The Normal Distribution

A continuous random variable needs a curve, not a list

The random variables in the previous topic took values you could count: the number of sixes in ten throws, the number of red counters drawn from a bag. You could list every possible value and attach a probability to each one. A continuous random variable is different. It measures something - a mass, a length, a waiting time - and between any two of its values there is always another value. The probability that it takes one exact value is zero, so every question has to be about an interval.

The model that replaces the list is a curve. The probability density function is drawn above a horizontal axis, and probability is the area under that curve: $P(a<X<b)$ is the area between $x=a$ and $x=b$. The whole area under the curve is 1, because the variable must take some value. That is why normal distribution questions are really area questions, and why a quick sketch is the fastest way to see which area the question is asking for.

A histogram of a large set of measurements is where this shape first appears. Weigh a thousand apples, draw the histogram with narrow classes, and the outline is a smooth symmetric hump: most apples are near the middle mass, fewer and fewer lie far away on either side, and the two sides mirror each other. The normal distribution is the curve that describes that outline, and Cambridge allows you to use it whenever the variable behaves that way.

The normal distribution has two parameters, and reading both of them correctly out of the notation is the first skill in the topic. $X~N(\mu ,{\sigma }^2)$ says that $X$ is normally distributed with mean $\mu$ and variance ${\sigma }^2$. The second number is the variance, not the standard deviation: $X~N(50,16)$ has $\sigma =4$. Dividing by 16 instead of 4 is the commonest single error in this topic and it destroys the answer, not just the presentation.

The shape of the normal curve

Every normal curve has the same bell shape, and the shape is completely fixed by $\mu$ and $\sigma$. The curve is symmetric about the vertical line $x=\mu$, so the mean, the median and the mode are all the same number. It reaches its maximum at $x=\mu$, and the highest part of the curve is where the measurements cluster.

The curve never touches the horizontal axis. It is asymptotic to it: the tails creep closer and closer as you move away from the mean but never reach zero, which matches the fact that a very large or small measurement is possible, merely unlikely. Where the curve is steepest is a genuine examinable detail. The points of inflection are at $x=\mu -\sigma$ and $x=\mu +\sigma$. Between those two points the curve bends downwards, and outside them it bends upwards.

Read as a probability, the area under the curve gives landmarks worth carrying in your head. Roughly 68% of the distribution lies within one standard deviation of the mean, about 95% within two, and about 99.7% within three. For $X~N(50,16)$ that means about two thirds of the values lie between 46 and 54, and almost all of them between 38 and 62. Treat these as sanity checks rather than answers: the tables give the exact figures, and $2\Phi (1)-1=0.6826$ is the precise version of the 68%.

Changing $\mu$ slides the whole curve along the axis without altering its shape, while changing $\sigma$ alters the shape itself.

• Symmetric about $x=\mu$, with its maximum at $x=\mu$
• Points of inflection at $x=\mu -\sigma$ and $x=\mu +\sigma$
• Asymptotic to the horizontal axis at both ends - it never touches it
• Total area 1, with about 68% of the area within one standard deviation of the mean

Standardising: one table for every normal distribution

There are infinitely many normal distributions, one for every pair of values of $\mu$ and $\sigma$, and printing a table for each of them is impossible. The way out is standardisation, which converts any normal variable into the same standard one. If $X~N(\mu ,{\sigma }^2)$ then $Z=\frac{X-\mu }{\sigma }$ has the standard normal distribution $Z~N(0,1)$, with mean 0 and variance 1. Subtracting $\mu$ moves the centre to 0 and dividing by $\sigma$ rescales the spread to 1.

Read $z=\frac{x-\mu }{\sigma }$ as a signed number of standard deviations. $z=2$ says that the value $x$ is two standard deviations above the mean; $z=-1.5$ says that it is one and a half standard deviations below. Because the transformation is only a shift and a scale, it preserves the shape of the curve and, crucially, it preserves areas. The area to the left of $x$ under the $X$ curve is exactly the area to the left of $z$ under the standard curve, so $P(X<x)=P(Z<z)$ and one printed table does all the work.

