S1.6 The Normal distribution

What does a Normal distribution look like?

Weigh a few hundred apples from one orchard, or time the same bus journey every morning for a year. The values pile up around a typical value and become rarer the further you go from it, in either direction, at about the same rate. A histogram of such data, drawn with narrow classes, looks like a bell. The Normal distribution is the smooth model of that bell. It is used for continuous quantities: lengths of manufactured parts, masses of fruit, journey times, reaction times.

Several facts follow straight from the shape, and you will use them in almost every question:

• The curve is symmetrical about \(\mu\), so mean = median = mode = \(\mu\), and \(P(X\lt\mu)=P(X\gt\mu)=0.5\).
• Probabilities are areas under the curve, and the total area is 1.
• The curve never touches the axis, but values more than 3 standard deviations from the mean are very rare.
• For a continuous variable \(P(X=a)=0\): a single value has no width, so no area. Hence \(P(X\lt a)\) and \(P(X\le a)\) are the same number.

Changing \(\mu\) slides the curve along the axis without changing its shape. Changing \(\sigma\) changes the spread: doubling \(\sigma\) makes the curve twice as wide and half as tall, so that the total area stays 1. This is why you cannot compare two Normal curves by height alone — a flatter curve is not "less likely", it is just more spread out.

Whatever the values of \(\mu\) and \(\sigma\), about 68.3% of the distribution lies within one standard deviation of the mean, 95.4% within two and 99.7% within three. These percentages are worth knowing as a sense check rather than as a method:

• A value 2 standard deviations above the mean has only about 2.3% of the distribution above it, so a tail probability beyond \(z=2\) should come out small.
• They also tell you when a Normal model is not suitable. If a quantity cannot be negative but \(\mu-2\sigma\) is well below zero, a Normal model gives a noticeable probability to impossible values.

How do you standardise a Normal variable?

There are infinitely many Normal distributions, one for each pair \(\mu,\sigma\), but only one of them has printed tables: the standard Normal distribution \(Z\sim N(0,1^2)\), with mean 0 and standard deviation 1. The trick is that every Normal variable can be turned into \(Z\). Subtract the mean, so the centre moves to 0; then divide by the standard deviation, so the spread becomes 1.

The \(z\)-value of \(x\) has a simple meaning: it is the number of standard deviations that \(x\) lies above the mean. A negative \(z\) means below the mean. Rearranged, the same link reads \(x=\mu+z\sigma\), which is the form you need when you work backwards from a probability.

Standardising relabels the axis without changing any area. Take cucumber lengths \(X\sim N(30,2.5^2)\) in cm. The value 33 is \(3\div2.5=1.2\) standard deviations above the mean, so \(P(X\gt33)\) is exactly the same area as \(P(Z\gt1.2)\). The two figures below have the same shape and the same shaded proportion; only the numbers on the axis differ.

How do you read the Φ(z) table?

The printed table gives \(\Phi(z)=P(Z\lt z)\), the area to the left of \(z\), for \(z\ge0\) in steps of 0.01. Round \(z\) to 2 d.p. before you look it up. Every other area comes from that one table by two ideas: the total area is 1, and the curve is symmetrical about 0. The method is always the same — draw the curve, shade the area you want, and decide which rule matches the picture.

How do you find probabilities for any Normal variable?

For a variable that is not already standard, add one step in front of the table work. The full method is:

1. Write the probability in terms of \(X\).
2. Sketch the curve, mark the mean and shade the area you want.
3. Standardise each boundary.
4. Use the table rules from the previous section.

Write the standardising step out in full, for example \(P(X\gt33)=P\!\left(Z\gt\frac{33-30}{2.5}\right)=P(Z\gt1.2)\). This line carries the method, and it shows anyone reading your work that you divided by \(\sigma\) and not by \(\sigma^2\).

Two further ideas from S1 probability often appear in the same question. For "at least one of …" or "all three …", find the single probability first and then combine with the multiplication rule for independent events. For a conditional probability use \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). With Normal tails, \(A\cap B\) is usually just the smaller tail: if a value is above 68, it is certainly above 60.

How do you find a value from a probability?

Now the area is given and the boundary is unknown. Reverse the steps: find the \(z\)-value first, then convert it to \(x\) with \(x=\mu+z\sigma\).

