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Contents · 0% complete
  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn9709Paper 5 · Probability & Statistics 1The normal distributionStandardising and normal probabilities
The normal distribution · Subtopic 1 of 3

Standardising and Normal Probabilities

Heights, masses and exam marks often follow a symmetric bell-shaped curve. Every normal distribution is the standard one, Z∼N(0,1)Z\sim N(0,1), shifted and stretched, so one table of Φ(z)\Phi(z) answers every question.

Cambridge 9709 Paper 5Edexcel YMA01 WST01Edexcel 9MA0 Paper 3AQA 7357 Paper 3≈ 35 minRevision 6 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Describe the shape and properties of the normal distribution N(μ,σ2)N(\mu,\sigma^2).
  • Standardise with z=x−μσz=\dfrac{x-\mu}{\sigma} and use Φ(z)=P(Z<z)\Phi(z)=P(Z<z).
  • Use symmetry: Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z).
  • Find probabilities for ranges, and expected numbers in a sample.
Pause and think

Adult heights have mean 170170 cm and standard deviation 88 cm. Which is more unusual: a height of 186186 cm, or an IQ of 130130 (mean 100100, standard deviation 1515)?

Section 2 of 14

Core idea

Start hereThe bell curve everywhere

Heights, exam marks, measurement errors, weights of apples: plot enough of them and the same bell shape appears. It's the normal distribution, and one table of values covers every normal curve, once you measure distances in standard deviations.

The big idea

X∼N(μ,σ2)X\sim N(\mu,\sigma^2) is symmetric about μ\mu with total area 11. Z=X−μσ∼N(0,1)Z=\frac{X-\mu}\sigma\sim N(0,1); Φ(z)=P(Z<z)\Phi(z)=P(Z<z) from tables. Use symmetry: Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z).

1. The shape

About 68% within 1 s.d., 95% within 2 s.d.
−3−2−11230.10.20.30.40z68%95% between −2 and 2
A rule of thumb: almost everything (99.7%) lies within 3 standard deviations of the mean.
X∼N(μ,σ2)X\sim N(\mu,\sigma^2)
where
μ\mu
mean (centre).
σ2\sigma^2
variance; the SD is σ\sigma. The second number is the variance.
  • About 68% within 1σ1\sigma, 95% within 2σ2\sigma, 99.7% within 3σ3\sigma.
  • Continuous: P(X=a)=0P(X=a)=0, so << and ≤\le are the same.

2. Standardising

Why Z=X−μσZ=\frac{X-\mu}\sigma is standard normal
  1. E(X−μ)=0E(X-\mu)=0
    Shifting by μ\mu centres it at 00.
  2. SD⁡(X−μσ)=σσ=1\operatorname{SD}\left(\tfrac{X-\mu}\sigma\right)=\tfrac\sigma\sigma=1
    Dividing scales the SD.
zz counts how many standard deviations xx is from the mean.
P(Z < z) is the area to the left of z
−3−2−11230.10.20.30.40zΦ(z)
The total area is 11. Φ(0)=0.5\Phi(0)=0.5 (half the area is left of the middle), Φ(1)=0.841\Phi(1)=0.841, Φ(2)=0.977\Phi(2)=0.977.
Worked exampleX∼N(50,16)X\sim N(50,16)

Find P(X<56)P(X<56) and P(46<X<56)P(46<X<56).

Goal
Two probabilities.
Strategy
Standardise; sketch; symmetry.
  1. σ=4; z=56−504=1.5\sigma=4;\ z=\tfrac{56-50}4=1.5
    Square root of 1616.
  2. P(X<56)=Φ(1.5)=0.9332P(X<56)=\Phi(1.5)=0.9332
    Table.
  3. z=46−504=−1; Φ(−1)=1−0.8413=0.1587z=\tfrac{46-50}4=-1;\ \Phi(-1)=1-0.8413=0.1587
    Symmetry.
  4. P(46<X<56)=0.9332−0.1587=0.7745P(46<X<56)=0.9332-0.1587=0.7745
    Subtract.
Answer0.93320.9332; 0.77450.7745.
✓ CheckBetween −1σ-1\sigma and +1.5σ+1.5\sigma: more than 68% ✓.
Common misconceptionVariance vs SD N(50,16)N(50,16) has SD 44, not 1616.
Pause and think

X∼N(100,225)X\sim N(100,225). P(X>130)P(X>130)?

