The normal distribution — Paper 6 · A Level Mathematics 9709

4 In a certain country the time taken for a common infection to clear up is normally distributed with mean µ days and standard deviation 2.6 days. 25% of these infections clear up in less than 7 days. (i) Find the value of µ. [4] In another country the standard deviation of the time taken for the infection to clear up is the same as in part (i), but the mean is 6.5 days. The time taken is normally distributed. (ii) Find the probability that, in a randomly chosen case from this country, the infection takes longer than 6.2 days to clear up. [3]

7 marks

Mark scheme: 4 (i) 7 − µ B1 ± .0674 seen only − .0 674 = M1 Standardising must have a recognisable z- 6.2 value, no cc and 2.6 M1 For solving their equation with recognisable z-value, µ and 2.6 not 1 – 0.674 or 0.326, allow cc µ = 8.75 A1 4 Correct answer  2.6 − 5.6  (ii) P(X > 6.2) = P  z >  M1 Standardising, no cc on the 6.2  6.2  M1 prob > 0.5 = P(z > –0.1154) = 0.546 A1 3 Correct answer GCE A/AS LEVEL – May/June 2008 9709 06

This question in 9709/61 May/June 2008

3 (i) The daily minimum temperature in degrees Celsius (◦C) in January in Ottawa is a random variable with distribution N(−15.1, 62.0). Find the probability that a randomly chosen day in January in Ottawa has a minimum temperature above 0 ◦C. [3] (ii) In another city the daily minimum temperature in ◦C in January is a random variable with distribution N(µ, 40.0). In this city the probability that a randomly chosen day in January has a minimum temperature above 0 ◦C is 0.8888. Find the value of µ. [3]

6 marks

Mark scheme:  0 151.  3 (i) P(X > 0) = 1 – Φ   M1 Standardising, sq rt, no cc  62  = 1 – Φ (1.918) M1 Prob < 0.5 after use of normal tables = 1 – 0.9724 = 0.0276 or answer rounding to A1 [3] Correct answer (ii) z = –1.22 B1 z = ±.1 22 0 − µ –1.22 = 40 M1 an equation in µ, recognisable z, 40 , no cc µ = .772 c.a.o A1 [3] correct answer c.w.o from same sign on both sides 12!

This question in 9709/61 Oct/Nov 2008

1 The volume of milk in millilitres in cartons is normally distributed with mean µ and standard deviation 8. Measurements were taken of the volume in 900 of these cartons and it was found that 225 of them contained more than 1002 millilitres. (i) Calculate the value of µ. [3] (ii) Three of these 900 cartons are chosen at random. Calculate the probability that exactly 2 of them contain more than 1002 millilitres. [2]

5 marks

Mark scheme: 1 (i) z = 0.674 B1 ± 0.674 or rounding to, seen, e.g. 0.6743 1002 − µ = 0.674 M1 Standardising and attempting to solve for µ, must 8 use recognisable z-value, no cc, no sq rt, no sq µ = 997 A1 [3] Correct answer rounding to 997 225 224 675 (ii) P(2) = 3 × × × M1 900 × 899 × 898 or 900C3 seen in denom 900 899 898 = 0.140 A1 [2] Correct answer not 0.141 or 0.14 225 C 2 × 675 C1 OR 900 C 3

This question in 9709/61 May/June 2009

3 On a certain road 20% of the vehicles are trucks, 16% are buses and the remainder are cars. (i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses. [3] (ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the probability that more than 73 are cars. [5]

8 marks

Mark scheme: 3 (i) P(X < 3) = P(0) + P(1) + P(2) M1 Binomial term with 11Cr pr (1–p)11–r seen = (0.84)11 + (0.16)(0.84)10 × 11C1 + M1 Correct expression for P(0, 1, 2) or P(0, 1, 2, 3) (0.16)2(0.84)9 × 11C2 Can have wrong p = 0.1469 + 0.30782 + 0.2931 = 0.748 A1 [3] Correct final answer. Normal approx M0 M0 A0 (ii) µ = 125 × 0.64 = 80 B1 80 and 28.8 or 5.37 seen σ2 = 125 × 0.64 × 0.36 = 28.8  735. − 80  P (X > 73) = 1 – Φ   M1 standardising, with or without cc, must have sq rt in  288.  denom M1 continuity correction 73.5 or 72.5 only = Φ (1.211) M1 correct region (> 0.5 if mean > 73.5, vv if mean < 73.5 = 0.887 A1 [5] correct answer GCE A/AS LEVEL – May/June 2009 9709 06 13 12 6 7

This question in 9709/61 May/June 2009

3 The times for a certain car journey have a normal distribution with mean 100 minutes and standard deviation 7 minutes. Journey times are classified as follows: ‘short’ (the shortest 33% of times), ‘long’ (the longest 33% of times), ‘standard’ (the remaining 34% of times). (i) Find the probability that a randomly chosen car journey takes between 85 and 100 minutes. [3] (ii) Find the least and greatest times for ‘standard’ journeys. [4]

7 marks

Mark scheme: 3 (i) P(85 < x < 100)   85 − 100   85 − 100 seen oe or ± .214 B1 ± z <   = 0.5 – P   7  7    = 0.5 – P (z < – 2.143) M1 Φ – 0.5 = 0.5 – (1 – Φ(2.143)) = 0.9839 – 0.5 = 0.484 A1 [3] Correct answer rounding to (ii) z = Φ–1 (0.67) = 0.44 B1 ± 0.44 seen a − 100 .044 = M1 Standardising, with or without sq rt, no cc, no 72 7 must be z-value e.g. could be 0.412 or 0.413 103.1 min (103) = upper limit A1 Correct upper or lower boundary allow even if obtained from z = 0.412 96.9 min = lower limit A1 [4] Correct other boundary GCE A/AS LEVEL – October/November 2009 9709 61

This question in 9709/61 Oct/Nov 2009

6 A box contains 4 pears and 7 oranges. Three fruits are taken out at random and eaten. Find the probability that (i) 2 pears and 1 orange are eaten, in any order, [3] (ii) the third fruit eaten is an orange, [3] (iii) the first fruit eaten was a pear, given that the third fruit eaten is an orange. [3] There are 121 similar boxes in a warehouse. One fruit is taken at random from each box. (iv) Using a suitable approximation, find the probability that fewer than 39 are pears. [5]

14 marks

Mark scheme: C 2 × C16 (i) = 0.255 M1 Using 2 combs mult for numerator and 1 comb for 11C 3 denom M1 Correct denom or num unsimplified A1 Correct answer 4 3 7 M1 Multiplying 3 correct probs OR × × × 3 11 10 9 M1 Mult by 3 or Σ their 3 options A1 [3] Correct answer = 0.255 (14/55) (42/165) (ii) P(3rd is orange) = P(P, P, O) + P(P, O, O) + P(O, P, O) + P(O, O, O) M1 Summing four 3-factor options with or without 4 3 7 4 7 6 replacement = × × + × × 11 10 9 11 10 9 7 4 6 7 6 5 + × × + × × A1 At least 3 correct unsimplified options 11 10 9 11 10 9  14 28 28 7  = + + +  165 165 165 33  = 7/11 (0.636 A1 Correct answer. Award B3 if the correct answer is stated with no working. OR using a tree diagram [3] P ( P ∩ O ) (iii) P(P O ) = M1 Substituting in cond prob formula with at least one P (O ) 3-factor product in num, and denom their (ii) or 7/11 P ( P , P , O ) + P ( P , O , O ) = M1 Summing exactly 2 three-factor products in num P (O ) 28 / 110 28 4 = = = 0.4 A1 [3] Correct answer 7 / 11 70 =10 4 (iv) µ = 121 × = 44 B1 44 and 28 or 5.29 seen 11 4 7 σ2 = 121 × × = 28 M1 Standardising, with or without cc, must have sq rt 11 11 on denom  385. − 44  P(X < 39) = Φ   M1 cc either 39.5 or 38.5  28  = Φ(–1.039) M1 Correct area “1 – Φ” seen = 1 – 0.8506 = 0.149 A1 [5] Correct answer

This question in 9709/61 Oct/Nov 2009

7 The weights, X grams, of bars of soap are normally distributed with mean 125 grams and standard deviation 4.2 grams. (i) Find the probability that a randomly chosen bar of soap weighs more than 128 grams. [3] (ii) Find the value of k such that P(k < X < 128) = 0.7465. [4] (iii) Five bars of soap are chosen at random. Find the probability that more than two of the bars each weigh more than 128 grams. [4]

11 marks

Mark scheme:  128 125  7 (i) P(X > 128) = P  z >  M1 Standardising, no cc, no sq rt  2.4  = P(z > 0.7143) = 1 – 0.7623 M1 Correct area of graph i.e. prob < 0.5 = 0.238 A1 [3] Correct answer, rounding to 0.238 (ii) P(X > k) = 0.7465 + 0.2377 = 0.9842 M1 Valid method to obtain P(X > k), no cc z = –2.15 A1 Answer rounding to ± 2.15 seen k − 125 –2.15 = M1 Solving equation with their z-value, k, 125 and 2.4 4.2 or 2.4 , no cc k = 116 A1 [4] Correct answer, rounding to 116 (iii) P(X > 2) = P(3, 4, 5) or 1 – P(0, 1, 2) M1 = 5C3(0.2377)3(0.7623)2 + 5C4(0.2377)4(0.7623)1 + 5C5(0.2377)5 M1 Binomial term of form 5Cx px (1 − p)5−x, x ≠ 0 = 0.07804 + 0.01216 + 0.0007588 A1 Sum of exactly 3 bin probs, any p = 0.0910 A1 [4] Correct unsimplified answer Correct answer accept 0.0909 and 0.091 from 0.0910

This question in 9709/62 Oct/Nov 2009

3 The random variable X is the length of time in minutes that Jannon takes to mend a bicycle puncture. X has a normal distribution with mean µ and variance σ2. It is given that P(X > 30.0) = 0.1480 and P(X > 20.9) = 0.6228. Find µ and σ. [5]

5 marks

Mark scheme: 3 (+/–) 1.045, (+/–) 0.313 B1, B1 1 correct z-value, the other correct z-value. 20.9 – µ = –0.313 σ M1 Valid attempt to solve 2 equations 30 – µ = 1.045 σ relating to µ, σ, 30, 20.9. No σ , σ2 σ = 6.70 A1 correct answer µ = 23.0 A1 correct answer [5]

This question in 9709/61 May/June 2010

5 In the holidays Martin spends 25% of the day playing computer games. Martin’s friend phones him once a day at a randomly chosen time. (i) Find the probability that, in one holiday period of 8 days, there are exactly 2 days on which Martin is playing computer games when his friend phones. [2] (ii) Another holiday period lasts for 12 days. State with a reason whether it is appropriate to use a normal approximation to find the probability that there are fewer than 7 days on which Martin is playing computer games when his friend phones. [1] (iii) Find the probability that there are at least 13 days of a 40-day holiday period on which Martin is playing computer games when his friend phones. [5]

8 marks

Mark scheme: 5 (i) P(X = 2)) = (0.25)2 × (0.75)6 × 8C2 M1 3 term binomial expression involving 8C something, powers summing to 8 = 0.311 A1 correct answer [2] (ii) 12 × 0.25 = 3, < 5 so not possible B1 [1] (iii) mean = 40 × 0.25 (= 10) variance = 40 × 0.25 × 0.75 ( = 7.5) B1 40 × 0.25 and 40 × 0.25 × 0.75 seen, o.e.  125. − 10  standardising, ±, with or without cc, must P(X at least 13) = P  z >  M1  5.7  have sq rt = P(z > 0.913) M1 continuity correction 12.5 or 13.5 = 1 – Φ(0.913) M1 correct area, i.e. < 0.5 legit = 1 – 0.8194 = 0.181 A1 correct answer [5] 10 10 10 10 10 10

This question in 9709/61 May/June 2010

7 The heights that children of a particular age can jump have a normal distribution. On average, 8 children out of 10 can jump a height of more than 127 cm, and 1 child out of 3 can jump a height of more than 135 cm. (i) Find the mean and standard deviation of the heights the children can jump. [5] (ii) Find the probability that a randomly chosen child will not be able to jump a height of 145 cm. [3] (iii) Find the probability that, of 8 randomly chosen children, at least 2 will be able to jump a height of more than 135 cm. [3]