There are two ways to find \(z\). The percentage points table gives the \(z\)-value with a stated probability \(p\) above it, \(P(Z\gt z)=p\), for \(p=0.5,\ 0.4,\ 0.3,\ 0.2,\ 0.15,\ 0.1,\ 0.05,\ 0.025,\ 0.01,\ 0.005,\ 0.001,\ 0.0005\). For example \(p=0.05\) gives \(z=1.6449\). If your probability is in that list, use it: the values are more accurate than reading the \(\Phi\) table backwards. Otherwise, read the \(\Phi\) table in reverse: find the probability inside the body of the table and read off \(z\). If it falls between two entries, use the nearer one — no interpolation is needed.

The sign of \(z\) is the step that goes wrong most often. A value in the lower part of the distribution has a negative \(z\). Sketch first: if the shaded area to the left of the boundary is less than 0.5, the boundary is below the mean and \(z\lt0\).

How do you find an unknown mean or standard deviation?

A statement such as \(P(X\gt92)=0.0228\) tells you how far 92 is from the mean in standard deviations: it gives the \(z\)-value of 92. Writing \(\dfrac{92-\mu}{\sigma}=z\) gives one equation, and one equation is enough for one unknown.

Two habits make this reliable. First, turn the given probability into "area to the left" before reading the \(\Phi\) table in reverse: \(P(X\gt92)=0.0228\) means \(\Phi(z)=0.9772\). Second, check whether the probability is in the percentage points table (0.1, 0.05, 0.01, …): if it is, use that exact \(z\)-value.

How do you find the mean and the standard deviation together?

With two unknowns you need two probability statements. Each one gives an equation of the form \(x-\mu=z\sigma\). Write both as \[\mu+z_1\sigma=x_1,\qquad \mu+z_2\sigma=x_2,\] then subtract to eliminate \(\mu\): \(\sigma=\dfrac{x_2-x_1}{z_2-z_1}\). There is a nice way to see this: the gap between the two values, measured in standard deviations, is \(z_2-z_1\), and the same gap in the units of \(x\) is \(x_2-x_1\). Dividing one by the other gives the size of one standard deviation.

Always check the result: \(\sigma\) must come out positive, and \(\mu\) should lie on the correct side of each given value.

Exam tips for S1.6

Common misconceptions

How this topic connects

S1 probability and discrete random variables

Conditional probability, "at least one" and independent repeats all reappear here, applied to Normal areas. The mean and variance of a Normal variable mean the same as \(E(X)\) and \(\text{Var}(X)\) for a discrete one.

S2 approximations and hypothesis tests

In S2 the Normal distribution approximates the binomial and the Poisson distributions for large parameters, with a continuity correction (S2.3), and is used to carry out the tests that follow (S2.4). The formula for the Normal probability density function is not needed in S1, and neither are continuous random variables defined by a density function (S2.2).

P1 transformations of graphs

\(X=\mu+\sigma Z\) corresponds to a stretch of the standard curve by factor \(\sigma\) parallel to the \(x\)-axis (with a stretch by factor \(1/\sigma\) parallel to the \(y\)-axis, so the area stays 1) followed by a translation of \(\mu\) — the change of shape seen in the second figure. In IAL Physics practical work, repeated readings scatter roughly symmetrically about a central value, which is why their mean is taken as the best estimate.

Key formulae

Check your understanding

Specification checklist

• (S1 6.1) I can describe the shape of a Normal distribution, use its symmetry about the mean, and explain what changing \(\mu\) or \(\sigma\) does to the curve.
• (S1 6.1) I know that in \(N(\mu,\sigma^2)\) the second parameter is the variance.
• (S1 6.1) I can standardise a Normal variable with \(Z=\dfrac{X-\mu}{\sigma}\).
• (S1 6.1) I can use the table of \(\Phi(z)\), with symmetry, to find probabilities below, above and between values, including for negative \(z\).
• (S1 6.1) I can use the percentage points table, or the \(\Phi\) table in reverse, to find a value with a given probability above or below it.
• (S1 6.1) I can find an unknown mean or an unknown standard deviation from one given probability.
• (S1 6.1) I can find both the mean and the standard deviation by solving simultaneous equations.
• (S1 6.1) I can combine Normal probabilities with conditional probability and with independent repeated observations.