Standardisation is a calculation, not a statement, and Cambridge says so in a note against this objective: for calculations involving standardisation, full details of the working should be shown. Write the formula, substitute the numbers and give the value of $z$. A bare probability with no $z$ in the working will not earn the method mark, however right the final answer happens to be.

Reading the normal distribution table

The table in the formula booklet gives $\Phi (z)=P(Z\le z)$, the area under the standard normal curve to the left of $z$. The row gives $z$ to one decimal place and the column gives the second decimal place, so $\Phi (1.23)=0.8907$. The smaller block of numbers headed ADD handles the third decimal place: the entry under the third digit is added in the fourth decimal place of $\Phi$. For $z=1.234$ the third digit is 4 and the ADD entry is 7, so $\Phi (1.234)=0.8907+0.0007=0.8914$.

The table prints values only for $z\ge 0$, which is not a limitation at all, because the curve is symmetric. The area to the left of a negative value equals the area to the right of the matching positive value: $\Phi (-z)=1-\Phi (z).$ So $\Phi (-1.5)=1-\Phi (1.5)=1-0.9332=0.0668$.

There is a second table in the booklet, headed Critical values for the normal distribution. It turns the question round: for a given probability $p$ it gives the value $z$ with $\Phi (z)=p$. The entries you will use most are $z=1.645$ for $p=0.95$, $z=1.960$ for $p=0.975$, $z=2.326$ for $p=0.99$ and $z=2.576$ for $p=0.995$. Reach for this table whenever you are going backwards from a probability to a value of $x$.

Four steps, every time

Every normal probability question in this topic is answered by the same four steps, and taking them in order removes most of the mistakes. Sketch the curve, mark $\mu$ on the axis and draw a vertical line at each value named in the question. Standardise each boundary with $z=\frac{x-\mu }{\sigma }$. Read the table for each $z$, adjusting for the tail you actually want. Answer in the form $P(X<x)=0.7257$, to four decimal places.

The tail rules cover every shape of question: $P(X<x)=\Phi (z),$ $P(X>x)=1-\Phi (z),$ $P(a<X<b)=\Phi (z_b)-\Phi (z_a).$ The middle rule is where the sketch pays for itself. The table gives you a left-hand area and nothing else, so $P(X>x)$ is the whole area 1 minus the part you can read, and $P(a<X<b)$ is the larger left-hand area minus the smaller one.

Worked through once. The masses of a batch of apples, in grams, are modelled by $M~N(150,400)$, so $\mu =150$ and $\sigma =20$. Find the probability that an apple has a mass between 130 g and 170 g. Standardising, $z_1=\frac{130-150}{20}=-1$ and $z_2=\frac{170-150}{20}=1$. From the table $\Phi (1)=0.8413$ and $\Phi (-1)=1-0.8413=0.1587$. So $P(130<M<170)=0.8413-0.1587=0.6826.$ The answer is a probability: it has no units and it must lie between 0 and 1.

Working backwards: from a probability to a value of $x$

Questions also run the other way: a probability is given and the value of $x$ that produces it is wanted. The method reverses the four steps. Find $z$ from the probability, using the critical values table or by reading the main table backwards, then undo the standardisation with $x=\mu +z\sigma .$ Sketch first, because the sketch is what tells you whether $z$ should be positive or negative.

Two cases cover everything. If $P(X<x)=p$ with $p>0.5$, then $x$ lies to the right of the mean and $z$ is the positive table entry for $p$: for $p=0.95$, $z=1.645$. If $p<0.5$, then $x$ lies to the left of the mean and $z$ is negative: for $P(X<x)=0.05$, $z=-1.645$. A right-hand probability has to be turned round before the table will help. $P(X>x)=0.025$ is the same statement as $P(X<x)=0.975$, so $z=1.960$.

Worked through. $X~N(30,16)$, so $\sigma =4$, and $P(X<x)=0.95$. The critical values table gives $z=1.645$ for $p=0.95$. Substituting into $z=\frac{x-\mu }{\sigma }$ gives $1.645=\frac{x-30}{4}$, so $x=30+4(1.645)=36.58.$ A student who stops at $z=1.645$ has found the standardised value, not the value of $X$, and has answered a different question.