11 marks

Mark scheme: 135 µ 7 (i) 0.431 = B1 One ±z-value correct, accept 0.430 σ B1 A second ±z-value correct 127 − µ –0.842 = M1 Solving two equations relating µ, σ, 135, σ 127 and their z-values (must be z-values) σ = 6.29 A1 Correct answer accept 6.28 µ = 132 A1 Correct answer [5]  145 − 1323.  (ii) P(X < 145) = P  z <  M1 Standardising no sq rt no cc  .6 284  =P(z < 2.023) M1 Correct use of normal tables = 0.978 A1 Answer rounding to 0.978 or 0.979 [3] (iii) p = 1/3 P(at least 2) = 1 – P(0, 1) M1 Binomial expression with powers summing to 8 and 8Csomething. (any p) = 1 – [ ( 2 / 3) 8 + 8 C1 × 1( / 3)1 ( 2 / 3) 7 ] A1 Correct unsimplified expression = 0.805 A1 Answer rounding to 0.805 [3]

This question in 9709/63 May/June 2010

3 The times taken by students to get up in the morning can be modelled by a normal distribution with mean 26.4 minutes and standard deviation 3.7 minutes. (i) For a random sample of 350 students, find the number who would be expected to take longer than 20 minutes to get up in the morning. [3] (ii) ‘Very slow’ students are students whose time to get up is more than 1.645 standard deviations above the mean. Find the probability that fewer than 3 students from a random sample of 8 students are ‘very slow’. [4]

7 marks

Mark scheme: 3 (i) P(X > 20) = P(z > –6.4/3.7) M1 Standardising no cc no sq rt = P(z > –1.730) = 0.9582 A1 Prob rounding to 0.958 Number of students = 335 or 336 A1ft Correct answer ft their prob, must be integer [3] (ii) P(very slow) = 0.05 B1 0.05 or 0.95 seen P(0, 1, 2) = M1 Binomial term with 8Cr pr (1 – p) 8 – r seen (0.95)8 + 8C1(0.05)1(0.95)7 + 8C2(0.05)2(0.95)6 any p M1 Correct expression for P(0, 1, 2), p close = 0.6634 + 0.2793 + 0.0515 to 0.05 = 0.994 A1 Answer rounding to 0.994 [4]

This question in 9709/61 Oct/Nov 2010

1 Name the distribution and suggest suitable numerical parameters that you could use to model the weights in kilograms of female 18-year-old students. [2]

2 marks

Mark scheme: 1 Normal B1 mean 60 kg, variance 90 kg2 B1 Any sensible values (mean 40–80 kg, variance 16–225 kg2), could give s.d. 4–15 kg [2]

This question in 9709/63 Oct/Nov 2010

7 The times spent by people visiting a certain dentist are independent and normally distributed with a mean of 8.2 minutes. 79% of people who visit this dentist have visits lasting less than 10 minutes. (i) Find the standard deviation of the times spent by people visiting this dentist. [3] (ii) Find the probability that the time spent visiting this dentist by a randomly chosen person deviates from the mean by more than 1 minute. [3] (iii) Find the probability that, of 6 randomly chosen people, more than 2 have visits lasting longer than 10 minutes. [3] (iv) Find the probability that, of 35 randomly chosen people, fewer than 16 have visits lasting less than 8.2 minutes. [5]

14 marks

Mark scheme: 7 (i) z = 0.807 B1 0.807 seen 10 − 2.8 0.807 = M1 standardising, must have σ, no sq rt, no σ cc and a z-value s = 2.23 A1 correct answer [3] 1 (ii) P(> 1 min from mean) = P(mod z > ) M1 standardising, their sd, no cc and adding .223 two areas = P( z > .04484 ) M1 using 1 – Φ(z) = (1 – 0.6729) × 2 = 0.654 A1 correct answer [3] (iii) P(> 2 longer) = 1 – P(0, 1, 2 longer) M1 binomial term 6Cxpx(1 − p)6 − x = 1 – {(0.79)6 + 6C1(0.21)(0.79)5 + A1 correct unsimplified answer 6C2(0.21)2(0.79)4} = 0.112 A1 correct answer [3] (iv) µ = 35 × 0.5 = 17.5 B1 17.5 and 8.75 or .8 75 seen σ2 = 35 × 0.5 × 0.5 = 8.75  15 5. − 175.  P(X < 16) = Φ   M1 standardising, with or without cc, must  .875  have sd in denom M1 continuity correction 15.5 or 16.5 only, seen = 1 – Φ(0.676) M1 using 1 – Φ(z) = 1 – 0.7505 = 0.2495 (0.249 or 0.250) A1 correct answer [5] OR 35C00.500.535 + 35C10.510.534 + 35C20.520.533 +... M1 binomial term 35Cx0.5x0.535 − x = 8582372584/235 = 0.250 A1 at least 2 correct terms (x Þ 0) seen M1 summing 16 or 17 terms A1 correct expression A1 correct answer

This question in 9709/63 Oct/Nov 2010

5 The weights of letters posted by a certain business are normally distributed with mean 20 g. It is found that the weights of 94% of the letters are within 12 g of the mean. (i) Find the standard deviation of the weights of the letters. [3] (ii) Find the probability that a randomly chosen letter weighs more than 13 g. [3] (iii) Find the probability that at least 2 of a random sample of 7 letters have weights which are more than 12 g above the mean. [3]

9 marks

Mark scheme: 5 (i) z = 1.882 or 1.881 B1 ±1.882 or ±1.881 seen 1.882 = (32 – 20) / σ M1 Equation using their z (must be a z-value) 32, 20 and s σ = 6.38 A1 [3] Correct answer  13 − 20  M1 Standardising (ii) P(x > 13) = P  z >   .6 376  = P(z > –1.0978) M1 Correct area > 0.5 = 0.864 A1 [3] Correct answer (iii) P(at least 2) = 1 – P(0, 1) M1 Using 0.03 and 0.97 or 0.06 and 0.94 in a binomial expression powers summing to 7 = 1 – (0.97)7 – (0.03)(0.97)67C1 M1 Correct unsimplified binomial expansion = 0.0171 A1 [3] Correct answer 12! M1 Dividing by 2! 3! 2!

This question in 9709/61 Oct/Nov 2011

7 (i) In a certain country, the daily minimum temperature, in ◦C, in winter has the distribution N(8, 24). Find the probability that a randomly chosen winter day in this country has a minimum temperature between 7 ◦C and 12 ◦C. [3] The daily minimum temperature, in ◦C, in another country in winter has a normal distribution with mean µ and standard deviation 2µ. (ii) Find the proportion of winter days on which the minimum temperature is below zero. [2] (iii) 70 winter days are chosen at random. Find how many of these would be expected to have a minimum temperature which is more than three times the mean. [3] (iv) The probability of the minimum temperature being above 6 ◦C on any winter day is 0.0735. Find the value of µ. [3]

11 marks

Mark scheme: 12 8 M1 Standardising any one, sq rt, no cc 7 (i) z1 = = 0.816 Ф1(0.816) = 0.7926 24 7 − 8 z2 = = –0.204 Ф2(–0.204) = 1 – 0.5808 M1 Correct area Ф1 + Ф2 – 1 24 Prob = 0.7926 – (1 – 0.5808) = 0.373 A1 [3] Correct answer 0 −µ (ii) z = = –0.5 M1 Standardising, no cc no sq rt, one 2µ variable P(z < –0.5) = 1 – 0.6915 = 0.309 or 30.9% A1 [2] Correct answer oe 3µ −µ M1 Standardising and eliminating µ (iii) z = = 1 2µ M1 Subt from 1 and multiplying by 70 P(z > 1) = 1 – 0.8413 = 0.1587 A1 [3] Correct answer accept 11 or 12 70 × 0.1587 = 11.1 (iv) z = 1.45 B1 ± 1.45 seen 6 −µ 1.45 = M1 Solving forµ with 6, 2µ , µ and their z 2µ µ = 1.54 A1 [3] Correct answer

This question in 9709/62 Oct/Nov 2011

1 The random variable X is normally distributed and is such that the mean µ is three times the standard deviation σ. It is given that P(X < 25) = 0.648. (i) Find the values of µ and σ. [4] (ii) Find the probability that, from 6 random values of X, exactly 4 are greater than 25. [2]

6 marks

Mark scheme: 1 (i) z = 0.38 B1 ± .0 38 (0) seen or implied 25 − µ M1 Standardising attempt resulting in z = ± = .038 µ / 3 some µ/σ/both, no continuity correction M1 Substituting to eliminate µ or σ and attempt to solve linear equation µ = 22.2, σ = 7.40 A1 [4] Both correct (ii) P(4) = 6C4(0.352)4(0.648)2 M1 6Cr × (p)r × (1 − p)6−r, r = 2 or 4 = 0.0967 A1 [2] Correct answer 12 12 16 5

This question in 9709/63 Oct/Nov 2011

3 Lengths of rolls of parcel tape have a normal distribution with mean 75 m, and 15% of the rolls have lengths less than 73 m. (i) Find the standard deviation of the lengths. [3] Alison buys 8 rolls of parcel tape. (ii) Find the probability that fewer than 3 of these rolls have lengths more than 77 m. [3]

6 marks

Mark scheme: 73 − 75 B1 ± correct z value accept ± .10373 (i) z = –1.036 = σ M1 Equation with 73, 75, σ and a z value σ = 1.93 A1 [3] Rounding to correct answer (ii) P(> 77) = 0.15 M1 Prob rounding to 0.15 and 0.85 P(< 3) = P(0, 1, 2) M1 8Cxpx(1–p)8–x seen any p, 0<p<1 = (0.85)8 + 8C1(0.15)(0.85)7 + 8C2(0.15)2(0.85)6 = 0.895 A1 [3] Correct answer

This question in 9709/61 Oct/Nov 2012

2 The random variable X is the daily profit, in thousands of dollars, made by a company. X is normally distributed with mean 6.4 and standard deviation 5.2. (i) Find the probability that, on a randomly chosen day, the company makes a profit between $10 000 and $12 000. [3] (ii) Find the probability that the company makes a loss on exactly 1 of the next 4 consecutive days. [4]

7 marks

Mark scheme: − 4.62 (i) z1 = 12 = .1 077 M1 Standardising, can be all in thousands, 2.5 no mix, no cc no sq rt no sq 10 − 4.6 z 2 = = .0 692 M1 Φ 2 – Φ 1, Φ 2 must be > Φ 1 2.5 Φ(z1) – Φ (z2) = 0.8593 – 0.7556 A1 [3] Correct answer = 0.104 (ii) P(loss) = P( z < 0 − 4.6 ) = P(z < –1.231) M1 2.5 Standardising using x = 0, accept = 1 – 0.8909 5.0 − 4.6 2.5 = 0.109 A1 Correct prob P(1) = (0.1091)1(0.8909)3 × 4C1 M1 Binomial term 4Cxpx(1–p)4-x any p x ≠ 0 = 0.309 or 0.308 A1 [4] Correct answer

This question in 9709/62 Oct/Nov 2012

4 The mean of a certain normally distributed variable is four times the standard deviation. The probability that a randomly chosen value is greater than 5 is 0.15. (i) Find the mean and standard deviation. [4] (ii) 200 values of the variable are chosen at random. Find the probability that at least 160 of these values are less than 5. [5]

9 marks

Mark scheme: 4 (i) z = 1.036 or 1.037 B1 ± .1 036 or ± .1 037 seen 5 − 4 s 5 − 4σ 5 − µ .1 036 = B1 seen or oe s σ µ / 4 s = 0.993 M1 One variable and sensible solving attempt z-value not nec Both answers correct ã = 3.97 A1 [4] (ii) p = 0.85 ã = 200 × 0.85 = 170, B1 200 × 0.85 (170) and 200 × 0.85 × 0.15 (25.5) seen var = 200 × 0.85 × 0.15 = 25.5 M1 Standardising, sq rt and must have  159 5. − 170  used 200 P(at least 160) = P z >    25 5.  M1 continuity correction 159.5 or 160.5 = P(z > -2.079) M1 correct area (> 0.5) must have used 200 = 0.981 A1 correct value [5]

This question in 9709/62 Oct/Nov 2012

4 (a) The random variable Y is normally distributed with positive mean - and standard deviation 12-. Find the probability that a randomly chosen value of Y is negative. [3] (b) The weights of bags of rice are normally distributed with mean 2.04 kg and standard deviation 3 kg. In a random sample of 8000 such bags, 253 weighed over 2.1 kg. Find the value of 3. [4]