Summary

  • z=(x−μ)/σz=(x-\mu)/\sigma.
  • Sketch; use symmetry.
Key words
Standard normal
N(0,1)N(0,1).
HistoryDe Moivre and Gauss. Abraham de Moivre found the bell curve in 1733 as an approximation to the binomial for large nn. Carl Friedrich Gauss used it in 1809 to describe errors in astronomical measurements, which is why it is often called the Gaussian distribution. Gauss and his curve appeared on the German 10-mark banknote until 2001.
Fun factThe 68–95–99.7 rule. Almost everything lies within 33 standard deviations of the mean: only about 33 in 10001000 values fall outside. The phrase "six sigma", used in manufacturing quality programmes, comes from aiming for defects so rare they sit 66 standard deviations away.
Section 3 of 14

Key formulae

StandardisingZ=X−μσ∼N(0,1)Z=\frac{X-\mu}{\sigma}\sim N(0,1)
Table valuesΦ(z)=P(Z<z),P(Z>z)=1−Φ(z)\Phi(z)=P(Z<z),\qquad P(Z>z)=1-\Phi(z)
SymmetryΦ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z)
A rangeP(a<X<b)=Φ ⁣(b−μσ)−Φ ⁣(a−μσ)P(a<X<b)=\Phi\!\left(\tfrac{b-\mu}{\sigma}\right)-\Phi\!\left(\tfrac{a-\mu}{\sigma}\right)
TrickSketch and shade every time. Draw a quick bell curve, mark the mean, mark your value, and shade the region you want. You will see at once whether the answer should be more or less than 0.50.5, which catches most sign and "1−1-" mistakes.
Section 4 of 14

Worked examples

Example 1 · Left tail0 / 2 steps
X∼N(20,25)X\sim N(20,25). Find P(X<27)P(X<27).
    Example 2 · A middle range0 / 3 steps
    Heights are N(170,64)N(170,64) in cm. Find the probability that a person's height is between 160160 and 180180 cm.
      Example 3 · Right tail in context0 / 2 steps
      Battery lifetimes are N(400,502)N(400,50^2) hours. Find the probability that a battery lasts more than 480480 hours.
        Section 5 of 14

        Explore the graph

        Area under the normal curve
        Loading the interactive graph…
        Try this
        1. The curve is N(m,s2)N(m,s^2). The shaded area is P(X<a)P(X<a).
        2. With m=50m=50, s=4s=4, a=56a=56, the shaded area is Φ(1.5)≈0.933\Phi(1.5)\approx0.933. Does it look like about 93%93\% of the area?
        3. Increase ss. The curve gets flatter and wider. Why must the total area stay the same?

        Probability is area; standardising measures aa in standard deviations from the mean.

        In the real worldSizing clothes. Clothing manufacturers model body measurements as normal. If chest sizes are N(98,64)N(98,64) cm, a size range covering 9090–106106 cm fits about 68%68\% of people. Extra sizes are added at the tails until the shop covers, say, 95%95\% of customers.
        Section 6 of 14

        Common mistakes

        IncorrectN(50,16)N(50,16): dividing by 1616
        CorrectDivide by σ=4\sigma=4
        The second parameter is the variance.
        IncorrectP(X>a)=Φ(z)P(X>a)=\Phi(z)
        Correct1−Φ(z)1-\Phi(z)
        Tables give the area to the left.
        IncorrectΦ(−1.25)\Phi(-1.25) looked up as Φ(1.25)\Phi(1.25)
        Correct1−Φ(1.25)1-\Phi(1.25)
        Negative zz is below the mean: less than 0.50.5.
        IncorrectWorrying about << versus ≤\le
        CorrectThey are the same here
        For a continuous variable P(X=a)=0P(X=a)=0.

