The normal distribution — Paper 7 · A Level Mathematics 9709

3 Weights of garden tables are normally distributed with mean 36 kg and standard deviation 1.6 kg. Weights of garden chairs are normally distributed with mean 7.3 kg and standard deviation 0.4 kg. Find the probability that the total weight of 2 randomly chosen tables is more than the total weight of 10 randomly chosen chairs. [5]

5 marks

Mark scheme:   1 A1 Correct integration ignore limits  3  1

This question in 9709/71 Oct/Nov 2008

3 The weights of pebbles on a beach are normally distributed with mean 48.5 grams and standard deviation 12.4 grams. (i) Find the probability that the mean weight of a random sample of 5 pebbles is greater than 51 grams. [3] (ii) The probability that the mean weight of a random sample of n pebbles is less than 51.6 grams is 0.9332. Find the value of n. [4]

7 marks

Mark scheme:  51 485. 3 (i) P(W > 51) = P  z >  M1 Standardising with 51 and mean 48.5  124. / 5  = 1 – Φ(0.451) M1 Standardising using √5 = 1 – 0.674 = 0.326 A1 [3] Correct answer (ii) z = 1.5 or 1.499 B1 1.5 or 1.499 seen 516. − 485. = 1.5 M1 Standardising must have n (no cc) (124. / n ) n = 6 M1 Attempt to solve equation with n , their z in n = 36 A1 [4] correct answer

This question in 9709/71 Oct/Nov 2009

7 The volume of liquid in cans of cola is normally distributed with mean 330 millilitres and standard deviation 5.2 millilitres. The volume of liquid in bottles of tonic water is normally distributed with mean 500 millilitres and standard deviation 7.1 millilitres. (i) Find the probability that 3 randomly chosen cans of cola contain less liquid than 2 randomly chosen bottles of tonic water. [5] (ii) A new drink is made by mixing the contents of 2 cans of cola with half a bottle of tonic water. Find the probability that the volume of the new drink is more than 900 millilitres. [4]

9 marks

Mark scheme: 7 (i) 3C ~ N(990, 5.22 × 3) (= N(990, 81.12)) B1 Correct mean for both 3 cans cola and 2 bottles water 2W ~ N(1000, 7.12 × 2 (= N(1000, 100.82)) B1 Correct variance for both 3 cans cola and 2 bottles water 3C – 2W ~ N(–10, 181.94) M1 Correct method for mean and variance for 3C – 2W or vice versa  0 − ( −10)  P((3C – 2W) < 0) = Φ   M1 Standardising and using tables, need the sq  181.94  root and area > 0.5 = Φ(0.741) = 0.771 A1 [5] Correct answer (ii) new drink ~ N(910, 2 × 5.22 + 0.52 × 7.12) B1 Correct mean for new drink ~ N (910, 66.68) B1 Correct variance for new drink  900 − 910  P(ND > 900) = 1 – P  z <  M1 Standardising with sq rt and using tables  66.68  = 1 – P(z < –1.225) = Φ (1.225) = 0.8897 (0.890) A1 [4] Correct answer

This question in 9709/71 Oct/Nov 2009

3 Metal bolts are produced in large numbers and have lengths which are normally distributed with mean 2.62 cm and standard deviation 0.30 cm. (i) Find the probability that a random sample of 45 bolts will have a mean length of more than 2.55 cm. [3] (ii) The machine making these bolts is given an annual service. This may change the mean length of bolts produced but does not change the standard deviation. To test whether the mean has changed, a random sample of 30 bolts is taken and their lengths noted. The sample mean length is m cm. Find the set of values of m which result in rejection at the 10% significance level of the hypothesis that no change in the mean length has occurred. [4]

7 marks

Mark scheme: .2 55 .2 62 3 (i) z = = –1.565 M1 Standardising no cc 3.0 / 45 M1 Dividing 0.3 by 45 as denominator P (z > –1.565) = 0.941 A1 Correct answer (Accept equivalent method using totals) [3] (ii) rejection region is m < 1a and m > a 2 a1 − .262 where = −.1645 B1 ±1.645 seen 3.0 / 30 a 2 − .262 and = .1645 M1 one correct unsimplified equation of correct form 3.0 / 30 M1 second unsimplified equation of correct form (or clear use of 1-tail test and ±1.282 used) m < 2.53 and m > 2.71 A1 correct answer [4] GCE AS/A LEVEL – May/June 2010 9709 71 2 2