7 marks

Mark scheme:  0 − µ  4 (a) P(y < 0) = P  z <  M1 Standardising containing 0 (can be  µ / 2  implied) and µ only = P ( z < – 2) A1 z < –2 seen = 1 – 0.9772 = 0.0228 A1 [3] Correct answer (b) P(x > 2.1 ) = 253/8000 = 0.031625 M1 1 – their 253/8000 used to obtain a P ( x < 2.1 ) = 0.968375 = Φ (z) z-value 1.2 − .204 z = 1.857 or 1.858 or 1.859 = A1 Rounded to 1.86 seen σ σ = 0.0323 M1 Solving for σ using their z val must be a z val A1 [4] Correct answer

This question in 9709/61 May/June 2013

1 The random variable Y is normally distributed with mean equal to five times the standard deviation. It is given that P Y > 20 = 0.0732. Find the mean. [3]

3 marks

Mark scheme: 1 z = 1.452 B1 Rounding to ± 1.45 20 − µ 20 − µ 20 − 5σ 1.452 = or seen oe B1 µ / 5 µ / 5 σ µ = 15.5 B1 [3] rounding to correct answer

This question in 9709/62 May/June 2013

3 Cans of lemon juice are supposed to contain 440 ml of juice. It is found that the actual volume of juice in a can is normally distributed with mean 445 ml and standard deviation 3.6 ml. (i) Find the probability that a randomly chosen can contains less than 440 ml of juice. [3] It is found that 94% of the cans contain between 445 −c ml and 445 + c ml of juice. (ii) Find the value of c. [3]

6 marks

Mark scheme: 3 (i) P( x < 440)  440 − 445  M1 Standardising no cc no sq or sq rt = P  z <  = 1 - Φ (1.389)  6.3  M1 Correct area (1 – Φ) oe (indep) = 1 – 0.9176 Ans = 0.0824 A1 Rounding to correct answer accept [3] 0.0825 (ii) z = 1.881 M1 ±1.88 or 1.881 or 1.882 or 1.555 seen± c = 1.881 M1 Equation with ± c/3.6 or 2c/3.6 only = 6.3 z or prob (can be implied) c = 6.77 A1 [3] Correct answer accept 6.78 5

This question in 9709/62 May/June 2013

3 Buildings in a certain city centre are classified by height as tall, medium or short. The heights can be modelled by a normal distribution with mean 50 metres and standard deviation 16 metres. Buildings with a height of more than 70 metres are classified as tall. (i) Find the probability that a building chosen at random is classified as tall. [2] (ii) The rest of the buildings are classified as medium and short in such a way that there are twice as many medium buildings as there are short ones. Find the height below which buildings are classified as short. [5]

7 marks

Mark scheme: 3 (i) P(tall) = P > z 70 50  = P(z > 1.25) M1 +ve/-ve Standardising no cc no sq rt no sq  16  = 1 – 0.8944 = 0.106 A1 [2] Correct answer (ii) P(short) = (1 – 0.1056)/3 M1 Subt their (i) from 1 or their (i) and multiplying 1 2 by or 3 3 (1 - (i)) = 0.2981 A1 ft Rounding to 0.298, only ft for 3 z = – 0.53 A1 ± z-value rounding to 0.53, condone ±0.24 x − 50 – 0.53 = M1 Standardising with their z value 16 (not a probability), no cc sq rt etc. x = 41.5 A1 [5] Correct answer GCE AS/A LEVEL – May/June 2013 9709 63 4 (i) (0.8)n < 0.001 M1 Eqn or inequ involving 0.8n or 0.2n and 0.001 or 0.999 n > 30.9 M1 Trial and error or logs (can be implied) n = 31 A1 [3] Correct answer MR 0.01, max available M1M1A0 (ii) µ = 120 × 0.2 = 24 B1 24 and 19.2 or 19 2. seen σ2 = 120 × 0.2 × 0.8 = 19.2 M1 Standardising with or without cc, must have sq rt in denom  32 5 −24 

This question in 9709/63 May/June 2013

5 Lengths of a certain type of carrot have a normal distribution with mean 14.2 cm and standard deviation 3.6 cm. (i) 8% of carrots are shorter than c cm. Find the value of c. [3] (ii) Rebekah picks 7 carrots at random. Find the probability that at least 2 of them have lengths between 15 and 16 cm. [6]

9 marks

Mark scheme: 5 (i) z = –1.406 B1 Rounding to ± .1 41 seen c − 142. = −.1406 M1 Standardising allow sq rt no cc 6.3 c = 9.14 A1 3 Correct answer  15 − 142.   16 − 142.  (ii) P   < z <   M1 2 attempts at standardising no cc no sq rt  6.3   6.3  = Φ(0.5) – Φ(0.222) M1 Subt two Φs (indep mark) = 0.6915 – 0.5879 = 0.1036 A1 Needn’t be entirely accurate, rounding to 0.10 P(at least 2) = 1 – P(0, 1) M1 Binomial term with 7Crpr(1–p)7–r seen r ≠ 0 = 1 – (0.8964)7 – (0.8964)6(0.1036)7C1 any p < 1 = 1 – 0.8413 M1 1 – P(0), 1 – P(1), 1 – P(0, 1) seen their p = 0.159 A1 6 Correct answer accept 3sf rounding to 0.16

This question in 9709/61 Oct/Nov 2013

3 The amount of fibre in a packet of a certain brand of cereal is normally distributed with mean 160 grams. 19% of packets of cereal contain more than 190 grams of fibre. (i) Find the standard deviation of the amount of fibre in a packet. [3] (ii) Kate buys 12 packets of cereal. Find the probability that at least 1 of the packets contains more than 190 grams of fibre. [2]

5 marks

Mark scheme: 3 (i) z = 0.878 B1 ± 0.878, 0.88, rounding to 0.88 seen 190 − 160 (190 – 160)/σ = something = .0878 M1 σ σ = 34.2 A1 [3] Correct answer (ii) P(at least 1) = 1 – P(0) M1 Using 1 – P(0), 1 – P(0, 1), P(1,2 … 12) or P(2, … 12) with p = 0.19 or 0.81, terms must be evaluated to get the M1 = 1 – (0.81)12 = 0.920 A1 [2] Correct answer accept 0.92

This question in 9709/62 Oct/Nov 2013

2 A factory produces flower pots. The base diameters have a normal distribution with mean 14 cm and standard deviation 0.52 cm. Find the probability that the base diameters of exactly 8 out of 10 randomly chosen flower pots are between 13.6 cm and 14.8 cm. [5]

5 marks

Mark scheme: 2 M1 Standardising 1 expression, no cc, no sq rt, no  13 6 − 14 14 8 − 14 

This question in 9709/63 Oct/Nov 2013

1 The petrol consumption of a certain type of car has a normal distribution with mean 24 kilometres per litre and standard deviation 4.7 kilometres per litre. Find the probability that the petrol consumption of a randomly chosen car of this type is between 21.6 kilometres per litre and 28.7 kilometres per litre. [4]

4 marks

Mark scheme: 1 P (21.6 < x < 28.7)    216. − 24   287. − 24   M1 Standardising; no cc, no sq rt = P   < z <      7.4   7.4   A1 One rounding to Φ (0.841 or 0.695) = P (–0.5106 < z < 1) = Φ (1) – Φ (–0.5106) M1 Φ1 + Φ2 – 1 = 0.8413 – (1 – 0.6953) = 0.537 (0.5366) A1 4 Correct answer 12 − µ

This question in 9709/61 May/June 2014

2 Lengths of a certain type of white radish are normally distributed with mean - cm and standard deviation 3 cm. 4% of these radishes are longer than 12 cm and 32% are longer than 9 cm. Find - and 3. [5]

5 marks

Mark scheme: 12 µ 2 1.751 = B1 Rounding to ±1.75 seen σ 9 − µ 0.468 = B1 ±0.468 seen σ M1 An eqn with a z-value, µ and σ no √σ, no σ2 σ = 2.34 M1 Sensible attempt to eliminate µ or σ by substitution or subtraction, need a value µ = 7.91 A1 5 correct answers

This question in 9709/61 May/June 2014

7 The time Rafa spends on his homework each day in term-time has a normal distribution with mean 1.9 hours and standard deviation 3 hours. On 80% of these days he spends more than 1.35 hours on his homework. (i) Find the value of 3. [3] (ii) Find the probability that, on a randomly chosen day in term-time, Rafa spends less than 2 hours on his homework. [2] (iii) A random sample of 200 days in term-time is taken. Use an approximation to find the probability that the number of days on which Rafa spends more than 1.35 hours on his homework is between 163 and 173 inclusive. [6]

11 marks

Mark scheme: 7 (i) z = –0.842 B1 ± rounding to 0.84 seen  .135 − 9.1  .135 − 9.1 P (x > 1.35) = P  z >  M1 ± = a prob or a z-value NOT 0.8 or 0.2  σ  σ allow a 1–... –0.842 = –0.55/σ σ = 0.653 A1 3 Correct answer from correct working  2 − 9.1  (ii) P(x < 2) = P  z <  M1 ± standardising no continuity correction their σ  .0 6532  = P ( z < 0.1531) = 0.561 A1 2 Correct answer (iii) X~N(160, 32) B1 Unsimplified 160 and 32 seen P(162.5 < x < 173.5) =  1625. − 160 1735. − 160  P  < z <  M1 Standardising need sq rt  32 32  P(0.442 < z < 2.386) M1 Any of 162.5, 163.5, 172.5, 173.5 seen = Φ(2.386) – Φ(0.442) M1 Φ2 – Φ1 oe = 0.9915 – 0.6707 A1 One correct Φ to 3sf = 0.321 A1 6 Correct answer accept 0.320

This question in 9709/62 May/June 2014

5 When Moses makes a phone call, the amount of time that the call takes has a normal distribution with mean 6.5 minutes and standard deviation 1.76 minutes. (i) 90% of Moses’s phone calls take longer than t minutes. Find the value of t. [3] (ii) Find the probability that, in a random sample of 9 phone calls made by Moses, more than 7 take a time which is within 1 standard deviation of the mean. [5]

8 marks

Mark scheme: 5 (i) z = –1.282 B1 Rounding to ± 1.28 seen t − 5.6 –1.282 = M1 Standardising, no cc, no sq or sq rt, z≠ ±0.9,±0.1 .176 t = 4.24 A1 3 Correct answer, accept 4.25 (ii) P(z < 1) = 0.8413 M1 z = 1 used to find a probability P(within 1sd of mean) = 2Φ – 1 B1 correct prob, accept answer rounding to 0.66, = 0.6826 0.67, 0.68, not from wrong working. If quoted, then implies first M1. P(8, 9) – r = 9C8(0.6826)8(0.3174)+ (0.6826)9 M1 Binomial term pr(1 – p)9 9Cr , 9Cr must be seen M1 Binomial expression for P(8)+P(9), any p = 0.167 A1 5 Correct ans

This question in 9709/63 May/June 2014

6 A farmer finds that the weights of sheep on his farm have a normal distribution with mean 66.4 kg and standard deviation 5.6 kg. (i) 250 sheep are chosen at random. Estimate the number of sheep which have a weight of between 70 kg and 72.5 kg. [5] (ii) The proportion of sheep weighing less than 59.2 kg is equal to the proportion weighing more than y kg. Find the value of y. [2] Another farmer finds that the weights of sheep on his farm have a normal distribution with mean kg and standard deviation 4.92 kg. 25% of these sheep weigh more than 67.5 kg. (iii) Find the value of . [3]

10 marks

Mark scheme: 70 664. 6 (i) z1 = = .06429 M1 Standardising one variable, no cc, no sq rt 6.5 725. − 664. z 2 = = .1089 M1 Correct area Φ2 – Φ1 6.5 Φ (.1 089 ) − Φ (.0 643 ) = .0 8620 − .0 7399 A1 Correct answer rounding to 0.12 = 0.1221 0.1221 × 250 = 30.5 M1 Mult by 250 30 or 31 sheep A1ft 5 Correct answer ft their 0.1221 (ii) 66.4 – 59.2 = 7.2 M1 Subt from 66.4 66.4 + 7.2 = 73.6 A1 2 Correct answer (iii) z = 0.674 B1 ± 0.674 or 0.675 seen 675. −µ = .0674 M1 Standardising with a z-value no cc no sq rt .492 µ = 64.2 A1 3 Correct answer

