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3.1StatisticsCore

The Normal Distribution

A normal model describes a continuous variable whose values cluster symmetrically about a mean. The diagram, notation and numerical examples below show how the mean and variance determine the shape, how area represents probability, and why a model must be checked against the situation rather than chosen just because a mean is known.

30 min The Normal Distribution

Before you start

  • Distinguish a mean, a variance and a standard deviation.
  • Interpret a probability as an area or proportion, between 0 and 1.
Edexcel A level Maths: 3.1 The Normal DistributionCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Read normal-distribution notation without confusing variance and standard deviation.
  • Relate location and spread to the density curve.
  • Interpret central and tail areas using symmetry.
  • Recognise when a continuous normal model is a poor fit.
1

Read the two parameters carefully

Write X∼N(μ,σ2)X\sim N(\mu,\sigma^2) for a normal model with mean μ\mu and variance σ2\sigma^2. The standard deviation is the positive square root of the variance. In X∼N(50,64)X\sim N(50,64) the mean is 50, the variance is 64 and the standard deviation is 8, not 64.

The mean locates the centre of the curve. The standard deviation controls its horizontal spread. If the variable is measured in grams, its mean and standard deviation are in grams, but its variance is in square grams. The notation describes a model for values; it does not assert that every measured value equals the mean.

X∼N(μ,σ2),σ=Var⁡(X)X\sim N(\mu,\sigma^2),\qquad \sigma=\sqrt{\operatorname{Var}(X)}
The second parameter in this convention is variance. Check the input labels when entering a normal calculation on a calculator.
2

Use the curve as a picture of probability

The curve is symmetric and has total area 1. The mean, median and mode coincide at its centre. Probabilities correspond to areas over intervals, not the height of the curve at one value. A taller density curve can describe a smaller spread without representing a total probability greater than 1.

For the continuous model, the probability of one exact value is zero. This does not mean a rounded reading can never equal that value: a reading recorded to the nearest unit corresponds to a small interval of underlying values.

Normal density with mean 50 and standard deviation 8. The interval from 34 to 66, two standard deviations either side of the mean, is shaded. Ticks also show 42, 50 and 58.
Scroll the diagram to see the full view.
For X∼N(50,64)X\sim N(50,64), the shaded area is P(34<X<66)≈0.9545P(34<X<66)\approx0.9545. The vertical axis is probability density; probability comes from an area, not an ordinate.
3

Estimate central areas and split the tails

Approximately 68.27% of a normal distribution lies within one standard deviation of the mean, 95.45% within two and 99.73% within three. The familiar 68-95-99.7 rule is rounded. Use it for interpretation and rough checks, rather than treating 95% as an exact calculator answer.

Symmetry divides the area outside a central interval equally between the two tails. Around 4.55% lies outside μ±2σ\mu\pm2\sigma, so approximately 2.275% lies above μ+2σ\mu+2\sigma and the same proportion below μ−2σ\mu-2\sigma.

Example 1Packet masses are modelled by X∼N(500,36)X\sim N(500,36), in grams. Find the mean and standard deviation, then an interval containing about 95% of modelled masses.
1The mean is 500 g500\,\mathrm{g} and the standard deviation is 36=6 g\sqrt{36}=6\,\mathrm{g}.
2Two standard deviations either side gives 500±2(6)500\pm2(6), or 488488 to 512 g512\,\mathrm{g}.
3This interval actually has area approximately 0.9545 under the normal model; calling it about 95% is an approximation.
Answer
Mean 500 g500\,\mathrm{g}, standard deviation 6 g6\,\mathrm{g}; approximately 95.45% between 488 g and 512 g.
Example 2For X∼N(50,64)X\sim N(50,64), estimate P(X>66)P(X>66) using the central area shown in the diagram.
1Here σ=8\sigma=8, and 66=50+2(8)66=50+2(8).
2The area outside the central interval is approximately 1−0.9545=0.04551-0.9545=0.0455.
3Divide equally between the two tails: P(X>66)≈0.0455/2=0.02275P(X>66)\approx0.0455/2=0.02275.
Answer
Approximately 0.022750.02275, or 2.275%. Do not use the full 4.55% outside the interval for just the upper tail.
4

Connect to a standard normal calculation

The transformation Z=(X−μ)/σZ=(X-\mu)/\sigma measures how many standard deviations a value lies above or below the mean. It produces a standard normal model with mean 0 and variance 1. Divide by the standard deviation, not the variance.

A calculator or table supplies the relevant areas. Always sketch the required tail or interval first: a correct numerical command for the wrong region still gives the wrong probability. The next lesson develops these calculations in more detail.

