Question Simulator

AS & A Level Mathematics (9709)

Normal distribution Banker

51 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 51 · Q6 (a, b, c)12 marksFinding mean/sd from probabilities · Normal approximation to binomial · Normal probabilities
  • 2026 May/June · Paper 52 · Q4 (b)5 marksNormal approximation to binomial
  • 2026 May/June · Paper 52 · Q7 (a, b)9 marksFinding mean/sd from probabilities · Normal probabilities
  • 2026 May/June · Paper 53 · Q5 (a, b, c)11 marksFinding mean/sd from probabilities · Normal approximation to binomial · Normal probabilities
  • 2026 May/June · Paper 55 · Q4 (b)5 marksNormal approximation to binomial
  • 2026 May/June · Paper 55 · Q6 (a, b, c)9 marksFinding mean/sd from probabilities · Normal probabilities

2025

  • 2025 Oct/Nov · Paper 51 · Q5 (b)5 marksNormal approximation to binomial
  • 2025 Oct/Nov · Paper 51 · Q6 (a, b)9 marksFinding mean/sd from probabilities · Normal probabilities
  • 2025 Oct/Nov · Paper 52 · Q3 (a, b)7 marksFinding mean/sd from probabilities · Normal probabilities
  • 2025 Oct/Nov · Paper 52 · Q6 (c)5 marksNormal approximation to binomial
  • 2025 Oct/Nov · Paper 53 · Q7 (a, b, c)10 marksNormal probabilities
  • 2025 Oct/Nov · Paper 55 · Q6 (a, b, c)11 marksFinding mean/sd from probabilities · Normal approximation to binomial · Normal probabilities
  • 2025 May/June · Paper 51 · Q13 marksFinding mean/sd from probabilities
  • 2025 May/June · Paper 51 · Q4 (a, c)9 marksNormal approximation to binomial · Normal probabilities
  • 2025 May/June · Paper 52 · Q25 marksNormal approximation to binomial
  • 2025 May/June · Paper 52 · Q7 (a, b)9 marksFinding mean/sd from probabilities · Normal probabilities
  • 2025 May/June · Paper 53 · Q6 (a, b, c)10 marksFinding mean/sd from probabilities · Normal probabilities
  • 2025 May/June · Paper 55 · Q2 (a, b)7 marksNormal probabilities
  • 2025 May/June · Paper 55 · Q3 (a)5 marksNormal approximation to binomial

2024

  • 2024 Oct/Nov · Paper 51 · Q5 (a, b)9 marksNormal approximation to binomial · Normal probabilities
  • 2024 Oct/Nov · Paper 51 · Q6 (a)5 marksFinding mean/sd from probabilities
  • 2024 Oct/Nov · Paper 52 · Q4 (a, b)7 marksFinding mean/sd from probabilities · Normal probabilities
  • 2024 Oct/Nov · Paper 52 · Q7 (d)5 marksNormal approximation to binomial
  • 2024 Oct/Nov · Paper 53 · Q1 (b)5 marksNormal approximation to binomial
  • 2024 Oct/Nov · Paper 53 · Q3 (a, b)7 marksNormal probabilities
  • 2024 May/June · Paper 51 · Q2 (a, b)7 marksOfficial paper ↗Finding mean/sd from probabilities · Normal probabilities
  • 2024 May/June · Paper 51 · Q5 (c)5 marksOfficial paper ↗Normal approximation to binomial
  • 2024 May/June · Paper 52 · Q3 (a, b)6 marksNormal probabilities
  • 2024 May/June · Paper 52 · Q6 (a)5 marksNormal approximation to binomial
  • 2024 May/June · Paper 53 · Q2 (a, b)8 marksFinding mean/sd from probabilities · Normal probabilities
  • 2024 May/June · Paper 53 · Q5 (d)5 marksNormal approximation to binomial

2023

  • 2023 Oct/Nov · Paper 51 · Q3 (a, b, c)11 marksNormal approximation to binomial · Normal probabilities
  • 2023 Oct/Nov · Paper 52 · Q35 marksNormal approximation to binomial
  • 2023 Oct/Nov · Paper 52 · Q5 (a(i), a(ii), b)8 marksFinding mean/sd from probabilities · Normal probabilities
  • 2023 Oct/Nov · Paper 53 · Q2 (a, b)6 marksFinding mean/sd from probabilities · Normal probabilities
  • 2023 Oct/Nov · Paper 53 · Q5 (c)5 marksNormal approximation to binomial
  • 2023 May/June · Paper 51 · Q4 (a, b)9 marksFinding mean/sd from probabilities · Normal probabilities
  • 2023 May/June · Paper 51 · Q6 (d)5 marksNormal approximation to binomial
  • 2023 May/June · Paper 52 · Q5 (a, b, c)12 marksFinding mean/sd from probabilities · Normal approximation to binomial · Normal probabilities
  • 2023 May/June · Paper 53 · Q25 marksNormal approximation to binomial
  • 2023 May/June · Paper 53 · Q6 (a, c)7 marksFinding mean/sd from probabilities · Normal probabilities

