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AS & A Level Mathematics (9709)

Linear combinations of random variables Banker

30 questions · about 2% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 61 · Q15 marksLinear combinations of normals
  • 2026 May/June · Paper 62 · Q25 marksLinear combinations of normals
  • 2026 May/June · Paper 63 · Q6 (a, b)10 marksLinear combinations of normals
  • 2026 May/June · Paper 65 · Q25 marksLinear combinations of normals

2025

  • 2025 Oct/Nov · Paper 61 · Q6 (a, b)10 marksLinear combinations of normals
  • 2025 Oct/Nov · Paper 62 · Q2 (a, b)6 marksLinear combinations of normals
  • 2025 Oct/Nov · Paper 63 · Q6 (a, b)10 marksLinear combinations of normals
  • 2025 Oct/Nov · Paper 65 · Q5 (a, b)10 marksLinear combinations of normals
  • 2025 May/June · Paper 61 · Q4 (a, b)6 marksLinear combinations of normals
  • 2025 May/June · Paper 62 · Q1 (b)5 marksLinear combinations of normals
  • 2025 May/June · Paper 63 · Q54 marksLinear combinations of normals
  • 2025 May/June · Paper 65 · Q55 marksLinear combinations of normals

2024

  • 2024 Oct/Nov · Paper 61 · Q25 marksLinear combinations of normals
  • 2024 Oct/Nov · Paper 63 · Q25 marksLinear combinations of normals
  • 2024 May/June · Paper 61 · Q25 marksOfficial paper ↗Linear combinations of normals
  • 2024 May/June · Paper 62 · Q4 (a, b)6 marksLinear combinations of normals
  • 2024 May/June · Paper 63 · Q35 marksLinear combinations of normals

2023

  • 2023 Oct/Nov · Paper 61 · Q4 (a, b)8 marksLinear combinations of normals
  • 2023 Oct/Nov · Paper 62 · Q67 marksLinear combinations of normals
  • 2023 Oct/Nov · Paper 63 · Q4 (a, b)8 marksLinear combinations of normals
  • 2023 May/June · Paper 61 · Q5 (a, b)7 marksLinear combinations of normals
  • 2023 May/June · Paper 62 · Q5 (a, b(i), b(ii))11 marksLinear combinations of normals
  • 2023 May/June · Paper 63 · Q46 marksLinear combinations of normals

2022

  • 2022 Oct/Nov · Paper 61 · Q4 (a, b)8 marksLinear combinations of normals
  • 2022 Oct/Nov · Paper 62 · Q6 (a, b)10 marksLinear combinations of normals
  • 2022 Oct/Nov · Paper 63 · Q4 (a, b)8 marksLinear combinations of normals
  • 2022 May/June · Paper 61 · Q35 marksLinear combinations of normals
  • 2022 May/June · Paper 61 · Q4 (a)5 marksLinear combinations of normals
  • 2022 May/June · Paper 62 · Q6 (a, b)10 marksLinear combinations of normals
  • 2022 May/June · Paper 63 · Q4 (a, b)10 marksLinear combinations of normals