Z=X−μσ∼N(0,1)Z=\frac{X-\mu}{\sigma}\sim N(0,1)
Subtract the mean, then divide by the standard deviation.
Example 1For X∼N(50,64)X\sim N(50,64), express P(X>62)P(X>62) as a standard normal probability.
1The boundary becomes z=(62−50)/8=1.5z=(62-50)/8=1.5.
2Because the transformation increases with X, the upper-tail direction stays the same: P(X>62)=P(Z>1.5)P(X>62)=P(Z>1.5).
3A normal cumulative-distribution calculation gives approximately 0.06680.0668. This is smaller than 0.5, as the boundary lies above the mean.
Answer
P(Z>1.5)≈0.0668P(Z>1.5)\approx0.0668. Using 64 as the divisor would misrepresent the spread.
5

Decide whether the model is useful

Symmetry and a roughly single-peaked shape can support a normal model, but they are not guarantees. Strong skew, outliers, a mixture of separate populations or values close to a physical boundary can make it unsuitable. A normal model assigns some probability to every real number, which may be unrealistic for a strictly nonnegative quantity if the spread is too large.

Use the variable and context in your conclusion. A count is discrete, while a normal variable is continuous; a normal approximation for a discrete distribution requires its own conditions and, where appropriate, a continuity correction. A sample mean being known does not by itself justify a normal model for individual observations.

Formula recap

X∼N(μ,σ2)X \sim N(\mu, \sigma^2)
Standard notation.
mean=median=mode=μ\text{mean} = \text{median} = \text{mode} = \mu
Symmetry.
≈95% within μ±2σ\approx 95\% \text{ within } \mu \pm 2\sigma
Spread.

Common mistakes to avoid

Reading N(50,64) as mean 50 and standard deviation 64.
The second parameter is variance, so the standard deviation is 8.
Using the curve height as P(X = 50).
Probability is an area. A single exact value has probability zero in the continuous model.
Putting the entire area outside a central interval into one tail.
Use symmetry to divide the outside area equally between the two tails.

Key takeaways

  • The normal model is continuous, symmetric and specified by mean and variance.
  • Use the square root of variance when you need standard deviation.
  • Probabilities are areas; 68-95-99.7 values are rounded checks.
  • Check the shape and context before accepting a normal model.

Explore further

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. For X∼N(18,25)X\sim N(18,25), what is the standard deviation?

    • 5
    • 25
    • 18
    • 625
    Show answer and explanation

    Answer: 5

    The variance is 25, so the standard deviation is its positive square root, 5. The mean is a separate parameter.

  2. For a continuous normal model, what represents P(a < X < b)?

    • The area between a and b
    • The curve height at b
    • The mean minus a
    • The total width of the graph
    Show answer and explanation

    Answer: The area between a and b

    Integrating the density over the interval gives its probability. A density height alone is not a probability.

  3. About 95.45% lies between mean minus 2 standard deviations and mean plus 2 standard deviations. Approximately what proportion lies above the upper boundary?

    • 0.02275
    • 0.0455
    • 0.47725
    • 0.9545
    Show answer and explanation

    Answer: 0.02275

    The outside area is 1 - 0.9545 = 0.0455. Symmetry assigns half of that to each tail.

  4. For X∼N(20,9)X\sim N(20,9), which standardises the value 26?

    • (26−20)/3=2(26-20)/3=2
    • (26−20)/9(26-20)/9
    • 26/2026/20
    • (20−26)/3=−2(20-26)/3=-2
    Show answer and explanation

    Answer: (26−20)/3=2(26-20)/3=2

    The standard deviation is the square root of 9, or 3. A value above the mean must have a positive standardised value.

  5. Which is a reason to question a normal model for individual waiting times?

    • Strong right skew in the observations
    • A mean is available
    • The sample contains real numbers
    • The measurement has units
    Show answer and explanation

    Answer: Strong right skew in the observations

    A normal model is symmetric, so strong right skew is a warning about fit. Merely knowing a mean does not establish normality.

  6. If two normal distributions have the same mean but one has half the standard deviation, how does its curve change?

    • It is narrower and taller, with total area still 1
    • Its total probability is halved
    • Its centre moves to zero
    • It becomes uniform
    Show answer and explanation

    Answer: It is narrower and taller, with total area still 1

    Reducing the spread concentrates the same total probability over a narrower range. The mean still sets the centre.

Next step

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Work through questions on The Normal Distribution. Use the feedback to decide what needs another look.

2 past-paper questions · 4 marks

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