        Every normal distribution, no matter its μ\mu and σ\sigma

        Z=X−μσZ=\dfrac{X-\mu}{\sigma}: "you're all the same bell to me"

        Section 7 of 14

        Exam tip

        Exam tipWrite the standardisation. Show P(X<56)=P ⁣(Z<56−504)=P(Z<1.5)P(X<56)=P\!\left(Z<\dfrac{56-50}{4}\right)=P(Z<1.5) before the table value. That line earns the M1 for standardising, even if a table look-up goes wrong. Keep zz to at least 33 decimal places when it is not exact.
        Section 8 of 14

        Guided practice

        SA-9709-P5-ND-G012 marksL1 FoundationCalculatorStandardising
        X∼N(50,16)X\sim N(50,16). Find P(X<56)P(X<56), correct to 3 significant figures.

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        Salam Academy Original Practice · Proposed mark scheme
        Section 9 of 14

        Practice questions

        SA-9709-P5-ND1-0012 marksL1 FoundationCalculatorStandardising
        X∼N(20,25)X\sim N(20,25). Find P(X<27)P(X<27), correct to 3 significant figures.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-ND-0012 marksL1 FoundationCalculatorStandardising
        X∼N(50,16)X\sim N(50,16). Find P(X>45)P(X>45), correct to 3 significant figures.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-ND-0131 markL1 FoundationCalculatorStandardising
        Z∼N(0,1)Z\sim N(0,1). Find P(Z<−0.5)P(Z<-0.5), correct to 3 significant figures.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-ND1-0022 marksL1 FoundationCalculatorStandardising
        X∼N(60,9)X\sim N(60,9). Find P(X>55)P(X>55), correct to 3 significant figures.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-ND-0023 marksL2 StandardCalculatorStandardising
        X∼N(50,16)X\sim N(50,16). Find P(47<X<53)P(47<X<53), correct to 3 significant figures.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-ND1-0033 marksL2 StandardCalculatorNormal in context
        Adult heights are normally distributed with mean 170170 cm and standard deviation 88 cm. Find the probability that a randomly chosen adult is between 160160 cm and 180180 cm tall, correct to 3 significant figures.

        Sign in or create a free account to check answers, get hints and track your progress.

        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-ND-0074 marksL2 StandardCalculatorNormal in context
        The masses of bags of flour are normally distributed with mean 250250 g and standard deviation 88 g. Find the probability that a bag weighs less than 240240 g, and the expected number of such bags in a batch of 500500, each correct to 3 significant figures.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-ND1-0042 marksL2 StandardCalculatorNormal in context
        Battery lifetimes are normally distributed with mean 400400 hours and standard deviation 5050 hours. Find the probability that a battery lasts more than 480480 hours, correct to 3 significant figures.

        Sign in or create a free account to check answers, get hints and track your progress.

        Salam Academy Original Practice · Proposed mark scheme

        More questions on every skill in this subtopic are in the question bank under Practise.

        Section 10 of 14

        Challenge

        SA-9709-P5-ND-0143 marksL4 ChallengingCalculatorNormal in context
        Using the flour bags of mean 250250 g and standard deviation 88 g, three bags are chosen at random. Find the probability that exactly two of them weigh less than 240240 g, correct to 3 significant figures.