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Linear combinations of normals · Core · Paper 6 style · 4 marks
  1. (a) The independent random variables X and Y have distributions N(20, 9) and N(15, 16) respectively. State the distribution of X + Y. [2]
  2. (b) Find P(X + Y > 40). [2]
Show answer
  1. (a) N(35, 25)
    B1 mean 35B1 variance 9 + 16 = 25
  2. (b) 0.159
    M1 z = (40 − 35)/5 = 1A1 1 − Φ(1) = 0.159
Q2Linear combinations of normals · Core · Paper 6 style · 4 marks
  1. (a) The independent random variables X and Y have distributions N(30, 5²) and N(24, 3²). Find P(X < Y). [4]
Show answer
  1. (a) 0.152
    M1 X − Y has mean 6M1 variance 25 + 9 = 34M1 P(X − Y < 0) = Φ(−6/√34) = 1 − Φ(1.029)A1 0.152
Q3Linear combinations of normals · Core · Paper 6 style · 3 marks
  1. (a) The independent random variables X and Y have E(X) = 10, Var(X) = 4, E(Y) = 6 and Var(Y) = 2.5. Find E(3X − 2Y) and Var(3X − 2Y). [3]
Show answer
  1. (a) E = 18, Var = 46
    B1 E = 30 − 12 = 18M1 Var = 9 × 4 + 4 × 2.5A1 46
Q4Linear combinations of normals · Core · Paper 6 style · 4 marks
  1. (a) The mass of an empty glass jar is normally distributed with mean 250 g and standard deviation 8 g. The mass of honey put into a jar is normally distributed with mean 500 g and standard deviation 12 g, independently of the jar. Find the probability that a filled jar has a total mass of more than 770 g. [4]
Show answer
  1. (a) 0.0828
    B1 mean 750B1 variance 64 + 144 = 208M1 z = 20/√208 = 1.387A1 0.0828
Q5Linear combinations of normals · Core · Paper 6 style · 4 marks
  1. (a) The masses of adults using a lift are normally distributed with mean 72 kg and standard deviation 11 kg. The lift's safe load is 600 kg. Eight adults, chosen at random, enter the lift. Find the probability that their total mass exceeds the safe load. [4]
Show answer
  1. (a) 0.220
    B1 mean 8 × 72 = 576B1 variance 8 × 121 = 968M1 z = 24/√968 = 0.771A1 0.220
Q6Linear combinations of normals · Core · Paper 6 style · 6 marks
  1. (a) The random variable X has the distribution N(12, 2²). X₁, X₂, …, X₅ are five independent observations of X. Find P(X₁ + X₂ + X₃ + X₄ + X₅ > 65). [3]
  2. (b) Find P(5X > 65) and explain why this differs from your answer to (a). [3]
Show answer
  1. (a) 0.132
    B1 sum ~ N(60, 20)M1 z = 5/√20 = 1.118A1 0.132
  2. (b) 0.309; 5X has variance 100, not 20
    B1 5X ~ N(60, 100)B1 1 − Φ(0.5) = 0.309B1 5X multiplies one observation, so variance is 25 × 4; the sum of five independent values has variance 5 × 4
Q7Linear combinations of normals · Core · Paper 6 style · 4 marks
  1. (a) A delivery driver's routes are of two types. The time for a type A route is distributed N(35, 4²) minutes and the time for a type B route is distributed N(22, 3²) minutes. On one day the driver completes 2 type A routes and 3 type B routes, all times being independent. Find the probability that the total time for the five routes is more than 140 minutes. [4]
Show answer
  1. (a) 0.301
    B1 mean 2 × 35 + 3 × 22 = 136B1 variance 2 × 16 + 3 × 9 = 59 (not 4 × 16 + 9 × 9)M1 z = 4/√59 = 0.521A1 1 − Φ(0.521) = 0.301
Q8Linear combinations of normals · Core · Paper 6 style · 5 marks
  1. (a) The random variable X has the distribution N(4, 0.5²). X₁, X₂, X₃ and X₄ are independent observations of X, and T = X₁ + X₂ + X₃ − 2X₄. Find the mean and variance of T. [3]
  2. (b) Find P(T < 3). [2]
Show answer
  1. (a) Mean 4, variance 1.75
    B1 mean 12 − 8 = 4M1 variance 3 × 0.25 + 4 × 0.25A1 1.75
  2. (b) 0.225
    M1 z = −1/√1.75 = −0.756A1 0.225
Q9Linear combinations of normals · Core · Paper 6 style · 4 marks