This question in 9709/71 May/June 2010

4 The weekly distance in kilometres driven by Mr Parry has a normal distribution with mean 512 and standard deviation 62. Independently, the weekly distance in kilometres driven by Mrs Parry has a normal distribution with mean 89 and standard deviation 7.4. (i) Find the probability that, in a randomly chosen week, Mr Parry drives more than 5 times as far as Mrs Parry. [5] (ii) Find the mean and standard deviation of the total of the weekly distances in miles driven by Mr Parry and Mrs Parry. Use the approximation 8 kilometres = 5 miles. [3]

8 marks

Mark scheme: 4 (i) Mr – 5Mrs ~ N(512 – 5×89, 622 + 25×7.42) B1 Correct unsimplified mean ~ N(67, 5213) B1 Correct unsimplified variance P(Mr > 5 Mrs) = P(Mr – 5 Mrs > 0) M1 Using distribution Mr – 5 Mrs  0 − 67  M1 Standardising and using tables = P  z >   5213  = P(z > –0.9280) = 0.823 A1 Correct answer [5] (ii) Mr + Mrs ~ N(601, 622 + 7.42) B1 Correct mean and variance E[5/8(Mr + Mrs)] = 376 miles B1 Correct answer 25 SR Two separate answers 320 and 55.6 B1 Var[5/8(Mr + Mrs)] = × 3898.76 64 = 1520 sd = 39.0 miles B1 Correct answer [3] 5 ∫ 0 2 t

This question in 9709/71 May/June 2010

3 Metal bolts are produced in large numbers and have lengths which are normally distributed with mean 2.62 cm and standard deviation 0.30 cm. (i) Find the probability that a random sample of 45 bolts will have a mean length of more than 2.55 cm. [3] (ii) The machine making these bolts is given an annual service. This may change the mean length of bolts produced but does not change the standard deviation. To test whether the mean has changed, a random sample of 30 bolts is taken and their lengths noted. The sample mean length is m cm. Find the set of values of m which result in rejection at the 10% significance level of the hypothesis that no change in the mean length has occurred. [4]

7 marks

Mark scheme: .2 55 .2 62 3 (i) z = = –1.565 M1 Standardising no cc 3.0 / 45 M1 Dividing 0.3 by 45 as denominator P (z > –1.565) = 0.941 A1 Correct answer (Accept equivalent method using totals) [3] (ii) rejection region is m < 1a and m > a 2 a1 − .262 where = −.1645 B1 ±1.645 seen 3.0 / 30 a 2 − .262 and = .1645 M1 one correct unsimplified equation of correct form 3.0 / 30 M1 second unsimplified equation of correct form (or clear use of 1-tail test and ±1.282 used) m < 2.53 and m > 2.71 A1 correct answer [4] GCE AS/A LEVEL – May/June 2010 9709 72 2 2

This question in 9709/72 May/June 2010

4 The weekly distance in kilometres driven by Mr Parry has a normal distribution with mean 512 and standard deviation 62. Independently, the weekly distance in kilometres driven by Mrs Parry has a normal distribution with mean 89 and standard deviation 7.4. (i) Find the probability that, in a randomly chosen week, Mr Parry drives more than 5 times as far as Mrs Parry. [5] (ii) Find the mean and standard deviation of the total of the weekly distances in miles driven by Mr Parry and Mrs Parry. Use the approximation 8 kilometres = 5 miles. [3]

8 marks

Mark scheme: 4 (i) Mr – 5Mrs ~ N(512 – 5×89, 622 + 25×7.42) B1 Correct unsimplified mean ~ N(67, 5213) B1 Correct unsimplified variance P(Mr > 5 Mrs) = P(Mr – 5 Mrs > 0) M1 Using distribution Mr – 5 Mrs  0 − 67  M1 Standardising and using tables = P  z >   5213  = P(z > –0.9280) = 0.823 A1 Correct answer [5] (ii) Mr + Mrs ~ N(601, 622 + 7.42) B1 Correct mean and variance E[5/8(Mr + Mrs)] = 376 miles B1 Correct answer 25 SR Two separate answers 320 and 55.6 B1 Var[5/8(Mr + Mrs)] = × 3898.76 64 = 1520 sd = 39.0 miles B1 Correct answer [3] 5 ∫ 0 2 t