This question in 9709/61 Oct/Nov 2014

1 Packets of tea are labelled as containing 250 g. The actual weight of tea in a packet has a normal distribution with mean 260 g and standard deviation 3 g. Any packet with a weight less than 250 g is classed as ‘underweight’. Given that 1% of packets of tea are underweight, find the value of 3. [3]

3 marks

Mark scheme: 1 z = –2.326 B1 ± 2.325 to ± 2.33 seen 250 − 260 = −.2326 M1 Standardising and = or < their z, no cc, sq, sq rt σ σ = .4 30 A1 3 Correct ans

This question in 9709/63 Oct/Nov 2014

1 The lengths, in metres, of cars in a city are normally distributed with mean - and standard deviation 0.714. The probability that a randomly chosen car has a length more than 3.2 metres and less than - metres is 0.475. Find -. [4]

4 marks

Mark scheme: 1 P(x < 3.273) = 0.5 – 0.475 = 0.025 M1 Attempt to find z-value using tables in reverse z = –1.96 A1 ±1.96 seen 2.3 −µ = − .1 96 M1 Solving their standardised equation .0 714 z-value not nec µ = 4.60s A1 [4] Correct ans accept 4.6

This question in 9709/61 May/June 2015

6 (i) In a certain country, 68% of households have a printer. Find the probability that, in a random sample of 8 households, 5, 6 or 7 households have a printer. [4] (ii) Use an approximation to find the probability that, in a random sample of 500 households, more than 337 households have a printer. [5] (iii) Justify your use of the approximation in part (ii). [1]

10 marks

Mark scheme: 6 (i) P(5, 6, 7) = 8C5(0.68)5(0.32)3 + M1 Binomial term 8Cx px(1–p)8-x seen 8C6(0.68)6(0.32)2 + 8C7(0.68)7(0.32) 0 < p < 1 M1 Summing 3 binomial terms A1 Correct unsimplified answer = 0.722 A1 [4] Correct answer (ii) np = 340, npq = 108.8 B1 Correct (unsimplified) mean and var  337 5. − 340  P(x > 337) = P  z >  M1 standardising with sq rt must have  108 8.  used 500 M1 cc either 337.5 or 336.5 = P(z > – 0.2396) M1 correct area (> 0.5) must have used = 0.595 500 A1 [5] correct answer (iii) np (340) > 5 and nq(160) > 5 B1 [1] must have both or at least the smaller, need numerical justification !9

This question in 9709/61 May/June 2015

1 The weights, in grams, of onions in a supermarket have a normal distribution with mean - and standard deviation 22. The probability that a randomly chosen onion weighs more than 195 grams is 0.128. Find the value of -. [3]

3 marks

Mark scheme: 1 z = 1.136 B1 ± 1.136 seen, not ±1.14, 195 − µ 1.136 = M1 Standardising, no cc no sq rt, 22 equated to their z not 0.128 or 0.872 µ = 170 A1 [3] Correct answer, nfww

This question in 9709/63 May/June 2015

4 The time taken for cucumber seeds to germinate under certain conditions has a normal distribution with mean 125 hours and standard deviation 3 hours. (i) It is found that 13% of seeds take longer than 136 hours to germinate. Find the value of 3. [3] (ii) 170 seeds are sown. Find the expected number of seeds which take between 131 and 141 hours to germinate. [4]

7 marks

Mark scheme: 4 (i) z = 1.127 B1 ± 1.127 seen accept rounding to ±1.13 136 − 125 .1127 = M1 Standardising no cc no sq rt, with attempt σ at z σ = 9.76 A1 3 (not ±0.8078, ±0.5517, ±0.13, ±0.87) Correct ans 131− 125 141− 125 (ii) P(131<x<141)= P ( <z< ) M1 Standardising once with their sd, no √,2, .976 .9 76 allow cc = Φ( 1.639) – Φ(0.6147) M1 Correct area Φ2 – Φ1 = 0.9493 – 0.7307 = 0.2186 M1 Mult by 170, P<1 Number = 0.2186 × 170 = 37 or 38 or awrt 37.2 A1 4 Correct answer, nfww

This question in 9709/63 Oct/Nov 2015

7 A factory makes water pistols, 8% of which do not work properly. (i) A random sample of 19 water pistols is taken. Find the probability that at most 2 do not work properly. [3] (ii) In a random sample of n water pistols, the probability that at least one does not work properly is greater than 0.9. Find the smallest possible value of n. [3] (iii) A random sample of 1800 water pistols is taken. Use an approximation to find the probability that there are at least 152 that do not work properly. [5] (iv) Justify the use of your approximation in part (iii). [1]

12 marks

Mark scheme: 7 (i) P(0, 1, 2) = M1 Binomial term 19Cxpx(1 – p)19-x seen 0<p<1 (0.92)19+19C1(0.08)(0.92)18+ 19C2(0.08)2(0.92)17 M1 Correct unsimplified expression = 0.809 A1 3 Correct answer (no working SC B2) (ii) P(at least 1) = 1 – P(0) = 1 – P(0.92)n > 0.90 M1 Eqn with their 0.92n, 0.9 or 0.1, 1 not nec 0.1 > (0.92)n M1 Solving attempt by logs or trial and error, n > 27.6 power eqn with one unknown power Ans 28 A1 3 Correct answer, not approx., ≈, ⩾, >, ⩽, < (iii) np = 1800 × 0.08 = 144 B1 correct unsimplified np and npq seen npq = 132.48 accept 132.5, 132, 11.5, awrt 11.51   1515. − 144   M1 standardising, with √ P( at least 152) = P  z >      132.48   M1 cont correction 151.5 or 152.5 seen = P(z > 0.6516) M1 correct area 1 – Φ (probability) = 1 – 0.7429 = 0.257 A1 5 correct answer (iv) Use because 1800 ×0.08 (and 1800 × 0.92 are B1 1 1800 ×0.08 > 5 is sufficient both) > 5 np>5 is sufficient if clearly evaluated in (iii) If npq>5 stated then award B0

This question in 9709/63 Oct/Nov 2015

1 The height of maize plants in Mpapwa is normally distributed with mean 1.62 m and standard deviation 3 m. The probability that a randomly chosen plant has a height greater than 1.8 m is 0.15. Find the value of 3. [3]

3 marks

Mark scheme: Question Answer Marks Guidance 1 z = 1.037 B1 Rounding to 1.04 1.8 − 1.62 1.037 = M1 Standardising attempt allow cc no sq rt σ must have a z-value i.e. not 0.8023 or 0.5596. σ = 0.18/1.037 = 0.174 A1 [3]

This question in 9709/61 May/June 2016

5 The heights of school desks have a normal distribution with mean 69 cm and standard deviation 3 cm. It is known that 15.5% of these desks have a height greater than 70 cm. (i) Find the value of 3. [3] When Jodu sits at a desk, his knees are at a height of 58 cm above the floor. A desk is comfortable for Jodu if his knees are at least 9 cm below the top of the desk. Jodu’s school has 300 desks. (ii) Calculate an estimate of the number of these desks that are comfortable for Jodu. [5]

8 marks

Mark scheme: 5 (i) z = 1.015 B1 Accept z between ±1.01 and 1.02 70 − 69 1.015 = M1 Standardising σ σ = 0.985 (200/203) A1 [3] (ii) 58 + 9 = 67 M1 58 + 9 seen or implied (or 69-58 or 69-9)  67 − 69  P ( > 67) = P  z >  M1 Standardising ± z no cc allow their sd  0.9852  (must be +ve)  9 − 11  Alt. 1 69-58 =11, P( >9)=P  z >   0.9852   58 − 60  Alt.2 69-9 =60, P( >58) =P  z >   0.9852  = P(z > – 2.03) M1 Correct prob area = 0.9788 Multiply their prob (from use of tables) by M1 300 300 × 0.9788 = 293.6 so 293 A1 [5] – accept 293 or 294 from fully correct working

This question in 9709/63 May/June 2016

4 Packets of rice are filled by a machine and have weights which are normally distributed with mean 1.04 kg and standard deviation 0.017 kg. (i) Find the probability that a randomly chosen packet weighs less than 1 kg. [3] (ii) How many packets of rice, on average, would the machine fill from 1000 kg of rice? [1] The factory manager wants to produce more packets of rice. He changes the settings on the machine so that the standard deviation is the same but the mean is reduced to - kg. With this mean the probability that a packet weighs less than 1 kg is 0.0388. (iii) Find the value of -. [3] (iv) How many packets of rice, on average, would the machine now fill from 1000 kg of rice? [1]

8 marks

Mark scheme:  1 − 1.04  4 (i) P(< 1) = P  z <  = P(z < –2.353) M1 Standardising no cc, no √ or sq  0.017  = 1 – 0.9907 M1 1 – Φ (final process) = 0.0093 A1 [3] (ii) expected number 1000 ÷ 1.04 = 961 or 962 B1 [1] Or anything in between (iii) z = –1.765 B1 ± 1.76 to 1.77 1 − µ −1.765 = M1 Standardising must have a z- 0.017 value, allow √ or sq [3] =1.03 A1 (iv) expected number = 1000 ÷ 1.03 = 971 or 970 B1 [1] Or anything in between, ft their (iii)

This question in 9709/61 Oct/Nov 2016

3 On any day at noon, the probabilities that Kersley is asleep or studying are 0.2 and 0.6 respectively. (i) Find the probability that, in any 7-day period, Kersley is either asleep or studying at noon on at least 6 days. [3] (ii) Use an approximation to find the probability that, in any period of 100 days, Kersley is asleep at noon on at most 30 days. [5]

8 marks

Mark scheme: 3 (i) Bin (7, 0.8) M1 7Cn pn(1–p)7–n seen P(6, 7) = 7C6 (0.8)6(0.2)1+ (0.8)7 M1 Correct unsimplified expression for P(6,7) = 0.577 A1 [3] (ii) mean = 100×0.2 = 20 B1 Correct unsimplified mean and var Var = 100×0.2×0.8 = 16  30.5 − 20  M1 Standardising must have sq rt, their µ, variance P(at most 30) = P z <   M1 cc either 29.5 or 30.5  16  M1 Correct area Φ , from final process = P(z < 2.625) = 0.996 A1 [5]

This question in 9709/62 Oct/Nov 2016

4 The time taken to cook an egg by people living in a certain town has a normal distribution with mean 4.2 minutes and standard deviation 0.6 minutes. (i) Find the probability that a person chosen at random takes between 3.5 and 4.5 minutes to cook an egg. [3] 12% of people take more than t minutes to cook an egg. (ii) Find the value of t. [3] (iii) A random sample of n people is taken. Find the smallest possible value of n if the probability that none of these people takes more than t minutes to cook an egg is less than 0.003. [3]

9 marks

Mark scheme: 4 (i)  4.5 − 4.2  P(< 4.5) = P  z <  = P(z < 0.5) M1 Standardising once no cc no sq no sq rt  0.6  = 0.6915  3.5 − 4.2  P(< 3.5) = P  z <  = P(z< -1.167)  0.6  M1 Φ1 – (1 – Φ2) [P1 – P2, 1>P1>0.5, 0.5>P2>0] oe = 1 – 0.8784 = 0.1216 A1 [3] 0.6915 – 0.1216 = 0.570 (ii) z = 1.175 B1 ±1.17 to 1.18 seen t − 4.2 1.175 = M1 Standardising no cc, allow sq, sq rt with z – value 0.6 (not ±0.8106, 0.5478, 0.4522, 0.1894, 0.175 etc.) t = 4.91 A1 [3] Correct answer from z = 1.175 seen (4sf) (iii) (0.88)n < 0.003 M1 Inequality or eqn in 0.88, power correctly placed using n or (n±1), 0.003 or (1 – 0.003) oe n > lg (0.003)/lg (0.88) M1 Attempt to solve by logs or trial and error n > 45.4 (may be implied by answer) A1 Correct integer answer n = 46 [3]

This question in 9709/62 Oct/Nov 2016

6 The weights of bananas in a fruit shop have a normal distribution with mean 150 grams and standard deviation 50 grams. Three sizes of banana are sold. Small: under 95 grams Medium: between 95 grams and 205 grams Large: over 205 grams (i) Find the proportion of bananas that are small. [3] (ii) Find the weight exceeded by 10% of bananas. [3] The prices of bananas are 10 cents for a small banana, 20 cents for a medium banana and 25 cents for a large banana. (iii) (a) Show that the probability that a randomly chosen banana costs 20 cents is 0.7286. [1] (b) Calculate the expected total cost of 100 randomly chosen bananas. [3]