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        Salam Academy Original Practice · Proposed mark scheme
        Section 11 of 14

        Summary

        • N(μ,σ2)N(\mu,\sigma^2): symmetric bell curve; the second number is the variance.
        • Standardise: z=x−μσz=\dfrac{x-\mu}{\sigma}; tables give Φ(z)=P(Z<z)\Phi(z)=P(Z<z).
        • Right tail 1−Φ(z)1-\Phi(z); negative zz: Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z).
        • Always sketch and shade.
        Section 12 of 14

        Quick quiz

        Q1 · SA-9709-P5-ND-Q011 markL1 FoundationCalculatorStandardising
        For a normal distribution, the mean, median and mode are
        Choose one option

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        Salam Academy Original Practice · Proposed mark scheme
        Q2 · SA-9709-P5-ND-Q021 markL1 FoundationCalculatorStandardising
        For X∼N(μ,σ2)X\sim N(\mu,\sigma^2), the standardised value of xx is
        Choose one option

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        Salam Academy Original Practice · Proposed mark scheme
        Q3 · SA-9709-P5-ND-0111 markL1 FoundationCalculatorStandardising
        Z∼N(0,1)Z\sim N(0,1). Find P(−1.96<Z<1.96)P(-1.96<Z<1.96), correct to 2 decimal places.

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        Salam Academy Original Practice · Proposed mark scheme
        Section 13 of 14

        Same topic, different board

        Specification
        9709 AS & A Level Mathematics
        Where
        Paper 5, Probability & Statistics 1 · 5.5 The normal distribution
        Expected
        Understand the use of a normal distribution to model a continuous random variable, and use normal distribution tables; solve problems concerning a variable X where X ~ N(μ, σ²).
        Calculator
        Permitted. Give non-exact answers to 3 significant figures; probabilities may be fractions or decimals.
        Typical style
        A context such as masses or times: find the probability of a range, then the expected number in a sample, or combine with the binomial.
        Open in
        9709 · Paper 5
        Section 14 of 14

        Past paper questions

        13 questions (120 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

        1. 9709/52 · Oct/Nov 2025Question 37 marks

          Normal distribution: expected numbers in a range and a standard deviation from a percentage.

          Question paper, page 4Mark scheme
        2. 9709/52 · May/June 2025Question 712 marks

          Normal probabilities, finding the mean and standard deviation from percentages, and a binomial probability.

          Question paper, page 10Mark scheme
        3. WST01/01 · June 2025Question 614 marks

          Normal distribution, a binomial probability, and σ from a percentage.

          Question paper, page 16Mark scheme
        4. 9709/52 · Feb/March 2025Question 59 marks

          Normal probabilities, a standard deviation from a percentage, and the first day an event happens.

          Question paper, page 10Mark scheme
        5. WST01/01 · January 2025Question 513 marks

          Normal probabilities, and a mean and standard deviation from two conditions.

          Question paper, page 14Mark scheme
        6. WST01/01 · January 2025Question 78 marks

          Read a box plot, interpolation and outliers, then a normal model.

          Question paper, page 22Mark scheme
        Show 7 more
        1. 9709/52 · Oct/Nov 2024Question 47 marks

          Normal distribution: expected count below a value and a standard deviation from a tail.

          Question paper, page 6Mark scheme
        2. 9709/52 · May/June 2024Question 36 marks

          Normal distribution: a percentage above a value and a boundary value.

          Question paper, page 4Mark scheme
        3. WST01/01 · June 2024Question 510 marks

          Normal distribution: standardising and inverse normal.

          Question paper, page 18Mark scheme
        4. 9709/52 · Oct/Nov 2023Question 58 marks

          Normal probabilities and a value from a tail probability.

          Question paper, page 6Mark scheme
        5. 9709/52 · May/June 2023Question 512 marks

          Normal probabilities, an unknown standard deviation and the normal approximation.

          Question paper, page 10Mark scheme
        6. 9709/52 · Oct/Nov 2022Question 27 marks

          Normal distribution: expected count and probability within half a standard deviation.

          Question paper, page 4Mark scheme
        7. 9709/52 · May/June 2022Question 47 marks

          Normal probabilities and an unknown standard deviation.

          Question paper, page 6Mark scheme

        All 9709 past paper questions by topicAll YMA01 past paper questions by topic

        Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9709, 2026–2027 (official specification).