This question in 9709/72 May/June 2010

5 Cans of drink are packed in boxes, each containing 4 cans. The weights of these cans are normally distributed with mean 510 g and standard deviation 14 g. The weights of the boxes, when empty, are independently normally distributed with mean 200 g and standard deviation 8 g. (i) Find the probability that the total weight of a full box of cans is between 2200 g and 2300 g. [6] (ii) Two cans of drink are chosen at random. Find the probability that they differ in weight by more than 20 g. [5]

11 marks

Mark scheme: 5 (i) W ~N(2240, 848) B2 B1 each parameter 2200 − 2240 (= –1.374) 848 Φ(“–1.374”) = 1– Φ(“1.374”) (= 0.0847) 2300 − 2240 (= 2.060) 848 Φ(“2.060”) (= 0.9803) M1A1 Standardise either value and evaluate correctly Φ(“2.060”) – (1 – Φ(“1.374”)) M1 Correct combination of Φ’s = 0.896 (3 sfs) A1 [6] (ii) X1 – X2 ~N(0, 392) B1 May be implied 20 − 0 (= 1.010) M1 392 (Φ(“1.010” = 0.8438) P(X > 20) = 1 – Φ(“1.010”) (= 0.1562) A1 2 × P(X > 20) M1 = 0.312 (3 sfs) A1 [5]

This question in 9709/71 May/June 2011

5 Each drink from a coffee machine contains X cm3 of coffee and Y cm3 of milk, where X and Y are independent variables with X ∼N(184, 152) and Y ∼N(50, 82). If the total volume of the drink is less than 200 cm3 the customer receives the drink without charge. (i) Find the percentage of drinks which customers receive without charge. [4] (ii) Find the probability that, in a randomly chosen drink, the volume of coffee is more than 4 times the volume of milk. [5]

9 marks

Mark scheme: 5 (i) E(T) = 234, Var(T) = 152 + 82 = 289 B1 200− 234 (= –2.000) M1 " 289" Φ(“–2.000”) = 1 – Φ (“2.000”) M1 1 – 0.9772 2.28% A1 [4] (ii) Require P(D > 0) where D = X – 4Y E(D) (= 184 – 4 × 50) = –16 B1 For –16 or +16 or ± (184 – 4 × 50) Var(D) (= 152 + 42 × 82) = 1249 B1 For 1249 or 152 + 42 × 82 0 − ( −16 ) (= 0.453) M1 "1249" 1 – Φ (“0.453”) M1 (= 1 – 0.6747) = 0.325 A1 [5] [Total: 9] 3 3 k ∫ ( 2 ) d ∫ ( 2 ) d

This question in 9709/73 May/June 2011

4 The masses of a certain variety of potato are normally distributed with mean 180 g and variance 1550 g2. Two potatoes of this variety are chosen at random. Find the probability that the mass of one of these potatoes is at least twice the mass of the other. [7]

7 marks

Mark scheme: 4 Use of X1 – 2X2 or similar Or use of ½ X1 – X2 E(X1–2X2) = 180 – 360 ( = –180 ) B1 E(2X1–X2) = 360 – 180 ( = 180 ) Or E(½ X1 – X2 ) = 90 – 180 = ( –90 ) Var(X1–2X2) = 5×1550 or 7750 M1 for 1550 + 4 × 1550 or ¼ × 1550 + 1550 A1 7750 or 1937.5 0 − ( −180) 0−180 M1 Allow incorrect var (dep > 0 & ≠ 1550), no or '7750' '7750' Standardising – no mixed methods (= ±2.045) Or ± (0 – –90)/ √1937.5 1 – c(‘2.045’) M1 For finding correct area (consistent with working) = 0.0205 or 0.0204 A1 Ans 0.041 (2 sf) B1ft [7] Allow double their prob Total [7]

This question in 9709/73 Oct/Nov 2012

3 Weights of cups have a normal distribution with mean 91 g and standard deviation 3.2 g. Weights of saucers have an independent normal distribution with mean 72 g and standard deviation 2.6 g. Cups and saucers are chosen at random to be packed in boxes, with 6 cups and 6 saucers in each box. Given that each empty box weighs 550 g, find the probability that the total weight of a box containing 6 cups and 6 saucers exceeds 1550 g. [5]