2022

  • 2022 Oct/Nov · Paper 51 · Q2 (b)5 marksNormal approximation to binomial
  • 2022 Oct/Nov · Paper 51 · Q4 (a, b, c)9 marksFinding mean/sd from probabilities · Normal probabilities
  • 2022 Oct/Nov · Paper 52 · Q2 (a, b)7 marksNormal probabilities
  • 2022 Oct/Nov · Paper 52 · Q6 (b, c)6 marksNormal approximation to binomial
  • 2022 Oct/Nov · Paper 53 · Q25 marksNormal approximation to binomial
  • 2022 Oct/Nov · Paper 53 · Q5 (b, c)6 marksNormal probabilities
  • 2022 May/June · Paper 51 · Q5 (a, b, c)11 marksFinding mean/sd from probabilities · Normal probabilities
  • 2022 May/June · Paper 52 · Q4 (a, b)7 marksFinding mean/sd from probabilities · Normal probabilities
  • 2022 May/June · Paper 52 · Q5 (b)5 marksNormal approximation to binomial
  • 2022 May/June · Paper 53 · Q5 (a, b, c)10 marksNormal probabilities

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.

Q1Normal probabilities · Core · Paper 5 style · 3 marks
  1. (a) The random variable X has the distribution N(50, 4²). Find P(X < 56). [3]
Show answer
  1. (a) 0.933
    M1 standardise: (56 − 50)/4 = 1.5M1 Φ(1.5)A1 0.933
Q2Normal probabilities · Core · Paper 5 style · 3 marks
  1. (a) The masses of Sindhri mangoes from an orchard are normally distributed with mean 320 g and standard deviation 25 g. Find the probability that a randomly chosen mango has a mass greater than 350 g. [3]
Show answer
  1. (a) 0.115
    M1 z = (350 − 320)/25 = 1.2M1 1 − Φ(1.2)A1 0.115
Q3Normal probabilities · Core · Paper 5 style · 4 marks
  1. (a) The time taken by Usman to drive to work is normally distributed with mean 42 minutes and standard deviation 6 minutes. Find the probability that on a randomly chosen day his journey takes between 35 and 50 minutes. [4]
Show answer
  1. (a) 0.787
    M1 standardise both: 1.333 and −1.167M1 Φ(1.333) − Φ(−1.167) = Φ(1.333) + Φ(1.167) − 1A1 0.9088 − 0.1216A1 0.787
Q4Normal probabilities · Core · Paper 5 style · 4 marks
  1. (a) A machine fills bags of rice. The mass of rice in a bag is normally distributed with mean 1005 g and standard deviation 3.5 g. Find the probability that a randomly chosen bag contains less than 1000 g. [3]
  2. (b) In a batch of 500 bags, find the expected number that contain less than 1000 g. [1]
Show answer
  1. (a) 0.0766
    M1 z = (1000 − 1005)/3.5 = −1.429M1 1 − Φ(1.429)A1 0.0766
  2. (b) 38.3 (about 38)
    B1 500 × 0.0766 = 38.3
Q5Normal probabilities · Standard · Paper 5 style · 4 marks
  1. (a) The heights of 300 students at a college in Peshawar are modelled by a normal distribution with mean 165 cm and standard deviation 8 cm. Estimate the number of students who are taller than 175 cm. [4]
Show answer
  1. (a) 32 students
    M1 z = 1.25M1 1 − Φ(1.25) = 0.106M1 300 × 0.106 = 31.7A1 32
Q6Normal probabilities · Standard · Paper 5 style · 3 marks
  1. (a) The lifetimes of a type of battery are normally distributed with mean 480 hours and standard deviation 35 hours. Find the lifetime that is exceeded by 90% of batteries. [3]
Show answer
  1. (a) 435.1 hours
    M1 Φ⁻¹(0.9) = 1.282M1 (t − 480)/35 = −1.282A1 t = 435.1 hours
Q7Normal probabilities · Standard · Paper 5 style · 3 marks
  1. (a) The random variable X is normally distributed with mean μ and standard deviation σ. Show that P(μ − 1.5σ < X < μ + 1.5σ) does not depend on μ or σ, and find its value. [3]
Show answer
  1. (a) 0.866
    M1 standardise: P(−1.5 < Z < 1.5)M1 2Φ(1.5) − 1A1 0.866
Q8Normal probabilities · Standard · Paper 5 style · 3 marks
  1. (a) The speeds of deliveries bowled by a fast bowler are normally distributed with mean 132 km/h and standard deviation 5.5 km/h. Find the value of k such that 20% of his deliveries are faster than k km/h. [3]
Show answer