  1. (a) The random variable X has the distribution N(3, 1.5²) and Y = 2X + 5. State the distribution of Y and find P(Y > 15). [4]
Show answer
  1. (a) Y ~ N(11, 9); 0.0912
    B1 mean 11B1 variance 4 × 2.25 = 9M1 z = 4/3A1 0.0912
Q10Linear combinations of normals · Standard · Paper 6 style · 4 marks
  1. (a) Zara's journey to university has three independent stages: walking to the bus stop takes N(12, 2²) minutes, waiting for the bus takes N(8, 3²) minutes and the bus ride takes N(25, 4²) minutes. Find the probability that her total journey time is less than 40 minutes. [4]
Show answer
  1. (a) 0.177
    B1 mean 45B1 variance 4 + 9 + 16 = 29M1 z = −5/√29 = −0.928A1 0.177
Q11Linear combinations of normals · Standard · Paper 6 style · 4 marks
  1. (a) The lifetimes of brand A light bulbs are distributed N(1200, 150²) hours and those of brand B bulbs are distributed N(1100, 100²) hours. One bulb of each brand is chosen at random. Find the probability that the brand A bulb lasts longer than the brand B bulb. [4]
Show answer
  1. (a) 0.710
    M1 A − B with mean 100M1 variance 22 500 + 10 000 = 32 500M1 P(A − B > 0) = Φ(100/√32 500)A1 0.710
Q12Linear combinations of normals · Standard · Paper 6 style · 5 marks
  1. (a) Rods of type A have lengths distributed N(42, 1.5²) cm and rods of type B have lengths distributed N(20, 0.8²) cm. One rod of each type is chosen at random. Find the probability that the type A rod is more than twice as long as the type B rod. [5]
Show answer
  1. (a) 0.819
    M1 consider A − 2BA1 mean 42 − 40 = 2M1 variance 2.25 + 4 × 0.64 = 4.81M1 Φ(2/√4.81) = Φ(0.912)A1 0.819
Q13Linear combinations of normals · Standard · Paper 6 style · 4 marks
  1. (a) The diameters of bolts are normally distributed with mean 9.8 mm and standard deviation 0.05 mm. The diameters of the holes in nuts are normally distributed with mean 10.0 mm and standard deviation 0.08 mm. A bolt fits a nut if the bolt's diameter is less than the hole's diameter. A bolt and a nut are chosen at random. Find the probability that the bolt fits the nut. [4]
Show answer
  1. (a) 0.983
    M1 H − B: mean 0.2M1 variance 0.0025 + 0.0064 = 0.0089M1 Φ(0.2/√0.0089) = Φ(2.120)A1 0.983
Q14Linear combinations of normals · Standard · Paper 6 style · 5 marks
  1. (a) Bags of flour have masses distributed N(2.02, 0.03²) kg. Twelve bags are packed into a box whose mass is distributed N(1.5, 0.1²) kg. All masses are independent. Find the probability that the total mass of the full box is between 25.6 kg and 26.0 kg. [5]
Show answer
  1. (a) 0.798
    B1 mean 24.24 + 1.5 = 25.74B1 variance 12 × 0.0009 + 0.01 = 0.0208M1 standardise both limits: −0.971 and 1.803M1 subtract Φ valuesA1 0.798
Q15Linear combinations of normals · Standard · Paper 6 style · 4 marks
  1. (a) In a 4 × 100 m relay the four runners' times, in seconds, are independent and distributed N(12.1, 0.2²), N(11.8, 0.15²), N(12.3, 0.25²) and N(11.9, 0.18²). Find the probability that the team's total time is less than 47.8 seconds. [4]
Show answer
  1. (a) 0.225
    B1 mean 48.1B1 variance 0.04 + 0.0225 + 0.0625 + 0.0324 = 0.1574M1 z = −0.3/√0.1574 = −0.756A1 0.225
Q16Linear combinations of normals · Standard · Paper 6 style · 5 marks
  1. (a) July rainfall in Lahore is modelled as N(190, 45²) mm and July rainfall in Islamabad as N(270, 60²) mm, independently. Find the probability that in a particular July Islamabad's rainfall is more than 1.5 times Lahore's. [5]
Show answer
  1. (a) 0.434
    M1 consider I − 1.5LA1 mean 270 − 285 = −15M1 variance 3600 + 2.25 × 2025 = 8156.25M1 1 − Φ(15/√8156.25) = 1 − Φ(0.1661)A1 0.434
Q17Linear combinations of normals · Standard · Paper 6 style · 4 marks