10 marks

Mark scheme: 95 150 6 (i) P(small) = P < z  M1 ± standardising using 95, no cc, no sq, no sq rt  50  = P(z < –1.1) = 1 – 0.8643 M1 1 – Φ ( in final answer) = 0.136 A1 [3] (ii) z = 1.282 B1 ± rounding to 1.28 x − 150 1.282 = M1 Standardised eqn in their z allow cc 50 x = 214 g A1 [3] (iii) P(small) = 0.1357, P(large) = 0.1357 symmetry P(medium) = 1 – 0.1357×2 = 0.7286 AG B1 [1] Correct answer legit obtained (b) Expected cost per banana = 0.1357×10 + *M1 Attempt at multiplying each ‘prob’ by a price 0.1357×25 + 0.7286×20 = 19.3215 cents and summing Total cost of 100 bananas DM1 Mult by 100 = 1930 (cents) ($19.30) A1 [3] 2 2

This question in 9709/63 Oct/Nov 2016

7 Each day Annabel eats rice, potato or pasta. Independently of each other, the probability that she eats rice is 0.75, the probability that she eats potato is 0.15 and the probability that she eats pasta is 0.1. (i) Find the probability that, in any week of 7 days, Annabel eats pasta on exactly 2 days. [2] (ii) Find the probability that, in a period of 5 days, Annabel eats rice on 2 days, potato on 1 day and pasta on 2 days. [3] (iii) Find the probability that Annabel eats potato on more than 44 days in a year of 365 days. [5]

10 marks

Mark scheme: 7 (i) P(2) = 7C2(0.1)2(0.9)5 M1 Bin term 7C2p2(1 – p)5 0 < p < 1 = 0.124 A1 [2] (ii) (0.15)1(0.1)2(0.75)2 × 5!/2!2! M1 Mult probs for options, (0.15)a(0.1)b(0.75)c where a + b + c sum to 5 M1 Mult by 5!/2!2! oe = 0.0253 or 81/3200 A1 [3] (iii) mean = 365×0.15 (= 54.75 or 219/4) B1 Correct unsimplified mean and var, oe Var = 365× 0.15×0.85 (= 46.5375 or 3723/80)  445. − 54.75  P(x > 44) = P  z >  M1 ± Standardising need sq rt  46.5375  M1 cc either 44.5 (or 43.5) = P(z > –1.5025) M1 Φ = 0.933 A1 [5] Correct answer accept 0.934

This question in 9709/63 Oct/Nov 2016

4 The weights in kilograms of packets of cereal were noted correct to 4 significant figures. The following stem-and-leaf diagram shows the data. 747 3 (1) 748 1 2 5 7 7 9 (6) 749 0 2 2 2 3 5 5 5 6 7 8 9 (12) 750 1 1 2 2 2 3 4 4 5 6 7 7 8 8 9 (15) 751 0 0 2 3 3 4 4 4 5 5 7 7 9 (13) 752 0 0 0 1 1 2 2 3 4 4 4 (11) 753 2 (1) Key: 748 5 represents 0.7485 kg. (i) On the grid, draw a box-and-whisker plot to represent the data. [5] (ii) Name a distribution that might be a suitable model for the weights of this type of cereal packet. Justify your answer. [2] … … … …

7 marks

Mark scheme: 4(i) LQ = 0.7495 Med = 0.7507 UQ = 0.7517 M1 Attempt to find all 3 quartiles can be implied, Condone LQ=0.7496, Med=0.7506, UQ=0.7515 B1 Correct median line in box using their scale A1 Correct quartiles in box B1 Correct end whiskers(not dots or boxes), lines not through box, B1 Correct uniform scale from at least 0.7473 to 0.7532, and label (wt) kg oe can be seen in title or scale Total: 5 0.747 0.748 0.749 0.750 0.751 0.752 0.753 Wt kg Question Answer Marks Guidance 4(ii) Normal B1 Symmetrical/peaks in middle or tails off quickly B1 Need symm + another reason Total: 2

This question in 9709/62 Feb/March 2017

7 (a) The lengths, in centimetres, of middle fingers of women in Raneland have a normal distribution with mean - and standard deviation 3. It is found that 25% of these women have fingers longer than 8.8 cm and 17.5% have fingers shorter than 7.7 cm. (i) Find the values of - and 3. [5] … … … … … … … … … … … … … The lengths, in centimetres, of middle fingers of women in Snoland have a normal distribution with mean 7.9 and standard deviation 0.44. A random sample of 5 women from Snoland is chosen. (ii) Find the probability that exactly 3 of these women have middle fingers shorter than 8.2 cm. [5] … … … … … … … … … … … … … … … … … … (b) The random variable X has a normal distribution with mean equal to the standard deviation. Find the probability that a particular value of X is less than 1.5 times the mean. [3] … … … … … … … … … … … … …

13 marks

Mark scheme: 7(a)(i) 8.8 0.674 σ = ⇒ 0.674σ = 8.8 – µ B1 ±0.674 seen 7.7 0.935 µ σ − − = ⇒-0.935σ = 7.7 – µ B1 ±0.935 seen (condone ±0.934) M1 An eqn with a z-value, µ and σ allow sq rt, sq cc M1 sensible attempt to eliminate µ or σ by substitution or subtraction σ = 0.684 µ = 8.34 A1 correct answers (from –0.935) Total: 5 7(a)(ii) P(< 8.2) = P 8.2 7.9 0.44 −   <     z M1 Standardising no cc no sq rt no sq M1 Correct area ie Φ, final solution = P(z < 0.6818) = 0.7524 A1 Correct prob rounding to 0.752 P(3) = 5C3 (0.7524)3(0.2476)2 M1 Binomial 5Cx powers summing to 5, any p, Σp = 1 = 0.261 A1 Total: 5 Question Answer Marks Guidance 7(b) P(< 1.5µ) = P 1.5µ µ µ   − <     z = P (z < 0.5) *M1 standardising with µ and σ (σ may be replaced by µ) DM1 just one variable = 0.692 A1 Total: 3

This question in 9709/62 Feb/March 2017

5 The lengths of videos of a certain popular song have a normal distribution with mean 3.9 minutes. 18% of these videos last for longer than 4.2 minutes. (i) Find the standard deviation of the lengths of these videos. [3] … … … … … … … … … … … … … (ii) Find the probability that the length of a randomly chosen video differs from the mean by less than half a minute. [4] … … … … … … … … … … … … … … … … … … … The lengths of videos of another popular song have a normal distribution with the same mean of 3.9 minutes but the standard deviation is twice the standard deviation in part (i). The probability that the length of a randomly chosen video of this song differs from the mean by less than half a minute is denoted by p. (iii) Without any further calculation, determine whether p is more than, equal to, or less than your answer to part (ii). You must explain your reasoning. [2] … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) 4.2 3.9 ( ) z σ = M1 z = 0.916 or 0.915 B1 Accept 0.915 ⩽ ± z ⩽0.916 seen σ = 0.328 A1 Correct final answer (allow 20/61 or 75/229) Total: 3 Question Answer Marks Guidance 5(ii) z = 4.4 – 3.9/their 0.328 or z = 3.4 – 3.9/their 0.328 = 1.5267 = –1.5267 M1 Standardising attempt with 3.4 or 4.4 only, allow square root of σ, or σ2 Φ = 0.9364 A1 0.936 ⩽ Φ ⩽ 0.937 or 0.063 ⩽ Φ ⩽ 0.064 seen Prob = 2Φ – 1 = 2(0.9364) – 1 M1 Correct area 2Φ – 1OE i.e. Φ = – (1 – Φ), linked to final solution = 0.873 A1 Correct final answer from 0.9363 ⩽ Φ ⩽0.9365 Total: 4 5(iii) dividing (0.5) by a larger number gives a smaller z-value or more spread out as sd larger or use of diagrams *B1 No calculations or calculated values present e.g. (σ = )0.656 seen Reference to spread or z value required Prob is less than that in (ii) DB1 Dependent upon first B1 Total: 2 6(i) EITHER: Route 1 A*********A in 9! / 2!2!5! = 756 ways (*M1 Considering AA and BB options with values B*********B in 9! / 4!5! = 126 ways A1 Any one option correct 756 + 126 DM1 Summing their AA and BB outcomes only Total = 882 ways A1) Question Answer Marks Guidance OR1: Route 2 A*********A in 9C5 × 4C2 = 756 ways (M1 Considering AA and BB options with values B*********B in 9C4 × 5C5= 126 ways A1 Any one option correct 756 + 126 DM1 Summing their AA and BB outcomes only Total = 882 A1) Total: 4 Question Answer Marks Guidance 6(ii) EITHER: (The subtraction method) As together, no restrictions 8! / 2!5! = 168 (*M1 Considering all As together – 8! seen alone or as numerator – condone × 4! for thinking A’s not identical As together and Bs together 7! / 5! = 42 M1 Considering all As together and all Bs together – 7! seen alone or numerator M1 Removing repeated Bs or Cs – Dividing by 5! either expression or 2! 1st expression only – OE Total 168 – 42 DM1 Subt their 42 from their 168 (dependent upon first M being awarded) = 126 A1) OR1: As together, no restrictions 8C5 x 3C1 = 168 (*M1 8C5 seen alone or multiplied M1 7C5 seen alone or multiplied As together and Bs together 7C5 x 2C1 = 42 M1 First expression x 3C1 or second expression x 2C1 Total 168 – 42 DM1 Subt their 42 from their 168 (dependent upon first M being awarded) = 126 A1) OR2: (The intersperse method ) (M1 Considering all “As together” with Cs – Mult by 6! (AAAA)CCCCC then intersperse B and another B M1 Removing repeated Cs – Dividing by 5!– [Mult by 6 implies M2] *M1 Considering positions for Bs – Mult by 7P2 oe –

This question in 9709/62 May/June 2017

5 Blank CDs are packed in boxes of 30. The probability that a blank CD is faulty is 0.04. A box is rejected if more than 2 of the blank CDs are faulty. (i) Find the probability that a box is rejected. [3] … … … … … … … … … … … … … … … … … … … … … … … … (ii) 280 boxes are chosen randomly. Use an approximation to find the probability that at least 30 of these boxes are rejected. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) EITHER: P(> 2) = 1 – P(0, 1, 2) (M1 = 1 – (0.96)30 – 30C1(0.04)(0.96)29 – 30C2(0.04)2(0.96)28 ( = 1 – 0.2938… – 0.3673… – 0.2219… ) A1 Correct unsimplified answer = 1-0.883103 = 0.117 (0.116896) A1) OR: P(> 2) = P(3,4,5,6,….30) (M1 Binomial term of form 30Cxpx(1 – p)30 – x , 0 < p < 1 any p = 30C3(0.04)3(0.96)27+ 30C4(0.04)4(0.96)26 + … +(0.04)30 A1 Correct unsimplified answer = 0.117 A1) 3 Question Answer Marks Guidance 5(ii) np = 280 × 0.1169 = 32.73, npq = 280 × 0.1169 × 0.8831 = 28.9 M1 FT Correct unsimplified np and npq, FT their p from (i), P(⩾ 30) = P 29.5 32.73 28.9 −   >     z = P(z > – 0.6008) M1 Substituting their µ and σ (√npq only) into the Standardisation Formula M1 Using continuity correction of 29.5 or 30.5 M1 Appropriate area Φ from standardisation formula P(z >….) in final solution = 0.726 A1 5 Question Answer Marks Guidance 6(a)(i) EITHER: 3**, 4**, 6**, 8** (M1 5P2 or 5C2 × 2! or 5 × 4 OE (considering final 2 digits) options 4 × 5 × 4 = 80 M1 Mult by 4 or summing 4 options (considering first digit) A1) Correct final answer OR: Total number of values: 6 × 5 × 4 = 120 (M1 Calculating total number of values (with subtraction seen) Number of values less than 300: 2 × 5 × 4 = 40 M1 Calculating number of unwanted values Number of evens = 120 – 40 = 80 A1) Correct final answer 3 Question Answer Marks Guidance 6(a)(ii) 3**, 4**, 6**, 8** EITHER: options 4 × 6 × 4 (last) (M1 6 linked to considering middle digit e.g. multiplied or in list M1 Multiply an integer by 4 × 4 (condone × 16) (No additional figures present for both M’s to be awarded) = 96 A1) OR: Total number of values 4 × 6 × 6 = 144 (M1 Calculating total number of values (with subtraction seen) Number of odd values 4 × 6 × 2 = 48 M1 Calculating number of unwanted values Number of evens = 144 – 48 = 96 A1) 3 6(b)(i) 252 B1 1