5 marks

Mark scheme: 3 Var(total) = 6(3.22 + 2.62) (+ 0)) (= 102) Total ~ N(1528, 102)) B1 B1 For mean (1528)oe and for variance (102) May be implied by use of N(1528, 10.12) 1550−"1528" (= 2.178) M1 For standardising. No SD/Var mix "102" 1 – Φ(“2.178”) M1 For correct area consistent with working = 0.0147 (3 sf) A1 [5]

This question in 9709/71 May/June 2013

4 The lengths, x m, of a random sample of 200 balls of string are found and the results are summarised by Σ x = 2005 and Σ x2 = 20 175. (i) Calculate unbiased estimates of the population mean and variance of the lengths. [3] (ii) Use the values from part (i) to estimate the probability that the mean length of a random sample of 50 balls of string is less than 10 m. [3] (iii) Explain whether or not it was necessary to use the Central Limit theorem in your calculation in part (ii). [2]

8 marks

Mark scheme: 4 (i) est(µ) = 2005/200 = (10.025) B1 1 20052 est(σ2) = 20175 – ) M1 Correct subst in correct formula 99 200 = 0.376 (3 sf) A1 [3] (ii) 10− '10. 025' (= –0.288) M1 Allow without √, but ÷√50 essential .0' 376256' 50 M1 1 – Φ(‘0.288’) A1 (Use of ‘biased’ variance can still score fully in (ii) ) = 0.387 (3 sf) [3] GCE AS/A LEVEL – May/June 2013 9709 71 (iii) Yes; (assumed distr of X normal) B1 although distr of X unknown B1 [2]

This question in 9709/71 May/June 2013

5 Packets of cereal are packed in boxes, each containing 6 packets. The masses of the packets are normally distributed with mean 510 g and standard deviation 12 g. The masses of the empty boxes are normally distributed with mean 70 g and standard deviation 4 g. (i) Find the probability that the total mass of a full box containing 6 packets is between 3050 g and 3150 g. [5] (ii) A packet and an empty box are chosen at random. Find the probability that the mass of the packet is at least 8 times the mass of the empty box. [5]

10 marks

Mark scheme: 5 (i) 6 × 510 + 70 = (3130) B1 6 × 122 + 42 = (880) B1 N(3130, 880) 3050−3130 (= – 2.697) 880 3150−3130 (= 0.674) M1 Both. With their mean and variance(≥0) Allow 880 without √ M1 Use of tables and attempt to find area consistent Φ(‘0.674’) – (1 – Φ(‘2.697’)) with their working (= 0.7499 – 0.0035) = 0.746 (3 sf) A1 [5] (ii) 510 – 8 × 70 =( – 50) B1 o.e. +50; 510/8 -70; – (510/8 – 70) 122 + 82 × 42 = (1168) B1 o.e. (12/8)² +4² P – 8C ~ N(– 50, 1168) 0 − ( − 50 ) (= 1.463) M1 For standardising with attempt “P-8C” oe with 1168 their mean and variance(≥0).Allow without √ M1 Use of tables and attempt to find area consistent 1 – Φ(‘1.463’) with their working = 0.0717 (3 s.f.) A1 [5] [Total: 10] GCE AS/A LEVEL – May/June 2013 9709 72

This question in 9709/72 May/June 2013

3 Each of a random sample of 15 students was asked how long they spent revising for an exam. The results, in minutes, were as follows. 50 70 80 60 65 110 10 70 75 60 65 45 50 70 50 Assume that the times for all students are normally distributed with mean - minutes and standard deviation 12 minutes. (i) Calculate a 92% confidence interval for -. [4] (ii) Explain what is meant by a 92% confidence interval for -. [1] (iii) Explain what is meant by saying that a sample is ‘random’. [1]

6 marks

Mark scheme: 3 (i) x = 930/15 =(62) B1 z = 1.751 B1 12 ‘62’ ± z × M1 Any z 15 = 56.6 to 67.4 (3 sf) A1 4 Must be an interval (ii) 92 % of such intervals will contain µ B1 1 Accept P(This interval contains µ) = 0.92 (iii) Each possible sample of this size is B1 1 Each member of pop equally likely to be equally likely chosen [Total: 6] GCE AS/A LEVEL – May/June 2013 9709 73