  1. (a) The midday temperature in a city in June, in °C, is distributed N(28, 4²). The temperature F in °F is given by F = 1.8C + 32. State the distribution of F. [2]
  2. (b) Find the probability that the midday temperature exceeds 90 °F. [2]
Show answer
  1. (a) N(82.4, 51.84)
    B1 mean 1.8 × 28 + 32 = 82.4B1 variance 1.8² × 16 = 51.84
  2. (b) 0.146
    M1 z = 7.6/7.2 = 1.056A1 0.146
Q18Linear combinations of normals · Standard · Paper 6 style · 5 marks
  1. (a) A household's daily electricity use, in kWh, is distributed N(14, 3²), independently from day to day. Electricity costs Rs 55 per kWh plus a fixed charge of Rs 100 per day. Find the mean and variance of the total cost for a week (7 days). [3]
  2. (b) Find the probability that the weekly cost exceeds Rs 6500. [2]
Show answer
  1. (a) Mean Rs 6090, variance 190 575
    B1 daily cost 55X + 100: mean 870B1 weekly mean 7 × 870 = 6090B1 variance 7 × 55² × 9 = 190 575
  2. (b) 0.174
    M1 z = 410/√190 575 = 0.939A1 0.174
Q19Linear combinations of normals · Standard · Paper 6 style · 4 marks
  1. (a) A machine fills bags labelled 1 kg with sugar. The mass in a bag is distributed N(1.008, 0.006²) kg. A shopkeeper buys 5 bags at random. Find the probability that the total mass of sugar in the 5 bags is less than 5 kg. [4]
Show answer
  1. (a) 0.00143
    B1 mean 5.04B1 variance 5 × 0.000036 = 0.00018M1 z = −0.04/√0.00018 = −2.981A1 0.00143
Q20Linear combinations of normals · Standard · Paper 6 style · 5 marks
  1. (a) In a class, Physics marks are distributed N(58, 12²) and Chemistry marks N(62, 10²). Assuming a student's two marks are independent, find the probability that a randomly chosen student's Physics mark is at least 10 more than their Chemistry mark. [4]
  2. (b) Comment on the independence assumption. [1]
Show answer
  1. (a) 0.185
    M1 P − C: mean −4M1 variance 144 + 100 = 244M1 1 − Φ(14/√244) = 1 − Φ(0.896)A1 0.185
  2. (b) Unlikely to be true: a student good at one science is likely to do well in the other.
    B1 sensible comment that the marks are probably positively related
Q21Linear combinations of normals · Standard · Paper 6 style · 5 marks
  1. (a) A river ferry carries 20 cars and 10 vans. The masses of cars are distributed N(1.4, 0.2²) tonnes and the masses of vans N(2.6, 0.3²) tonnes, all independently. The ferry's maximum safe load is 56 tonnes. Find the probability that the total mass of the vehicles exceeds the maximum safe load. [5]
Show answer
  1. (a) 0.0625
    B1 mean 28 + 26 = 54B1 variance 20 × 0.04 + 10 × 0.09 = 1.7M1 z = 2/√1.7 = 1.534M1 1 − Φ(1.534)A1 0.0625
Q22Linear combinations of normals · Standard · Paper 6 style · 4 marks
  1. (a) Sacks of basmati rice have masses distributed N(25.2, 0.4²) kg. A restaurant buys 4 sacks chosen at random. Find the probability that their total mass is within 0.5 kg of 100.8 kg. [4]
Show answer
  1. (a) 0.468
    B1 total ~ N(100.8, 0.64)M1 P(|Z| < 0.5/0.8) = P(|Z| < 0.625)M1 2Φ(0.625) − 1A1 0.468
Q23Linear combinations of normals · Challenge · Paper 6 style · 5 marks
  1. (a) Boxes have masses distributed N(15, 2²) kg independently. A trolley can safely carry 250 kg. Find the largest number of boxes, n, that can be loaded so that the probability that the total mass exceeds 250 kg is less than 0.05. [5]
Show answer
  1. (a) n = 15
    M1 total ~ N(15n, 4n)M1 (250 − 15n)/(2√n) > 1.645M1 test n = 15: z = 25/√60 = 3.227 > 1.645M1 test n = 16: z = 10/8 = 1.25 < 1.645A1 n = 15
Q24Linear combinations of normals · Challenge · Paper 6 style · 5 marks
  1. (a) The independent random variables X and Y have distributions N(50, 4) and N(49, 5). Find P(|X − Y| < 3). [5]
Show answer