This question in 9709/62 Oct/Nov 2017

7 In Jimpuri the weights, in kilograms, of boys aged 16 years have a normal distribution with mean 61.4 and standard deviation 12.3. (i) Find the probability that a randomly chosen boy aged 16 years in Jimpuri weighs more than 65 kilograms. [3] … … … … … … … … (ii) For boys aged 16 years in Jimpuri, 25% have a weight between 65 kilograms and k kilograms, where k is greater than 65. Find k. [4] … … … … … … … … … … … … … In Brigville the weights, in kilograms, of boys aged 16 years have a normal distribution. 99% of the boys weigh less than 97.2 kilograms and 33% of the boys weigh less than 55.2 kilograms. (iii) Find the mean and standard deviation of the weights of boys aged 16 years in Brigville. [5] … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 7(i) P(> 65) = P 65 61.4 12.3 −   >     z = P (z > 0.2927) M1 condone ± standardisation formula M1 Correct area (< 0.5) = 1 – 0.6153 = 0.385 A1 3 Question Answer Marks Guidance 7(ii) P (< 65) = 0.6153 so P(< k) = 0.25 + 0.6153 = 0.8653 B1 z = 1.105 B1 z = ± 1.105 seen or rounding to 1.1 1.105 = 61.4 12.3 − k M1 standardising allow ±, cc, sq rt, sq. Need to see use of tables backwards so must be a z-value, not 1 – z value. k = 75.0 A1 Answers which round to 75.0. Condone 75 if supported. 4 7(iii) 2.326 = B1 ± 2.326 seen (Use of critical value) –0.44 = B1 ± 0.44 seen M1 An equation with a z-value, µ, σ and 97.2 or 55.2, allow √σ or σ2 M1 Algebraic elimination µ or σ from their two simultaneous equations µ = 61.9 σ = 15.2 A1 both correct answers 5 σ µ − 2. 97 σ µ − 2. 55

This question in 9709/62 Oct/Nov 2017

7 Josie aims to catch a bus which departs at a fixed time every day. Josie arrives at the bus stop T minutes before the bus departs, where T ∼N 5.3, 2.12 . (i) Find the probability that Josie has to wait longer than 6 minutes at the bus stop. [3] … … … … … … … … … … On 5% of days Josie has to wait longer than x minutes at the bus stop. (ii) Find the value of x. [3] … … … … … … … … … … … (iii) Find the probability that Josie waits longer than x minutes on fewer than 3 days in 10 days. [3] … … … … … … … … … … … … (iv) Find the probability that Josie misses the bus. [3] … … … … … … … … … … … …

12 marks

Mark scheme: 7(i) P(t > 6) = P 6 5.3 2.1 −   >     z = P(z > 0.333) M1 = 1 – 0.6304 M1 Correct area 1 – Φ (< 0.5), final solution = 0.370 or 0.369 A1 3 7(ii) z = 1.645 B1 ± 1.645 1.645 = 5.3 2.1 x − M1 Standardising, no continuity correction, allow sq, sq rt. Must be equated to a z-value x = 8.75 or 8.755 or 8.7545 A1 3 7(iii) n = 10, p = 0.05 M1 Bin term 10Cx p x(1–p)10–x P(0, 1, 2) = (0.95)10 + 10C1(0.05)(0.95)9 + 10C2(0.05)2(0.95)8 M1 Correct unsimplified answer = 0.988 (0.9885 to 4 sf) A1 3 7(iv) P(misses bus) = P(t < 0) *M1 Seeing t linked to zero = P 0 5.3 2.1 −   <     z = P(z < –2.524) = 1 – Φ(2.524) = 1 – 0.9942 DM1 Standardising with t = 0, no continuity correction, no sq, no sq rt = 0.0058 A1 3

This question in 9709/63 Oct/Nov 2017

7 The weights of packets of a certain type of biscuit are normally distributed with mean 400 grams and standard deviation 3 grams. (i) In a random sample of 6000 packets of this type of biscuit, 225 packets weighed more than 410 grams. Find the value of 3. [4] … … … … … … … … … … … … … … … … … … … … … … … (ii) In a random sample of 500 packets of this type of biscuit, how many packets would you expect to find with weights that are more than 1.5 standard deviations from the mean? [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(i) P(X > 410) = 225/6000 = 0.0375 P 410 400 σ −   >     Z = 0.0375: 0.9625 z value = ±1.78 A1 z value: ± 1.78 10 1.78 σ = M1 (410-400)/σ = their z (must be a z value) 5.62 σ = A1 4 7(ii) We need P(Z < −1.5) and P(Z > 1.5) M1 Attempt at P(Z < −1.5) or P(Z > 1.5) 1 – Φ(1.5) seen ( ) ( ) Φ 1.5 1 Φ 1.5 − + − = ( ) 2 2Φ 1.5 − M1 Or equivalent expression with values =2 – 2 × 0.9332 = 0.1336 (0.134) A1 Correct to 3sf Number expected = 500 × 0.1336 = 66.8: 66 or 67 packets B1ft 0.1336 used or FT their 4sf probability times 500, (not 0.9625 or 0.0375) rounded or truncated 4

This question in 9709/62 Feb/March 2018

8 The results of a survey at a certain large college show that the proportion of students who own a car is 14. (i) Five students at the college are chosen at random. Find the probability that at least four of these students own a car. [3] … … … … … … … … … … … … … … (ii) For a random sample of n students at the college, the probability that at least one of the students owns a car is greater than 0.995. Find the least possible value of n. [3] … … … … … … … … … … … … … (iii) For a random sample of 160 students at the college, use a suitable approximate distribution to find the probability that fewer than 50 own a car. [4] … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 8(i) P(4) + P(5) = 4 1 5 0 5 5 4 5 1 3 1 3 C C 4 4 4 4       +             = 0.014648.. + 0.00097656.. M1 Add 2 correct unsimplified binomial terms = 0.0156 or 1 64 A1 3 8(ii) 1 −P(0) > 0.995: 0.75 0.005 n < M1 Equation or inequality involving 0.75n and 0.005 or 0.25n and 0.995 log0.75 log0.005 n < n > 18.4: M1 Attempt to solve their exponential equation using logs, or trial and error May be implied by their answer n = 19 A1 3 8(iii) p = 0.25, n = 160: mean = 160 x 0.25 (= 40) variance = 160 x 0.25 x 0.75 (=30) B1 Correct unsimplified mean and variance P(X < 50) = P 49.5 40 30 −   <     Z M1 Use standardisation formulae must include square root. M1 Use continuity correction ±0.5 (49.5 or 50.5) = P(Z < 1.734) = 0.959 A1 Correct final answer 4

This question in 9709/62 Feb/March 2018

5 In Pelmerdon 22% of families own a dishwasher. (i) Find the probability that, of 15 families chosen at random from Pelmerdon, between 4 and 6 inclusive own a dishwasher. [3] … … … … … … … … … … … … … … … … … … … … … … … … (ii) A random sample of 145 families from Pelmerdon is chosen. Use a suitable approximation to find the probability that more than 26 families own a dishwasher. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) M1 15C6(0.22)6(0.78)9 A1 Correct unsimplified expression = 0.398 A1 Correct answer 3 5(ii) µ = 145 × 0.22 = 31.9 σ2 = 145 × 0.22 × 0.78 = 24.882 B1 Correct unsimplified mean and variance P(x > 26) = P 26.5 31.9 24.882 z −   >     = P(z > –1.08255) M1 Standardising must have sq rt M1 25.5 or 26.5 seen as a cc = Φ(1.08255) M1 Correct area Φ, must agree with their µ = 0.861 A1 Correct final answer accept 0.861, or 0.860 from 0.8604 not from 0.8599 5

This question in 9709/61 May/June 2018

3 (i) The volume of soup in Super Soup cartons has a normal distribution with mean - millilitres and standard deviation 9 millilitres. Tests have shown that 10% of cartons contain less than 440 millilitres of soup. Find the value of -. [3] … … … … … … … … … … … … (ii) A food retailer orders 150 Super Soup cartons. Calculate the number of these cartons for which you would expect the volume of soup to be more than 1.8 standard deviations above the mean. [3] … … … … … … … … … …

6 marks

Mark scheme: 3(i) z = –1.282 B1 –1.282 = 440 9 µ − M1 ±Standardisation equation with 440, 9 and µ, equated to a z-value, (not 1 – z-value or probability e.g. 0.1841, 0.5398, 0.6202, 0.8159) µ = 452 A1 Correct answer rounding to 452, not dependent on B1 3 3(ii) P(z > 1.8) = 1 – 0.9641 = 0.0359 B1 Number = 0.0359 × 150 = 5.385 M1 p × 150, 0 < p < 1 (Number of cartons = ) 5 A1FT Accept either 5 or 6, not indicated as an approximation, e.g. ~, about FT their p × 150, answer as an integer 3

This question in 9709/62 May/June 2018

2 The random variable X has the distribution N −3, 32 . The probability that a randomly chosen value of X is positive is 0.25. (i) Find the value of 3. [3] … … … … … … … … … … … (ii) Find the probability that, of 8 random values of X, fewer than 2 will be positive. [3] … … … … … … … … … … …

6 marks

Mark scheme: 2(i) z = 0.674 B1 z value ±0.674 0.674 = σ 3 0 − − M1 ±Standardising with 0 and equating to a z-value σ = 4.45 A1 Correct answer www ie not ignoring a minus sign Total: 3 2(ii) P(0, 1) = (0.75)8 + 8C1(0.25)(0.75)7 M1 Any bin of form 8Cx(0.75)x (0.25)8–xany x M1 Correct unsimplified answer, may be implied by numerical values 0.1001+ 0.2670 = 0.367 A1 Correct answer Method 2 1 – P(8,7,6,5,4,3,2) = 1 – (0.25)8 – 8C1(0.75)(0.25)7 – … – 8C2(0.75)6 (0.25)2 = 0.367 M1 Any bin of form 8Cx(0.75)x (0.25)8–xany x M1 Correct unsimplified answer A1 Correct answer Total: 3

This question in 9709/63 May/June 2018

6 The diameters of apples in an orchard have a normal distribution with mean 5.7 cm and standard deviation 0.8 cm. Apples with diameters between 4.1 cm and 5 cm can be used as toffee apples. (i) Find the probability that an apple selected at random can be used as a toffee apple. [3] … … … … … … … … … … … … … … … … … … … … … … … … (ii) 250 apples are chosen at random. Use a suitable approximation to find the probability that fewer than 50 can be used as toffee apples. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(ii) B1ft Correct unsimplified mean and var – ft their prob for (i) providing (0 < p < 1) Implied by 34.944 5.911 σ = = P(< 50) = P       − < 944 . 34 42 5. 49 z = P(z < 1.2687) M1 ± Standardising using 50, their mean and sd; must have sq rt. M1 49.5 or 50.5 seen as a cc = Φ(1.2687) M1 Correct area Φ(> 0.5 for + z and < 0.5 for –z)in their final answer = 0.898 A1 Correct final answer Total: 5

This question in 9709/63 May/June 2018

5 At the Nonland Business College, all students sit an accountancy examination at the end of their first year of study. On average, 80% of the students pass this examination. (i) A random sample of 9 students who will take this examination is chosen. Find the probability that at most 6 of these students will pass the examination. [3] … … … … … … … … … … … … … … … (ii) A random sample of 200 students who will take this examination is chosen. Use a suitable approximate distribution to find the probability that more than 166 of them will pass the examination. [5] … … … … … … … … … … … … … … … … … … … … … … … … … (iii) Justify the use of your approximate distribution in part (ii). [1] … … … … …

9 marks

Mark scheme: 5(i) 1 – (P(7) + P(8) + P(9)) = 1 – ( 9C7 7 2 0.8 0.2 × + 9C8 8 1 0.8 0.2 × + 9C9 9 0 0.8 0.2 ) × M1 Any binomial term of form 9Cxpx(1 – p)9 – x, x ≠ 0 M1 Correct unsimplified expression = 1 – (0.3019899 + 0.3019899 + 0.1342177) = 0.262 A1 Correct answer 3 Question Answer Marks Guidance 5(ii) Mean = 200 × 0.8 = 160: var = 200 × 0.8 × 0.2 = 32 B1 Both unsimplified P(X > 166) = P( 166.5 160 32 Z − > ) M1 Standardise, 1 60 32 x their z their − = ± with square root M1 166.5 or 165.5 seen in attempted standardisation expression = P(Z > 1.149) = 1 – 0.8747 M1 1 – a Φ -value, correct area expression, linked to final answer = 0.125 A1 Correct final answer 5 5(iii) np = 160, nq = 40: both > 5 (so normal approx. holds) B1 Both parts required 1