This question in 9709/73 May/June 2013

1 The masses, in grams, of apples of a certain type are normally distributed with mean 60.4 and standard deviation 8.2. The apples are packed in bags, with each bag containing 8 randomly chosen apples. The bags are checked by Quality Control and any bag containing apples with a total mass of less than 436 g is rejected. Find the proportion of bags that are rejected. [4]

4 marks

Mark scheme: 2.8 2.8 21 N(483.2, 537.92) or N(483.2, 23.22) B1 or or seen or implied 8 8 436 − 604. 436− 483 2. 436− 483 2. 8 or (= – M1 or standardising (no mixed methods) 23 2. 537 .92 2.8 / 8 2.035) M1 Correct area consistent with their working Φ(“–2.035”) = 1 – Φ(“2.035”) A1 [4] = 0.021 or 2.1% [Total: 4] 70

This question in 9709/71 May/June 2014

8 In an examination, the marks in the theory paper and the marks in the practical paper are denoted by the random variables X and Y respectively, where X ∼N 57, 13 and Y ∼N 28, 5 . You may assume that each candidate’s marks in the two papers are independent. The final score of each candidate is found by calculating X + 2.5Y. A candidate is chosen at random. Without using a continuity correction, find the probability that this candidate (i) has a final score that is greater than 140, [5] (ii) obtains at least 20 more marks in the theory paper than in the practical paper. [5]

10 marks

Mark scheme: 8 (i) X + 2.5Y ~ N(127, 44.25) B1 B1 for 127 Allow at early stage (57 + 2.5 × 28) B1 B1 for 44.25 or 6.65 Allow at early stage (13 + 2.52 × 5) May be implied by next line 140 − "127 " (±) M1 For standardising " 44 . 25" = ±(1.954) M1 For area consistent with their working 1 – Φ(“1.954”) = 0.0254/0.0253 (3 s.f.) A1 [5] (ii) X – Y ~ N(29, 18) B1 B1 for 29 Give at early stage (57 – 28) B1 B1 for 18 Give at early stage (13 + 5) May be implied by next line 20 − " 29 " (= –2.121) M1 For Standardising "18" 1 – Φ(“–2.121”) = Φ(“2.121”) M1 For area consistent with their working = 0.983 (3 s.f.) A1 [5]

This question in 9709/73 May/June 2014

2 The mean and standard deviation of the time spent by people in a certain library are 29 minutes and 6 minutes respectively. (i) Find the probability that the mean time spent in the library by a random sample of 120 people is more than 30 minutes. [4] (ii) Explain whether it was necessary to assume that the time spent by people in the library is normally distributed in the solution to part (i). [2]

6 marks

Mark scheme: 62 (i) B1 Or 6²/120 oe seen oe seen 120 30 − 29 M1 ± (= 1.826)  6  Allow without √120. No sd/var mix    120  M1 P(z > ‘1.826’) = 1 – Φ(‘1.826’) Correct tail consistent with their A1 [4] working = 0.034 (2 sf) 0.0339 (ii) No B1 1st B1 for either comment n is large (⩾30) Sample mean is (appr) normally distrib 2nd B1 for’No’with 2nd comment or The CLT applies oe B1 [2] (No mark for ‘No’ alone) Total: 6 3420

This question in 9709/71 Oct/Nov 2015

6 The weights, in kilograms, of men and women have the distributions N 78, 72 and N 66, 52 respectively. (i) The maximum load that a certain cable car can carry safely is 1200 kg. If 9 randomly chosen men and 7 randomly chosen women enter the cable car, find the probability that the cable car can operate safely. [5] (ii) Find the probability that a randomly chosen woman weighs more than a randomly chosen man. [4]

9 marks

Mark scheme: 6 (i) E(T) = 9 × 78 + 7 × 66 (= 1164) B1 Or 9 × 78 + 7 × 66 – 1200 B1 Var(T) = 9 × 7² + 7 × 5² (= 616) M1 ± Allow without √ 1200− '1164 ' (= 1.450) '616 ' P(z < 1.450) = Φ (1.450) M1 = 0.927 (3 sf) A1 [5] Correct tail consistent with their mean (ii) E(D) = 66 – 78 (= –12) B1 Both needed Var(D) = 7² + 5² (= 74) 0 − ('−12' ) M1 ± Allow without √ (= 1.395) 74 P(D > 0) = 1 – Φ (‘1.395’) M1 Correct tail consistent with their mean A1 [4] Similar scheme for P(M – W) < 0 0.0815 (3 sf) Total: 9