  1. (a) k = 136.6
    M1 z = 0.842M1 (k − 132)/5.5 = 0.842A1 k = 136.6
Q9Normal probabilities · Standard · Paper 5 style · 5 marks
  1. (a) At a petrol station the waiting time of a customer is normally distributed with mean 6.5 minutes and standard deviation 2 minutes. Find the probability that a randomly chosen customer waits more than 9 minutes. [3]
  2. (b) Three customers are chosen at random. Find the probability that at least one of them waits more than 9 minutes. [2]
Show answer
  1. (a) 0.106
    M1 z = 1.25M1 1 − Φ(1.25)A1 0.106
  2. (b) 0.285
    M1 1 − (1 − 0.106)³A1 0.285
Q10Normal probabilities · Challenge · Paper 5 style · 7 marks
  1. (a) The lengths of bolts made by a machine are normally distributed with mean 40 mm and standard deviation 0.6 mm. A bolt is acceptable if its length is between 39 mm and 41.2 mm. Find the probability that a randomly chosen bolt is acceptable. Give your answer correct to 4 significant figures. [4]
  2. (b) Bolts are packed in boxes of 10. Find the probability that a box contains at most one bolt that is not acceptable. [3]
Show answer
  1. (a) 0.9295
    M1 z values −1.667 and 2M1 Φ(2) − Φ(−1.667) = Φ(2) + Φ(1.667) − 1A1 0.9772 + 0.9522 − 1A1 0.9295
  2. (b) 0.846 (accept 0.847 from p = 0.9295)
    M1 binomial with p = 0.9295M1 p¹⁰ + 10p⁹(1 − p)A1 0.846 (0.847 if the rounded p = 0.9295 is used)
Q11Normal probabilities · Challenge · Paper 5 style · 4 marks
  1. (a) The scores in an aptitude test are normally distributed with mean 70 and standard deviation 10. A candidate is known to have scored more than 75. Find the probability that the candidate scored more than 85. [4]
Show answer
  1. (a) 0.217
    M1 P(X > 85 | X > 75) = P(X > 85)/P(X > 75)M1 1 − Φ(1.5) = 0.0668M1 1 − Φ(0.5) = 0.3085A1 0.217
Q12Normal probabilities · Challenge · Paper 5 style · 6 marks
  1. (a) A fruit seller has a large crate of apples, 60% of variety A and 40% of variety B. The masses of variety A apples are distributed N(150, 12²) grams and those of variety B are distributed N(165, 15²) grams. An apple is chosen at random. Find the probability that its mass is more than 160 g. [4]
  2. (b) Given that the apple has a mass of more than 160 g, find the probability that it is variety A. [2]
Show answer
  1. (a) 0.374
    M1 P(A > 160) = 1 − Φ(0.833) = 0.2023M1 P(B > 160) = Φ(0.333) = 0.6306M1 0.6 × 0.2023 + 0.4 × 0.6306A1 0.374
  2. (b) 0.325
    M1 0.6 × 0.2023 ÷ 0.3736A1 0.325
Q13Normal approximation to binomial · Core · Paper 5 style · 5 marks
  1. (a) The random variable X has the distribution B(80, 0.3). Use a suitable approximation to find P(X ≤ 20). [5]
Show answer
  1. (a) 0.197
    B1 mean 24, variance 16.8M1 continuity correction 20.5M1 standardise (20.5 − 24)/√16.8 = −0.854M1 1 − Φ(0.854)A1 0.197
Q14Normal approximation to binomial · Core · Paper 5 style · 5 marks
  1. (a) It is known that 15% of the population of a town are left-handed. A random sample of 120 people is taken. Use a suitable approximation to find the probability that more than 22 of them are left-handed. [5]
Show answer
  1. (a) 0.125
    B1 mean 18, variance 15.3M1 P(X ≥ 23) → continuity correction 22.5M1 z = (22.5 − 18)/√15.3 = 1.150M1 1 − Φ(1.150)A1 0.125
Q15Normal approximation to binomial · Core · Paper 5 style · 5 marks
  1. (a) The probability that a seed of a certain variety of wheat germinates is 0.85. A packet contains 200 seeds. Use a suitable approximation to find the probability that at least 165 of the seeds germinate. [5]
Show answer
  1. (a) 0.862
    B1 mean 170, variance 25.5M1 continuity correction 164.5M1 z = (164.5 − 170)/√25.5 = −1.089M1 Φ(1.089)A1 0.862
Q16Normal approximation to binomial · Standard · Paper 5 style · 6 marks