This question in 9709/61 Oct/Nov 2018

5 The weights of apples sold by a store can be modelled by a normal distribution with mean 120 grams and standard deviation 24 grams. Apples weighing less than 90 grams are graded as ‘small’; apples weighing more than 140 grams are graded as ‘large’; the remainder are graded as ‘medium’. (i) Show that the probability that an apple chosen at random is graded as medium is 0.692, correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … (ii) Four apples are chosen at random. Find the probability that at least two are graded as medium. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) 1 90 120 24 z = ± = – 5 4 , 2 140 120 24 z = ± = 5 6 and either 90 or 140 = 20 30 Φ Φ 24 24     − −         A1 –5/4 and 5/6 unsimplified = ( ) ( ) Φ 0.8333 (1 Φ 1.25 ) − − = 0.7975 – (1 – 0.8944) or 0.8944 – 0.2025 = 0.6919 M1 Correct area Φ – Φ legitimately obtained and evaluated from phi(their z2) – phi (their z1) = 0.692 AG A1 Correct answer obtained from 0.7975 and 0.1056 oe to 4sf or 0.6919 seen www 4 Question Answer Marks Guidance 5(ii) Method 1 Probability = P(2, 3, 4) = 0.6922(1 – 0.692)2 × 4C2 + 0.6923(1 – 0.692) × 4C3 + 0.6924 M1 Any binomial term of form ( ) 4 4 1 x x x C p p − − , x ≠ 0 or 4 B1 One correct bin term with 4 n = and 0.692 p = , = 0.27256 + 0.40825 + 0.22931 M1 Correct unsimplified expression using 0.692 or better = 0.910 A1 Correct answer Method 2: 1 – P(0, 1) = M1 Any binomial term of form ( ) 4 4 1 x x x C p p − − , x≠0 or 4 1 − 0.6920(1 – 0.692)4 × 4C0 − 0.6921(1 – 0.692)3 × 4C1 B1 One correct bin term with 4 n = and 0.692 p = = 1 – 0.00899 – 0.0808757 M1 Correct unsimplified expression using 0.692 or better = 0.910 A1 Correct answer 4

This question in 9709/63 Oct/Nov 2018

6 The lifetimes, in hours, of a particular type of light bulb are normally distributed with mean 2000 hours and standard deviation 3 hours. The probability that a randomly chosen light bulb of this type has a lifetime of more than 1800 hours is 0.96. (i) Find the value of 3. [3] … … … … … … … … … … … … … … … … … … … … … … … New technology has resulted in a new type of light bulb. It is found that on average one in five of these new light bulbs has a lifetime of more than 2500 hours. (ii) For a random selection of 300 of these new light bulbs, use a suitable approximate distribution to find the probability that fewer than 70 have a lifetime of more than 2500 hours. [4] … … … … … … … … … … … … … … … … … (iii) Justify the use of your approximate distribution in part (ii). [1] … … … … …

8 marks

Mark scheme: 6(i) P(X >1800) = 0.96, so P( 1800 2000 Z σ > ) = 0.96 B1 ± 1.75 seen 200 Φ( ) σ = 0.96 200 1.751 σ = M1 1800 2000 z σ − = ± , allow cc, allow sq rt, allow sq equated to a z-value 114 σ = A1 Correct final answer www 3 6(ii) Mean = 300 × 0.2 = 60 and variance = 300 × 0.2 × 0.8 = 48 B1 Correct unsimplified mean and variance P(X < 70) = P( 69.5 60 48 Z − > ) M1 Z = ± 60 48 x their their − = ( ) Φ 1.371 M1 69.5 or 70.5 seen in an attempted standardisation expression as cc =0.915 A1 Correct final answer 4 6(iii) np = 60, nq = 240: both > 5, (so normal approximation holds) B1 Both parts evaluated are required 1

This question in 9709/63 Oct/Nov 2018

3 The times taken, in minutes, for trains to travel between Alphaton and Beeton are normally distributed with mean 140 and standard deviation 12. (i) Find the probability that a randomly chosen train will take less than 132 minutes to travel between Alphaton and Beeton. [3] … … … … … … … … … … (ii) The probability that a randomly chosen train takes more than k minutes to travel between Alphaton and Beeton is 0.675. Find the value of k. [3] … … … … … … … … … … …

6 marks

Mark scheme: 3(i) P(X< 132) = P 132 140 P( 0.6667) 12 −   < = < −     Z Z √σ = 1 – 0.7477 M1 Appropriate area Φ from standardisation formula P(z<….) in final solution = 0.252 awrt A1 Condone linear interpolation = 0.25243 3 3(ii) P(time>k) = 0.675, z = –0.454 B1 ±0.454 seen 140 0.454 12 − = − k M1 An equation using the standardisation formula with a z-value (not 1 – z), condone σ2 or √σ k = 135, 134.6, 134.55 A1 B0M1A1 max from –0.45 3

This question in 9709/62 Feb/March 2019

6 The results of a survey by a large supermarket show that 35% of its customers shop online. (i) Six customers are chosen at random. Find the probability that more than three of them shop online. [3] … … … … … … … … … … … (ii) For a random sample of n customers, the probability that at least one of them shops online is greater than 0.95. Find the least possible value of n. [3] … … … … … … … … … … … (iii) For a random sample of 100 customers, use a suitable approximating distribution to find the probability that more than 39 shop online. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 6(i) A1 Correct unsimplified answer = 0.117 A1 3 6(ii) 1 0.65 0.95 − > n 0.65 0.05 < n M1 Equation or inequality involving ‘0.65n or 0.35n’ and ‘0.95 or 0.05’ log0.05 6.95 log0.65 > = n M1 Attempt to solve their exponential equation using logs or Trial and Error. n = 7 A1 CAO 3 6(iii) Mean = 0.35 100 35 × = Variance = 0.35 0.65 100 22.75 × × = B1 Correct unsimplified np and npq, P ( ) 39.5 35 0.943 22.75 −   > = >     z P z M1 Substituting their µ and σ (condone σ2) into the ±Standardisation Formula with a numerical value for ‘39.5’. M1 Using continuity correction 39.5 or 40.5 = 1 0.8272 − M1 Appropriate area Φ from standardisation formula P(z>….) in final solution, (>0.5 if z is -ve, <0.5 if z is +ve) = 0.173 A1 Final answer 5

This question in 9709/62 Feb/March 2019

5 In a certain country the probability that a child owns a bicycle is 0.65. (i) A random sample of 15 children from this country is chosen. Find the probability that more than 12 own a bicycle. [3] … … … … … … … … … … … … … … … … … … … … … … … … (ii) A random sample of 250 children from this country is chosen. Use a suitable approximation to find the probability that fewer than 179 own a bicycle. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(i) (P > 12) = P(13, 14, 15) = 15C13(0.65)13(0.35)2 + 15C14(0.65)14(0.35)1 + (0.65)15 A1 Correct unsimplified answer = 0.0617 A1 SC if use np and npq with justification give (12.5 – 9.75)/√3.41 M1 1–F(1.489) A1 0.0681 A0 3 5(ii) mean = 250 × 0.65 = 162.5 variance = 250 × 0.65 × 0.35 = 56.875 B1 Correct unsimplified np and npq P(< 179) = P(z <178.5 162.5 56.875 − ) = P(z < 2.122) M1 Substituting their µ and σ (condone σ2) into the Standardisation Formula with a numerical value for ‘178.5’. Continuity correct not required for this M1. Condone ± standardisation formula Using continuity correction 178.5 or 179.5 M1 = 0.983 A1 Correct final answer 4

This question in 9709/61 May/June 2019

2 The volume of ink in a certain type of ink cartridge has a normal distribution with mean 30 ml and standard deviation 1.5 ml. People in an office use a total of 8 cartridges of this ink per month. Find the expected number of cartridges per month that contain less than 28.9 ml of this ink. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 P( < 28.9) = P 28.9 30 1.5 −   <     z B1 = P(z < –0.733) = 1 – 0.7682 M1 Appropriate area Φ from standardisation formula P(z <….) in final probability solution, Must be a probability, e.g. 1 – 0.622 is M0 = 0.2318 A1 Correct final probability rounding to 0.232. (Only requires M1 not B1 to be awarded Number of cartridges is their 0.2318 × 8 = 1.85, so 2 (Also accept 1 but not both) B1 FT using their 4 SF (or better) value, ans. rounded or truncated to integer, no approximation indicated. 4

This question in 9709/62 May/June 2019

4 It is known that 20% of male giant pandas in a certain area weigh more than 121 kg and 71.9% weigh more than 102 kg. Weights of male giant pandas in this area have a normal distribution. Find the mean and standard deviation of the weights of male giant pandas in this area. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 z = 0.842 = 121 µ σ −       so 0.842σ = 121 – µ B1 M1 One appropriate standardisation equation with a z-value, µ, σ and 121 or 102, condone continuity correction. Not 0.158, 0.42,… z = –0.58 = 102 µ σ −       so –0.58σ = 102 – µ B1 ± 0.58(0) seen but B0 if 1 ± 0.58 oe seen Solving M1 Correct algebraic elimination of µ or σ from their two simultaneous equations to form an equation in one variable, condone 1 numerical slip σ = 13.4 µ = 110 A1 If M0A0 scored (i.e. no algebraic elimination seen), SC B1 can be awarded for both answers correct Consistent use of σ 2 or √σ throughout apply MR penalty to A mark or SC B mark. 5

This question in 9709/62 May/June 2019

1 The time taken, in minutes, by a ferry to cross a lake has a normal distribution with mean 85 and standard deviation 6.8. (i) Find the probability that, on a randomly chosen occasion, the time taken by the ferry to cross the lake is between 79 and 91 minutes. [3] … … … … … … … … … … (ii) Over a long period it is found that 96% of ferry crossings take longer than a certain time t minutes. Find the value of t. [3] … … … … … … … … … … …

6 marks

Mark scheme: 1(i) P(79 < X < 91) = P 79 85 91 85 6.8 6.8 − −   < <     Z = P(–0.8824 < Z < 0.8824) correction = ( ) ( ) Φ 0.8824 Φ 0.8824 − − = 0.8111 – (1 – 0.8111) M1 Correct area ( Φ Φ − ) with one +ve and one –ve z-value or 2 Φ – 1 or 2(Φ 0.5) − = 0.622 A1 Correct answer 3 1(ii) z = –1.751 B1 ± 1.751 seen –1.751 = 85 6.8 − t M1 An equation using ± standardisation formula with a z-value, condone σ2 or √σ t = 73.1 A1 Correct answer 3

This question in 9709/63 May/June 2019

5 On average, 34% of the people who go to a particular theatre are men. (i) A random sample of 14 people who go to the theatre is chosen. Find the probability that at most 2 people are men. [3] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Use an approximation to find the probability that, in a random sample of 600 people who go to the theatre, fewer than 190 are men. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) = 0.0029758 + 0.02146239 + 0.071866 A1 Correct unsimplified answer = 0.0963 A1 Correct answer 3 5(ii) Mean =600 × 0.34 = 204, Var = 600 × 0.34 × 0.66 = 134.64 B1 Correct unsimplified np and npq (or sd = 11.603 or Variance = 3366/25) P(< 190) = P 189.5 204 134.64 −   <     z = P(z < –1.2496) M1 Substituting their µ and σ, (no σ2 or √σ) into the Standardisation Formula with a numerical value for ‘189.5’. Condone ± standardisation formula M1 Using continuity correction 189.5 or 190.5 within a Standardisation formula = 1 – Φ (1.2496) M1 Appropriate area Φ from standardisation formula P(z<….) in final solution, (<0.5 if z is –ve, >0.5 if z is +ve) = 1 – 0.8944 = 0.106 A1 Correct final answer 5