This question in 9709/72 Oct/Nov 2015

7 Bags of sugar are packed in boxes, each box containing 20 bags. The masses of the boxes, when empty, are normally distributed with mean 0.4 kg and standard deviation 0.01 kg. The masses of the bags are normally distributed with mean 1.02 kg and standard deviation 0.03 kg. (i) Find the probability that the total mass of a full box of 20 bags is less than 20.6 kg. [5] (ii) Two full boxes are chosen at random. Find the probability that they differ in mass by less than 0.02 kg. [5]

10 marks

Mark scheme: 7 (i) E(T) =20.8 B1 Var(T)= 20 × 0.032 + 0.012(= 0.0181) B1 or √(20 × 0.032 + 0.012 ) = 0.135 (3sf) 20.6 − 20.8 (= –1.487) "0.0181" M1 For standardising (σ must come from combination) 1 – Φ(“1.487”) M1 Area consistent with their working = 0.0684 to 0.686 A1 [5] Any answer within range (ii) E(D) = 0 Var(D) = 2 × 0.0181(= 0.0362) B1 Both (Seen or implied) 0.02 − 0 ( = 0.105) M1 Allow without √ "0.0362) A1 Allow to 3sf Φ(“0.105”) = 0.5418 or 1-Φ(0.015) =0.4582 M1 or 1 – 2(1 – Φ(“0.105”)) Φ(“0.105”) – (1 – Φ(“0.105”) (= 1 – 2 × 0.4582) (= 0.5418 – 0.4582) A1 [5] = 0.0836/0.0837

This question in 9709/73 May/June 2016

5 Large packets of sugar are packed in cartons, each containing 12 packets. The weights of these packets are normally distributed with mean 505 g and standard deviation 3.2 g. The weights of the cartons, when empty, are independently normally distributed with mean 150 g and standard deviation 7 g. (i) Find the probability that the total weight of a full carton is less than 6200 g. [5] … … … … … … … … … … … … … … … … … … … … … … … Small packets of sugar are packed in boxes. The total weight of a full box has a normal distribution with mean 3130 g and standard deviation 12.1 g. (ii) Find the probability that the weight of a randomly chosen full carton is less than double the weight of a randomly chosen full box. [5] … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(i) W ~ N(6210, 171.88) B2 seen or implied. B1 each parameter 6200 "6210" "171.88" − (= – 0.763) M1 Standardising with their values. No sd / var mix 1 – Φ(“0.763”) M1 For area consistent with their mean = 0.223 (3 sfs) A1 Total: 5 Question Answer Marks Guidance 5(ii) E(C – 2B) = ̶ 50 M1 “6210”–2(3130) (or E(2B–C)=50 Var(C – 2B) = "171.88" + 22 × 12.12 (= 757.52) M1 0 ( 50) "757.52" −− (= 1.817) M1 Standardising with their values Φ(“1.817”) M1 For area consistent with their mean = 0.965 (3 sfs) A1 Total: 5

This question in 9709/71 May/June 2017

3 The mass, in tonnes, of iron ore produced per day at a mine is normally distributed with mean 7.0 and standard deviation 0.46. Find the probability that the total amount of iron ore produced in 10 randomly chosen days is more than 71 tonnes. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 10 × 0.462 (= 2.116) or 0.46 10 B1 Total mass of ore ~ N(70, 2.116) or ~N 2 0.46 7, 10               B1 71 "70" "2.116" − ± or 7.1 "7.0" 0.46 / 10 − ± (= 0.687) M1 correct, using their sd or √(their var) e.g. allow 71 "70" 4.6 − for M1 1 – ɸ("0.687") M1 for correct area consistent with their working = 0.246 (3 sf) A1 Total: 5

