 probs correct for box B d B1 All correct probs for box B Red Blue Red Blue Blue 3 6(b) 7 5 1 9 M1 Two 2 factor terms added, correct or FT their 6(a). × + × 8 15 8 15 44 11 A1 OE = or 0.367 120 30 2 6(c) P ( A blue B ∩ blue ) M1 1 6 P(A blue |B blue) = their × seen as numerator or denom of fraction P ( B blue ) 8 15 1 6 1 × = 8 15 = 20 7 5 1 6 41 × + × 8 15 8 15 120 M1 7 5 1 6 their × + × seen 8 15 8 15 M1 7 5 1 6 their × + × seen as denominator 8 15 8 15 6 A1 = or 0.146 41 4
Question
Helen measures the lengths of 150 fish of a certain species in a large pond. These lengths, correct to the nearest centimetre, are summarised in the following table. Length (cm) 0−9 10 − 14 15 − 19 20 − 30 Frequency 15 48 66 21 (a) Draw a cumulative frequency graph to illustrate the data. [4] (b) 40% of these fish have a length of d cm or more. Use your graph to estimate the value of d . [2] The mean length of these 150 fish is 15.295 cm. (c) Calculate an estimate for the variance of the lengths of the fish. [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) 15, 63, 129, 150 B1 Correct cumulative frequencies seen (may be on graph) B1 0 ⩽ Horizontal axis ⩽ 30, 0 ⩽ vertical axis ⩽ 150 Labels correct: length cm, cf M1 At least 3 points plotted at upper end points (e.g. allow 9, 9.5, 10) with a linear horizontal scale. A1 Linear vertical scale, all points at correct upper end points (9.5 etc.), curve drawn accurately, joined to (0,0) (condone (–0.5, 0)) 4 7(b) 60% of 150 = 90 M1 90 seen or implied by use on graph Approx. 16.5 [cm] A1FT FT their increasing cumulative frequency graph, Use of graph must be seen. If no clear evidence of use of graph SCB1FT correct value from their graph 2 7(c) Midpoints: 4.75, 12, 17, 25 M1 At least 3 correct midpoints used (39449.4375 implies M1) 4.752 × 15 + 122 × 48 + 17 2 × 66 + 252 × 21 M1 Using midpoints ±0.5 in correct var formula, including Var = − 15.2952 subtraction of their µ2. 150 = 29.1 A1 3
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Mathematics 9709/52 for February/March 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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