  1. (a) A survey suggests that 60% of commuters in Karachi travel to work by bus. A random sample of 50 commuters is taken. Show that a normal approximation is appropriate. [1]
  2. (b) Use the normal approximation to find the probability that between 25 and 35 (inclusive) of the commuters travel by bus. [5]
Show answer
  1. (a) np = 30 > 5 and nq = 20 > 5
    B1 both np and nq calculated and compared with 5
  2. (b) 0.888
    B1 mean 30, variance 12M1 continuity corrections 24.5 and 35.5M1 z = ±1.588M1 2Φ(1.588) − 1A1 0.888
Q17Normal approximation to binomial · Standard · Paper 5 style · 5 marks
  1. (a) A fair six-sided die is thrown 300 times. Use a suitable approximation to find the probability that the number of sixes obtained is between 45 and 55 inclusive. [5]
Show answer
  1. (a) 0.606
    B1 mean 50, variance 250/6 = 41.67M1 continuity corrections 44.5 and 55.5M1 z = ±0.852M1 2Φ(0.852) − 1A1 0.606
Q18Normal approximation to binomial · Standard · Paper 5 style · 5 marks
  1. (a) The random variable X has the distribution B(90, 1/3). Use a normal approximation to estimate P(X = 30). [4]
  2. (b) The exact value of P(X = 30) is 0.0889. Comment on the accuracy of the approximation. [1]
Show answer
  1. (a) 0.0890
    B1 mean 30, variance 20M1 P(29.5 < Y < 30.5)M1 2Φ(0.1118) − 1A1 0.0890
  2. (b) Very close (differs only in the 4th decimal place), so the approximation is good.
    B1 sensible comparison, e.g. close because np and nq are large
Q19Normal approximation to binomial · Standard · Paper 5 style · 5 marks
  1. (a) At an airport, 8% of flights depart late. In a particular week 250 flights are scheduled. Use a suitable approximation to find the probability that fewer than 15 flights depart late. [5]
Show answer
  1. (a) 0.0999
    B1 mean 20, variance 18.4M1 P(X ≤ 14), continuity correction 14.5M1 z = (14.5 − 20)/√18.4 = −1.282M1 1 − Φ(1.282)A1 0.0999
Q20Normal approximation to binomial · Standard · Paper 5 style · 6 marks
  1. (a) 5% of the light bulbs made in a factory are faulty. Explain why a normal approximation is not suitable for the number of faulty bulbs in a sample of 40. [1]
  2. (b) A sample of 200 bulbs is taken. Use a normal approximation to find the probability that more than 12 bulbs are faulty. [5]
Show answer
  1. (a) np = 2, which is less than 5
    B1 np = 40 × 0.05 = 2 < 5
  2. (b) 0.209
    B1 mean 10, variance 9.5M1 P(X ≥ 13), continuity correction 12.5M1 z = 2.5/√9.5 = 0.811M1 1 − Φ(0.811)A1 0.209
Q21Normal approximation to binomial · Challenge · Paper 5 style · 5 marks
  1. (a) In a large school 70% of students pass a mathematics test. A group of 150 students is chosen at random. Use a suitable approximation to find the probability that more than 100 but fewer than 115 of them pass. [5]
Show answer
  1. (a) 0.743
    B1 mean 105, variance 31.5M1 P(101 ≤ X ≤ 114): use 100.5 and 114.5M1 z = −0.802 and 1.693M1 Φ(1.693) − Φ(−0.802)A1 0.743
Q22Normal approximation to binomial · Challenge · Paper 5 style · 7 marks
  1. (a) The daily demand for naan at a bakery in Rawalpindi is normally distributed with mean 380 and standard deviation 40. The bakery makes 420 naan each day. Find the probability that on a randomly chosen day the demand is at least 420, so that the bakery sells out. Give your answer to 4 decimal places. [2]
  2. (b) Use a suitable approximation to find the probability that the bakery sells out on at least 40 days out of 200. [5]
Show answer
  1. (a) 0.1587
    M1 z = (420 − 380)/40 = 1A1 1 − Φ(1) = 0.1587
  2. (b) 0.0666 (accept 0.0663 from unrounded p)
    B1 X ~ B(200, 0.1587): mean 31.74, variance 26.70M1 continuity correction 39.5M1 z = (39.5 − 31.74)/5.167 = 1.502M1 1 − Φ(1.502)A1 0.0666 (0.0663 with unrounded p = 0.15866)