This question in 9709/73 May/June 2017

4 The volume, in millilitres, of a small cup of coffee has the distribution N 103.4, 10.2 . The volume of a large cup of coffee is 1.5 times the volume of a small cup of coffee. (i) Find the mean and standard deviation of the volume of a large cup of coffee. [3] … … … … … … … (ii) Find the probability that the total volume of a randomly chosen small cup of coffee and a randomly chosen large cup of coffee is greater than 250 ml. [4] … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) mean= 155.1 B1 var = 1.52 × 10.2 ( = 22.95) M1 or 1.5 × √10.2 sd = √"22.95" = 4.79 A1 3 4(ii) mean = 103.4 + “155.1” (= 258.5) B1ft Both. ft their 155.1 and 22.95. Accept var = 10.2 + “22.95” (=33.15) sd. 250 −"258.5" (= –1.476) M1 Standardising – no sd/var mix. Their "33.15" mean/sd must be from an attempt at combination 1– ɸ(–1.476) = ɸ(1.476) M1 For area consistent with their working = 0.930 (3 sf) A1 Allow 0.93 4

This question in 9709/71 May/June 2018

5 The mass, in kilograms, of rocks in a certain area has mean 14.2 and standard deviation 3.1. (i) Find the probability that the mean mass of a random sample of 50 of these rocks is less than 14.0 kg. [3] … … … … … … … … … … … … … … … … … … (ii) Explain whether it was necessary to assume that the population of the masses of these rocks is normally distributed. [1] … … … … (iii) A geologist suspects that rocks in another area have a mean mass which is less than 14.2 kg. A random sample of 100 rocks in this area has sample mean 13.5 kg. Assuming that the standard deviation for rocks in this area is also 3.1 kg, test at the 2% significance level whether the geologist is correct. [5] … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) 14 − 14.2 M1 For stand'n; must have √50 (= – 0.456) 3.1 50 1 – Φ(“0.456”) M1 for area consistent with their working = 0.324 (3 sfs) A1 3 5(ii) No because n large B1 Accept n > 30 1 5(iii) H0: µ = 14.2 B1 H1: µ < 14.2 or ‘pop mean’, but not just ‘mean’ 13.5 − 14.2 M1 For stand'n; must have √100 3.1 100 = –2.258 A1 comp –2.054 (or –2.055) M1 Valid comparison of z values or areas (0.0119 < 0.02) There is evidence (at 2% level) that mean A1ft Ft their z. Correct conclusion no mass in this area < 14.2 contradictions 5

This question in 9709/71 May/June 2018

4 The mean mass of packets of sugar is supposed to be 505 g. A random sample of 10 packets filled by a certain machine was taken and the masses, in grams, were found to be as follows. 500 499 496 495 498 490 492 501 494 494 (i) Find unbiased estimates of the population mean and variance. [3] … … … … … … … … … … The mean mass of packets produced by this machine was found to be less than 505 g, so the machine was adjusted. Following the adjustment, the masses of a random sample of 150 packets from the machine were measured and the total mass was found to be 75 660 g. (ii) Given that the population standard deviation is 3.6 g, test at the 2% significance level whether the machine is still producing packets with mean mass less than 505 g. [5] … … … … … … … … … … … … … … … … … … … … … … … … (iii) Explain why the use of the normal distribution is justified in carrying out the test in part (ii). [1] … … … … … … … … …

9 marks

Mark scheme: 4(i) Est(µ) = 495.9 B1 Accept 496 Est(σ2) = 2 10 2459283 9 10 ( "495.9" ) − M1 Attempt Σx2 and subst in correct formula (1/9(“2459283” – “4959”2/10)). May be implied by correct answer = 12.8 (3 sf) or 383/30 A1 (Note: Biased var “11.49” scores M0 A0) 3 4(ii) H0: µ = 505 H1: µ < 505 75660 505 150 3.6 150 − ÷ B1 Allow ‘Pop mean’ but not just ‘mean’ = –2.04 M1 Correct stand'n; must have √150. No sd/var mixes. Condone sample SD (3.58/3.39) Accept standardisation of totals ((75660-75750)/44.091) Accept CV method A1 Accept +2.04 (Note: if valid area comparison done 0.0207/0.0206 or 0.979 needed for A1) comp z = –2.054 M1 Valid comparison of z’s or area (0.0207/6>0.02; 0.979(3)<0.98) No evidence (at 2%) that machine pkts mean mass < 505 A1ft oe No contradictions. SC Two tail test can score B0 M1 A1 M1 for comparison with 2.326 A0 (max 3/5) 5 Question Answer Marks Guidance 4(iii) Large sample, so sample mean approx normally distr'd B1 Allow just ‘Sample is large’ or ‘n is large’ n>30 1