Finding $\mu$ and $\sigma$ from two probabilities

The syllabus asks you to find a relationship between $x$, $\mu$ and $\sigma$ given a probability, and the standard version of that question supplies two probabilities and asks for both parameters. Each probability standardises into an equation connecting $\mu$ and $\sigma$, so two probabilities give two simultaneous equations, solved in the usual way.

The $z$-values come from the tables and must be read in the right direction. If $P(X<x)=0.1587$ then $\Phi (z)=0.1587$, and because that is less than 0.5 the value of $z$ is $-1$. If $P(X>x)=0.0228$ then $\Phi (z)=1-0.0228=0.9772$, so $z=2$. Turning every tail into a left-hand area before looking anything up is what keeps the signs right.

Worked through. $X~N(\mu ,{\sigma }^2)$, with $P(X<32)=0.0228$ and $P(X>44)=0.1587$. From the tables $\Phi (-2)=0.0228$ and $\Phi (1)=0.8413=1-0.1587$. So $\frac{32-\mu }{\sigma }=-2$ and $\frac{44-\mu }{\sigma }=1$. Multiplying out, $32=\mu -2\sigma$ and $44=\mu +\sigma$. Subtracting the first from the second eliminates $\mu$: $12=3\sigma$, so $\sigma =4$, and then $\mu =44-4=40$. Check the first equation: $\mu -2\sigma =40-8=32$.

The normal approximation to the binomial

When $n$ is large a binomial distribution is well approximated by a normal one. The syllabus states the condition: $n$ must be large enough to ensure that both $np>5$ and $nq>5$, where $q=1-p$. Both, not either. When that holds, the binomial probability histogram traces out a bell shape. The approximating normal distribution matches the mean and the variance of the binomial: $\mu =np,$ ${\sigma }^2=npq.$

The one thing that must be dealt with before standardising is the continuity correction. A binomial variable is discrete and can take only whole numbers, while a normal variable is continuous and can take any value at all. Each whole number $a$ in the binomial model becomes the interval from $a-0.5$ to $a+0.5$ in the normal model, so every boundary moves half a unit away from the region you are counting. Getting this wrong shifts the answer by about the same amount as rounding $z$ to one decimal place, and it is the most heavily penalised error in the topic.

The rules follow from that picture: $X\ge a$ becomes $Y>a-0.5$; $X>a$ becomes $Y>a+0.5$; $X\le a$ becomes $Y<a+0.5$; and $X<a$ becomes $Y<a-0.5$. The pattern is that the boundary always moves away from the region being counted, so that the half-unit added or removed covers exactly the bar at the edge of that region.

Worked through. A fair coin is tossed 100 times and $X$ is the number of heads. Then $X~B(100,0.5)$ with $np=50>5$ and $nq=50>5$, so the approximation is valid with $\mu =50$ and ${\sigma }^2=100(0.5)(0.5)=25$, giving $\sigma =5$. For $P(X\ge 60)$ the continuity correction moves the boundary to 59.5, so $z=\frac{59.5-50}{5}=1.9$ and $P(X\ge 60)\approx 1-\Phi (1.9)=1-0.9713=0.0287.$

Sketches, and the marks they earn

The syllabus note for this topic says that sketches of normal curves may be required, and a sketch is also the fastest way to avoid a wrong tail. A sketch that earns its mark has five features: a symmetric bell shape; a horizontal axis drawn and labelled with the variable; the mean marked at the centre with a vertical line; the boundary value or values from the question marked with a vertical line on the correct side of the mean and at a sensible distance from it; and the required region shaded, with the shading between the two lines when the question asks for a region rather than a tail.

Do not try to scale the sketch precisely. It is a diagram, not a graph, and the marks are for the features and the labelling rather than for plotting. What must be right is the position of each marked value relative to the mean and which side is shaded. If your value lies to the right of the mean then the shaded tail is the small one on the right, and a sketch that shades the left of a right-hand value is worse than no sketch at all, because it will lead you to $\Phi (z)$ when the question wanted $1-\Phi (z)$.