  1. (a) 0.656
    B1 X − Y ~ N(1, 9)M1 P(−3 < X − Y < 3)M1 z values −4/3 and 2/3M1 Φ(0.667) − (1 − Φ(1.333))A1 0.656
Q25Linear combinations of normals · Challenge · Paper 6 style · 5 marks
  1. (a) The masses of men in a town are distributed N(78, 10²) kg and the masses of women N(62, 8²) kg. Three men and four women are chosen at random. Find the probability that the total mass of the three men is greater than the total mass of the four women. [5]
Show answer
  1. (a) 0.276
    M1 consider M₁ + M₂ + M₃ − (W₁ + … + W₄)A1 mean 234 − 248 = −14M1 variance 300 + 256 = 556M1 1 − Φ(14/√556) = 1 − Φ(0.594)A1 0.276
Q26Linear combinations of normals · Challenge · Paper 6 style · 4 marks
  1. (a) The independent random variables X and Y have distributions N(μ, 3²) and N(2μ, 4²). Given that P(X + Y > 60) = 0.0228, find μ. [4]
Show answer
  1. (a) μ = 50/3 (16.7)
    B1 X + Y ~ N(3μ, 25)M1 z = 2M1 (60 − 3μ)/5 = 2A1 μ = 50/3
Q27Linear combinations of normals · Challenge · Paper 6 style · 5 marks
  1. (a) The independent random variables X and Y have distributions N(10, σ²) and N(15, 9). Given that P(2X > Y) = 0.9, find σ. [5]
Show answer
  1. (a) σ = 1.25
    B1 2X − Y has mean 5B1 variance 4σ² + 9M1 5/√(4σ² + 9) = 1.282M1 4σ² + 9 = 15.22A1 σ = 1.25
Q28Linear combinations of normals · Challenge · Paper 6 style · 5 marks
  1. (a) A wall is built from 12 courses of bricks. The height of each course is distributed N(7.5, 0.2²) cm and between consecutive courses there is a layer of mortar whose thickness is distributed N(1.0, 0.15²) cm. All heights are independent. Find the probability that the height of the wall is between 100 cm and 102.5 cm. [5]
Show answer
  1. (a) 0.840
    B1 11 layers of mortar; mean 90 + 11 = 101B1 variance 12 × 0.04 + 11 × 0.0225 = 0.7275M1 z values −1.172 and 1.759M1 subtract Φ valuesA1 0.840
Q29Linear combinations of normals · Challenge · Paper 6 style · 6 marks
  1. (a) In a garment factory, the numbers of shirts finished in an hour by machines A, B and C are modelled by independent normal distributions N(120, 9²), N(95, 7²) and N(30, 5²). Find the probability that in a given hour machine A finishes more shirts than machines B and C together. [5]
  2. (b) Explain why a normal model for these counts is only an approximation. [1]
Show answer
  1. (a) 0.344
    M1 consider A − B − CA1 mean −5M1 variance 81 + 49 + 25 = 155M1 1 − Φ(5/√155) = 1 − Φ(0.402)A1 0.344
  2. (b) The numbers of shirts are discrete (whole numbers), while the normal distribution is continuous.
    B1 discrete versus continuous
Q30Linear combinations of normals · Challenge · Paper 6 style · 6 marks
  1. (a) Floor tiles have lengths distributed N(30, 0.4²) cm. A row of 10 tiles is laid with a fixed gap of 0.3 cm between neighbouring tiles. The row must fit along a wall of length 303.5 cm. Find the probability that a row of 10 randomly chosen tiles, with gaps, fits. [5]
  2. (b) Explain why it would be wrong to use 10X, where X is the length of one tile, for the total length of the tiles. [1]
Show answer
  1. (a) 0.736
    B1 9 gaps: 2.7 cmB1 mean 300 + 2.7 = 302.7B1 variance 10 × 0.16 = 1.6M1 Φ(0.8/√1.6) = Φ(0.632)A1 0.736
  2. (b) 10X assumes all ten tiles have the same length; the tiles vary independently, so the variance is 10 × 0.16, not 100 × 0.16.
    B1 independent tiles: add variances, not multiply one tile

Similar questions

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References only: find each question in its paper (official link where available, otherwise through your school).

Linear combinations of normals Banker30 questions · 9/9 sessions

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