This question in 9709/72 May/June 2018

6 The times, in minutes, taken to complete the two parts of a task are normally distributed with means 4.5 and 2.3 respectively and standard deviations 1.1 and 0.7 respectively. (i) Find the probability that the total time taken for the task is less than 8.5 minutes. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the probability that the time taken for the first part of the task is more than twice the time taken for the second part. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) E(T) = 4.5 + 2.3 (= 6.8) Var(T) = 1.12 + 0.72 (= 1.7) M1 8.5 "6.8" "1.7" − (= 1.304) M1 Correct stand'n using their µ and σ 2 must be a combination of the two variables ɸ("1.304") M1 Area consistent with their working = 0.904 (3 sf) A1 4 6(ii) E(D) = 4.5 – 2 × 2.3 or –0.1 M1 Var(D) = 1.12 + 22×0.72 or 3.17 M1 Both can seen or implied 0 (' 0.1') '3.17' −− (= 0.056) M1 Correct stand'n using their µ and σ 2 must be a Combination of the two variables 1 – ɸ("0.056") M1 Area consistent with their working = 0.478 (3 sf) A1 5

This question in 9709/73 May/June 2018

5 The times, in months, taken by a builder to build two types of house, P and Q, are represented by the independent variables T1 ∼N 2.2, 0.42 and T2 ∼N 2.8, 0.52 respectively. (i) Find the probability that the total time taken to build one house of each type is less than 6 months. [4] … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the probability that the time taken to build a type Q house is more than 1.2 times the time taken to build a type P house. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) T1 + T2 ~ N( 5, 0.42 + 0.52 ) B1 or N( 5, 0.41 ) 6 − 5 (= 1.562) M1 Allow cc '0.41' Φ(‘1.562’) M1 Correct area consistent with their working = 0.941 A1 4 5(ii) Var(T2 - 1.2T1) = 0.52 + 1.22 × 0.42 B1 Or similar using 1.2T1 – T2 (= 0.4804) T2 – 1.2T1 – N(0.16, 0.4804) B1 ft Only ft attempt at combination. no ft for neg var. 0 − '0.16' M1 Standardise with their mean and variance. (= -0.231) Allow cc '0.4804' P(T2 – 1.2T1) > 0 = Φ(‘0.231’) M1 Correct area consistent with their working = 0.591 (3 sfs) A1 5

This question in 9709/73 Oct/Nov 2018

4 The heights of a certain variety of plant are normally distributed with mean 110 cm and variance 1050 cm2. Two plants of this variety are chosen at random. Find the probability that the height of one of these plants is at least 1.5 times the height of the other. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4 Use of 1.5X1 – X2 or similar B1 E(1.5X1–X2) =1.5(110) –110 (= 55) B1 or E(X1–1.5X2) =110 – 1.5(110) (= –55) Var(1.5X1–X2) =1.52×1050+1050 (or 3412.5) M1 Correct expression or result 0 55 '3412.5' − or 0 ( 55) '3412.5' −− (= ± 0.942) M1 Their ‘55’. Allow incorrect var (dep > 0 and ≠ 1050) 1 – Φ(‘0.942’) M1 Area consistent with their working = 0.173 A1 Ans 0.346 (3 sf) B1 FT double their prob (must be <1) 7

This question in 9709/72 May/June 2019

5 The masses, in grams, of large boxes of chocolates and small boxes of chocolates have the distributions N 325, 6.1 and N 167, 5.6 respectively. (i) Find the probability that the total mass of 10 randomly chosen large boxes of chocolates is less than 3240 g. [4] … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the probability that the mass of a randomly chosen large box of chocolates is more than twice the mass of a randomly chosen small box of chocolates. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) B1 Or mean = 325 var. 6.1 10 = ( ) 3240 3250 1.280 61 − = − M1 Standardise with their values (no mixed methods) φ('–1.280') = 1 – φ('1.280) M1 Area consistent with their figures 0.100 A1 Allow 0.1 4 5(ii) E(D) = 325 – 2 × 167 = –9 B1 Accept ±9 Var(D) = 6.1 + 22 × 5.6 (= 28.5) B1 ( )( ) 0 9 1.686 28.5 −− = M1 Standardising with their values. Must have a combination attempt on denominator and 1 −φ('1.686') M1 Area consistent with their figures 0.0459 A1 5

This question in 9709/72 Oct/Nov 2019