Q23Normal approximation to binomial · Challenge · Paper 5 style · 5 marks
  1. (a) The random variable X has the distribution B(400, 0.25). Using a normal approximation, find the smallest integer k such that P(X ≤ k) > 0.9. [5]
Show answer
  1. (a) k = 111
    B1 mean 100, variance 75M1 P(X ≤ k) ≈ Φ((k + 0.5 − 100)/√75)M1 (k + 0.5 − 100)/√75 > 1.282A1 k > 110.60A1 k = 111
Q24Finding mean/sd from probabilities · Core · Paper 5 style · 3 marks
  1. (a) The random variable X has the distribution N(μ, 5²). Given that P(X < 62) = 0.8, find μ. [3]
Show answer
  1. (a) μ = 57.79
    M1 z = 0.842M1 (62 − μ)/5 = 0.842A1 μ = 57.79
Q25Finding mean/sd from probabilities · Core · Paper 5 style · 3 marks
  1. (a) The random variable X has the distribution N(30, σ²). Given that P(X > 36) = 0.1, find σ. [3]
Show answer
  1. (a) σ = 4.68
    M1 z = 1.282M1 6/σ = 1.282A1 σ = 4.68
Q26Finding mean/sd from probabilities · Standard · Paper 5 style · 5 marks
  1. (a) The lengths of fish of a certain species caught in Keenjhar Lake are normally distributed. 15% of the fish are longer than 48 cm and 25% are shorter than 36 cm. Find the mean and standard deviation of the lengths. [5]
Show answer
  1. (a) μ = 40.73 cm, σ = 7.01 cm
    B1 z = 1.036 and B1 z = −0.674M1 μ + 1.036σ = 48, μ − 0.674σ = 36A1 σ = 7.01A1 μ = 40.73
Q27Finding mean/sd from probabilities · Standard · Paper 5 style · 4 marks
  1. (a) The masses of watermelons sold at a market are normally distributed. The mean mass is four times the standard deviation, and 6.68% of the watermelons have a mass less than 2.5 kg. Find the mean and standard deviation. [4]
Show answer
  1. (a) μ = 4 kg, σ = 1 kg
    B1 z = −1.5M1 (2.5 − 4σ)/σ = −1.5A1 σ = 1A1 μ = 4
Q28Finding mean/sd from probabilities · Standard · Paper 5 style · 5 marks
  1. (a) The times taken by candidates to complete an online test are normally distributed with standard deviation 8 minutes. 20% of candidates take more than 75 minutes. Find the mean time. [3]
  2. (b) Find the percentage of candidates who take less than 60 minutes. [2]
Show answer
  1. (a) 68.27 minutes
    M1 z = 0.842M1 (75 − μ)/8 = 0.842A1 μ = 68.27
  2. (b) 15.1%
    M1 Φ((60 − 68.27)/8)A1 15.1%
Q29Finding mean/sd from probabilities · Challenge · Paper 5 style · 8 marks
  1. (a) Marks in an examination are normally distributed. The top 10% of candidates, who scored 78 or more, receive a distinction, and the bottom 20%, who scored 45 or less, fail. Find the mean and standard deviation of the marks. [5]
  2. (b) Find the proportion of candidates scoring between 50 and 70. [3]
Show answer
  1. (a) μ = 58.08, σ = 15.5
    B1 z = 1.282, B1 z = −0.842M1 μ + 1.282σ = 78, μ − 0.842σ = 45A1 σ = 15.5A1 μ = 58.08
  2. (b) 0.477
    M1 standardise both with the values from (a)M1 Φ(0.767) − Φ(−0.520)A1 0.477
Q30Finding mean/sd from probabilities · Challenge · Paper 5 style · 6 marks
  1. (a) The random variable X is normally distributed with P(X < 20) = 0.12 and P(X > 44) = 0.12. Find the mean and standard deviation of X. [4]
  2. (b) Find P(X > 40). [2]
Show answer
  1. (a) μ = 32, σ = 10.2
    B1 μ = 32 by symmetryM1 z = 1.175M1 12/σ = 1.175A1 σ = 10.2
  2. (b) 0.217
    M1 1 − Φ(8/10.2)A1 0.217

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

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

Normal probabilities Banker30 questions · 9/9 sessions
Normal approximation to binomial Banker26 questions · 9/9 sessions
Finding mean/sd from probabilities Banker22 questions · 9/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Normal probabilities Banker1479/9
Normal approximation to binomial Banker1319/9
Finding mean/sd from probabilities Banker869/9

← All Mathematics topics