This question in 9709/63 May/June 2019

7 The shortest time recorded by an athlete in a 400 m race is called their personal best (PB). The PBs of the athletes in a large athletics club are normally distributed with mean 49.2 seconds and standard deviation 2.8 seconds. (i) Find the probability that a randomly chosen athlete from this club has a PB between 46 and 53 seconds. [4] … … … … … … … … … … … … … … (ii) It is found that 92% of athletes from this club have PBs of more than t seconds. Find the value of t. [3] … … … … … … … … … … … … … … Three athletes from the club are chosen at random. (iii) Find the probability that exactly 2 have PBs of less than 46 seconds. [3] … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(i) P(46 < X < 53) = P 46 49.2 53 49.2 2.8 2.8 Z − −   < <     continuity correction, σ2 or √σ P( 1.143 1.357) Z − < < A1 Both standardisations correct unsimplified ( ) ( ) Φ 1.357 Φ 1.143 1 + − = 0.9126 + 0.8735 – 1 M1 Correct final area 0.786 A1 Final answer 4 Question Answer Marks Guidance 7(ii) 49.2 1.406 2.8 t − = − B1 ±1.406 seen M1 An equation using ± standardisation formula with a z-value, condone σ2 or √σ 45.3 A1 3 7(iii) P(X < 46) = 0.1265 M1 Calculated or ft from (i) P(2PB < 46) = ( ) 2 3 1 0.1265 0.1265 − M1 3(1-p)p2, 0<p<1 0.0419 A1 3

This question in 9709/61 Oct/Nov 2019

4 In Quarendon, 66% of households are satisfied with the speed of their wificonnection. (i) Find the probability that, out of 10 households chosen at random in Quarendon, at least 8 are satisfied with the speed of their wificonnection. [3] … … … … … … … … … … … … … … … … … … … … … … … … (ii) A random sample of 150 households in Quarendon is chosen. Use a suitable approximation to find the probability that more than 84 are satisfied with the speed of their wificonnection. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(i) a+b = 10, 0 < a,b < 10 A1 Correct unsimplified expression 0.284 B1 CAO 3 Question Answer Marks Guidance 4(ii) 0.66 150 99 np = × = ( ) 0.66 1 0.66 150 33.66 npq = × − × = B1 Accept evaluated or unsimplified µ, σ2 numerical expressions, condone 33.66 5.8017 or 5.802 σ = = CAO P(X > 84) = P 84.5 99 33.66 Z −   >     M1 ± Standardise, 99 33.66 x their their − , condone σ2, x a value M1 84.5 or 83.5 used in their standardisation formula (= P( ) 2.499 Z > − ) M1 Correct final area 0.994 A1 Final answer (accept 0.9938) SC if no standardisation formula seen, B2 P(Z > -2.499) = 0.994 5

This question in 9709/62 Oct/Nov 2019

6 The heights, in metres, of fir trees in a large forest have a normal distribution with mean 40 and standard deviation 8. (i) Find the probability that a fir tree chosen at random in this forest has a height less than 45 metres. [2] … … … … … … … … … … (ii) Find the probability that a fir tree chosen at random in this forest has a height within 5 metres of the mean. [2] … … … … … … … … … … … In another forest, the heights of another type of fir tree are modelled by a normal distribution. A scientist measures the heights of 500 randomly chosen trees of this type. He finds that 48 trees are less than 10 m high and 76 trees are more than 24 m high. (iii) Find the mean and standard deviation of the heights of trees of this type. [5] … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) P(X < 45) = P 45 40 8 Z −   <     = P(Z < 0.625) M1 ± Standardise, no continuity correction, σ2 or σ , formula must be seen 0.734(0) A1 CAO 2 6(ii) ( ) ( ) ( ) 1 2 1 2 1 i i − − = − = 2((i) – 0.5) M1 Use result of part (i) or recalculated to find area OE 0.468 A1ft 0 < FT from (i) < 1 or correct. 2 6(iii) P(X < 10) = 48/500 = 0.096 z = –1.305 B1 z = ± 1.305 P(X > 24) = 76/500 = 0.152 z = 1.028 B1 z = ± 1.028 10 1.305 µ σ − = − 24 1.028 µ σ − = M1 Form 1 equation using 10 or 24 with ߤ, ߪ, ݖ–value. Allow continuity correction, not ߪଶ, √ߪ 14 = 2.333σ M1 OE Solve two equations in σ and µ to form equation in one variable [ ] [ ] 6. 00 , 17.8 3 σ µ = = A1 CAO, WWW 5

This question in 9709/62 Oct/Nov 2019

4 The heights of students at the Mainland college are normally distributed with mean 148 cm and standard deviation 8 cm. (i) The probability that a Mainland student chosen at random has a height less than h cm is 0.67. Find the value of h. [3] … … … … … … … … … … … … … … … … … … … … … … … 120 Mainland students are chosen at random. (ii) Find the number of these students that would be expected to have a height within half a standard deviation of the mean. [4] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) P(h < 148) = 0.67 B1 z = ±0.44 seen 148 8 − h = 0.44 M1 z-value = ± ( 148) 8 − h 151.52 ≈ 152 A1 CAO 3 4(ii) P(144 < X < 152) = P 144 148 152 148 8 8 − −   < <     Z M1 Using ± standardisation formula for either 144 or 152, µ = 148, σ = 8 and no continuity correction, allow σ2 or √σ = P 1 1 2 2   − < <     Z = 0.6915 – (1 – 0.6915) = 2 × 0.6915 – 1 M1 Correct final area legitimately obtained from phi(their z2) – phi(their z1) = 0.383 A1 Final probability answer 0.383 × 120 = 45.96 Accept 45 or 46 only B1FT Their prob (to 3 or 4 sf) × 120, rounded to a whole number or truncated 4

This question in 9709/63 Oct/Nov 2019

6 The volumes, in millilitres, of large and small cups of tea are modelled by the distributions N 200, 30 and N 110, 20 respectively. (a) Find the probability that the total volume of a randomly chosen large cup of tea and a randomly chosen small cup of tea is less than 300 ml. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the probability that the volume of a randomly chosen large cup of tea is more than twice the volume of a randomly chosen small cup of tea. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) N(310, 50) B1 SOI ' 50 ' ' 310 ' 300− (= –1.414) M1 Standardise using their values Φ(‘–1.414’) = 1 – ɸ(‘1.414’) M1 Area consistent with their values = 0.0786 or 0.0787 (3 sf) A1 As final answer 4 Question Answer Marks Guidance 6(b) P(L – 2S > 0 ) M1 OE SOI E(X) = 200-2x110 or = – 20 B1 OE seen Var = 30 + 22 × 20 or = 110 B1 Seen N(–20, 110) ' 110 ' )' 20 (' 0 − − (= 1.907) M1 Standardising with their values. Mean and variance must come from a combination attempt. 1 – Φ(‘1.907’) M1 Correct area consistent with their working = 0.0283 (3 sf) A1 Final answer 6

This question in 9709/62 Feb/March 2020

2 The time, in minutes, taken by students to complete a test has the distribution N 125, 36 . (a) Find the probability that the mean time taken to complete the test by a random sample of 40 students is less than 123 minutes. [3] … … … … … … … … … … … … … … … … (b) Explain whether it was necessary to use the Central Limit theorem in the solution to part (a). [1] … … … … … …

4 marks

Mark scheme: 2(a) ±123 125 6 40 − [= –2.108...] No standard deviation/variance mix. Ignore any continuity correction attempts for this mark. P(z < ‘–2.108’) = 1 – Φ(‘2.108’) M1 For correct probability area consistent with their working. = 0.0175 or 0.0176 (3 sf) A1 3 2(b) No, population is normal B1 Need both. 1

This question in 9709/61 May/June 2021

5 The heights of buildings in a large city are normally distributed with mean 18.3m and standard deviation 2.5m. (a) Find the probability that the total height of 5 randomly chosen buildings in the city is more than 95m. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the probability that the difference between the heights of two randomly chosen buildings in the city is less than 1m. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) B1 SOI 95 '91.5 '31.25' − [= 0.626] M1 FT their mean and variance 1 – Φ(‘0.626’) M1 For finding area consistent with their values 0.266 (3 sf) A1 4 5(b) E(D) = 0 B1 Or E(D ‒ 1) = ‒1 Var(D) = 2.52 × 2 [= 12.5] B1 1 0 '12.5' − [= 0.283] or 1 0 '12.5' −− [= –0.283] M1 FT their E and Var Φ(‘0.283’) – (1 – ɸ(0.283)) [= 0.6115 – 0.3885] M1 For finding area consistent with their values 0.223 (3 sf) A1 5

This question in 9709/62 Feb/March 2022

4 A population is normally distributed with mean 35 and standard deviation 8.1 . A random sample of size 140 is chosen from this population and the sample mean is denoted by X . (a) Find P ( X 2 36) . [3] … … … … … … … … … … (b) It is given that P ( X 1 a) = 0. 986 . Find the value of a. [3] … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(a) 36 −35 M1 Ignore inclusion of cc for M1. Must have √140. [= 1.461] 8.1 140 1 − Φ(‘1.461’) M1 For area consistent with their values. = 0.0720 (3 sf) A1 Allow 0.072. 3 4(b) [Φ−1(0.986)] = 2.197 to 2.198 B1 Seen. Note: 2.2 and nothing better seen scores B0 ± a −35 = ± ‘2.198’ M1 Must be a z value. 8.1 140 a = 36.5 (3 sf) A1 CWO Note: use of 2.2 scores A1 so 2/3. But e.g. 2.196 gives 36.5 but scores B0 M1 A0 so 1/3 3

This question in 9709/62 Oct/Nov 2024

3 The times, T minutes, taken by a random sample of 75 students to complete a test were noted. The results were summarised by / t = 230 and /t 2 = 930 . (a) Calculate unbiased estimates of the population mean and variance of T. [3] … … … … … … … … … … … You should now assume that your estimates from part (a) are the true values of the population mean and variance of T. (b) The times taken by another random sample of 75 students were noted, and the sample mean, T , was found. Find the value of a such that P ( T 2 a) = 0. 234 . [3] … … … … … … … … … … …

6 marks

Mark scheme: 3(a) t = 23075 [= 3.0666… or 3.07 (3 sf)] [ 0r 46/15 ] B1 s2 = 74 75 ( 93075 − ( 23075 ) 2 ) or 1/74(930 – 2302/75 ) M1 Use of correct formula. = 3.0360… or 3.04 (3 sf) or = 337/111 A1 3 3(b) [ Φ−1(1 − 0.234) ] = 0.726 B1 a − '3.0667' M1 Ft their 0.726 but must be a z value. ± = ± ‘0.726’ Note using 0.766 is M0. '3.04'/75 Must have sqrt 75. a = 3.21 (3 sf) A1 CWO 3

This question in 9709/63 Oct/Nov 2024

3 2 The random variable X has the distribution Bb,8 l. A random sample of 100 values of X is chosen, and 4 the sample mean, X , is found. (a) Find P ( X 2 6 .2) . You are not expected to use a continuity correction. [6] … … … … … … … … … … … … … … … … … … … … (b) State why the Central Limit Theorem was needed in the calculation in part (a). [1] … … … … …

7 marks

Mark scheme: 2(a) N (6, ….) B1 Normal used with a mean of 6. 2 3 6 M1 Var of 1.5 (or sd found).  = = 1.5 (or = 1.5 or or 1.22 (3sf) 2 2 6.2 - 6 M1 For standardising. Must have 100 . Ignore 1.5 incorrect continuity correction (correct cc if used is 100 1/2n). = 1.633 A1 Or accept 1.674 with correct cc. M1 P X  6.2 = 1 −Φ(‘1.633’) ( ) = 0.0512 or 0.0513 (3sf) A1 Or accept 0.0471 with correct cc (scores full marks). 6 2(b) Population (X) is not normally distributed / Population (X) is Binomial B1 OE. Accept underlying distribution. 1

This question in 9709/61 May/June 2025

5 In Urberia, the masses, in kilograms, of men have the distribution N(70.3, .592). A certain footbridge in Urberia can take a maximum safe load of 1500 kg. When n men stand on the bridge, the probability that the bridge is unsafe is less than 0.01. Stating a necessary assumption, find the maximum value of n. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 5 Assume the men on the bridge are a random sample B1 OE. Allow: assume that all the men on the bridge are independent of one another. 1500 − 70.3n M1 Must have square root n. = 2.326 Allow wrong z but must be a z value. 5.9 n A1 Correct quadratic equation or inequality in square 70.3n + 13.7234 n − 1500 = 0 root n. Or 4942.09 n2 - 211088.33 n + 2250000 = 0 OE. [ n = 4.5226 (3 sf)] Max value of n is 20 A1 CWO. 4

This question in 9709/63 May/June 2025