Mathematics/Lessons/The Normal Distribution/The Normal Distribution
Statistics · The Normal Distribution · Year 2

The Normal Distribution

From bar charts to a smooth curve — the shape behind almost every naturally occurring measurement.

7 min read 1 worked example

Continuous random variables

A continuous random variable can take any one of infinitely many values within a range. Because there are infinitely many possible values, the probability that a continuous random variable takes any one exact value is $0$ — instead, you write the probability that it falls within a given range.

Contrast this with a discrete random variable, such as the number of double-sixes rolled in 20 throws of two dice: it can only take the distinct values $0,1,2,\ldots,20$, and each of those individual values can have a non-zero probability. A continuous random variable, such as the exact time taken to run a 400m race, could in principle take any real value in a range — there's no meaningful "next" value to count.

Key fact

A continuous random variable has a continuous probability distribution, shown as a curve on a graph. The area under this curve is always equal to $1$, and the probability that the variable lies in some interval is the area under the curve over that interval.

The normal distribution

Continuous variables that occur naturally are often more likely to take values grouped around a central value than to take extreme values. The normal distribution is a continuous probability distribution that models many such variables well. Examples include:

  • the resting heart rate of adults in a population
  • the wingspan of a particular species of bird
  • errors in a scientific instrument's readings
  • the diameters of ball bearings produced by a lathe

As you collect more data and use narrower and narrower class widths, a histogram of a naturally occurring variable like these tends to smooth out into a symmetric, bell-shaped curve.

48 56 64 72 80 88 96 X ~ N(72, 8²)
The resting heart rate of adults in a city, modelled with a normal distribution with mean $72$ bpm and standard deviation $8$ bpm.
Notation

If $X$ is a normally distributed random variable, you write $X\sim \mathrm{N}(\mu,\sigma^2)$, where $\mu$ is the population mean and $\sigma^2$ is the population variance.

  • The normal distribution has parameters $\mu$, the population mean, and $\sigma^2$, the population variance.
  • It is symmetrical: mean $=$ median $=$ mode.
  • It has a bell-shaped curve with asymptotes at each end — the curve gets arbitrarily close to the horizontal axis but never quite touches it.
  • The total area under the curve is equal to $1$.
  • It has points of inflection at $\mu+\sigma$ and $\mu-\sigma$.

For a normally distributed variable:

  • approximately $68\%$ of the data lies within one standard deviation of the mean
  • approximately $95\%$ of the data lies within two standard deviations of the mean
  • approximately $99.7\%$ of the data lies within three standard deviations of the mean
Watch out

Although a normal random variable could in theory take any value, in practice observations more than about $5$ standard deviations from the mean have probabilities so close to $0$ that they can be treated as impossible.

Worked example

Reading probabilities directly from the shape of the curve
The volume of paint, $V$ ml, in tins filled by a machine is modelled as $V\sim \mathrm{N}(500, 8^2)$. Find: a $\mathrm{P}(V>500)$   b $\mathrm{P}(492<V<508)$

a

500 V ~ N(500, 8²)
$\mathrm{P}(V>500)=0.5$. The curve is symmetrical about the mean, so exactly half the area lies on each side of it.
1
$500$ is the mean of the distribution.
The normal distribution is symmetrical, so for any normally distributed variable $\mathrm{P}(X>\mu)=0.5$.
▪ solved

b

492 500 508 V ~ N(500, 8²)
$\mathrm{P}(492<V<508)=0.6827$. The bounds $492$ and $508$ are each exactly one standard deviation from the mean.
1
$492 = 500-8$ and $508=500+8$, so both bounds are exactly one standard deviation from the mean.
$\sigma=8$ here.
2
$\mathrm{P}(492<V<508) = 0.6827$.
Approximately $68\%$ of the data lies within one standard deviation of the mean: $\mathrm{P}(\mu-\sigma<V<\mu+\sigma)=0.6827$.
▪ solved

Check yourself

The length, $L$ mm, of nails produced by a machine is modelled as $L\sim \mathrm{N}(50, 0.4^2)$. Using the properties of the normal distribution, find: i $\mathrm{P}(L>50)$   ii $\mathrm{P}(49.6<L<50.4)$.

i
$\mathrm{P}(L>50) = 0.5$, by symmetry about the mean.
ii
$49.6$ and $50.4$ are each one standard deviation ($0.4$) from the mean, so $\mathrm{P}(49.6<L<50.4) = 0.6827$.
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