Choosing a probability distribution

Choosing a distribution means matching a random variable to a mathematical model. At A level, the central choice is usually between the binomial distribution, which counts successes, and the Normal distribution, which models continuous measurements.

The context alone does not determine the distribution. First define exactly what the random variable measures, then check the model’s assumptions.

Prerequisites

You should be able to:

• interpret a discrete random variable;
• use probability notation and independence from probability;
• recognise the conditions for a binomial distribution;
• interpret the parameters of a Normal distribution;
• distinguish a model from reality using probability modelling.

The first question: what does the variable record?

A random variable is discrete if its possible values are separated, usually because it counts. It is continuous if it can take any value in an interval, usually because it measures.

Random variable	Type	Possible starting model
Number of defective bulbs among $40$	Discrete count	Binomial
Mass of a randomly selected apple	Continuous measurement	Normal, if the shape is plausible
Number of calls until the first sale	Discrete count	Neither binomial nor Normal in the usual A-level setting
Time taken to complete a race	Continuous measurement	Possibly Normal, but only if the distribution is approximately symmetric

The words number of often signal a count, while mass, height, length and time usually signal a measurement. This is a useful clue, not a proof.

The same experiment can produce different variables

From $20$ light bulbs, let $X$ be the number that fail and let $Y$ be the lifetime of the first bulb tested. The experiment concerns bulbs in both cases, but $X$ is a discrete count while $Y$ is a continuous measurement. A distribution models a variable, not a topic word.

A reliable decision process

Use these questions in order.

1. Define the random variable. State what it records and its units, if any.
2. Determine its support. Is it an integer count, or a continuous measurement?
3. Test the assumptions. Do not choose a model merely because the variable is discrete or continuous.
4. Identify the parameters. Give $n$ and $p$ for a binomial model, or $\mu$ and $\sigma^2$ for a Normal model.
5. Check the conclusion in context. A probability must lie between $0$ and $1$, and an extreme result should have a suitably small probability.

In compact form,

$$\boxed{\text{define} \longrightarrow \text{classify} \longrightarrow \text{check} \longrightarrow \text{parameterise}}.$$

When to choose a binomial distribution

If $X$ counts successes, write

$$X\sim\operatorname{B}(n,p)$$

only when all four conditions hold:

• there is a fixed number $n$ of trials;
• each trial is classified into success or failure;
• trials are independent;
• the success probability $p$ is constant.

The word success is only a label for the outcome counted. It need not be desirable.

Worked example 1: choosing and parameterising a binomial model

A factory’s production process makes a defective component with probability $0.018$. Defects occur independently. A quality controller examines the next $80$ components. Let $D$ be the number that are defective.

$D$ is a discrete count. There are $80$ fixed trials, each component is defective or not defective, independence is stated, and the defect probability is constant. Therefore

$$\boxed{D\sim\operatorname{B}(80,0.018)}.$$

Its mean is

$$\operatorname{E}(D)=np=80(0.018)=1.44.$$

The non-integer mean is not a possible observed count. It is the long-run average number of defects per batch of $80$.

Worked example 2: a count that is not binomial

Five cards are drawn from a standard pack without replacement. Let $A$ count the aces drawn.

$A$ is discrete and there is a fixed number of draws. Each draw can be classified as ace or not ace. However, without replacement:

• the draws are dependent;
• the probability of an ace changes after each draw.

For example, if the first card is an ace, the next ace probability is $3/51$, not $4/52$. Thus

$$\boxed{A\text{ is not exactly binomial}.}$$

Being a bounded count is necessary for a binomial variable, but it is not sufficient.

Self-check 1

A survey asks $200$ independently selected voters whether they support a proposal. Suppose every selected voter has probability $0.63$ of answering yes. Let $S$ count yes responses. Choose a model and state its parameters.

Answer

There are $200$ fixed, independent trials, two classifications per response, and constant probability $0.63$ of a yes response. Hence

$$\boxed{S\sim\operatorname{B}(200,0.63)}.$$

When to choose a Normal distribution

A Normal random variable is continuous, symmetric and bell-shaped. Write

$$X\sim\operatorname{N}(\mu,\sigma^2),$$

where

$$\operatorname{E}(X)=\mu, \qquad \operatorname{Var}(X)=\sigma^2, \qquad \operatorname{SD}(X)=\sigma.$$

At A level, a question may state that a variable is Normally distributed. In data-based modelling, a Normal model is plausible when observations are approximately symmetric, have one central peak, and show no severe outliers or long tail. Many biological and manufacturing measurements are approximately Normal because numerous small influences combine, but the context by itself never guarantees Normality.

Worked example 3: choosing a Normal model

The fill volume $V$ ml of cartons from a calibrated machine is Normally distributed with mean $1002$ ml and standard deviation $4.5$ ml.

$V$ is a continuous measurement, and Normality is stated. Since the second parameter is the variance,

$$\sigma^2=4.5^2=20.25.$$

Therefore

$$\boxed{V\sim\operatorname{N}(1002,20.25)}.$$

A common error is to write $\operatorname{N}(1002,4.5)$. In the notation used here, the second parameter is $\sigma^2$, not $\sigma$.

Worked example 4: continuous does not mean Normal

Let $T$ be the waiting time from a randomly chosen moment until the next bus. Buses arrive exactly every $12$ minutes.

$T$ is continuous and, under an idealised random-arrival assumption, every waiting time from $0$ to $12$ minutes is equally likely. The distribution is flat rather than bell-shaped. Hence a Normal model is unsuitable.

There is another warning sign: a Normal distribution has a positive, though perhaps tiny, probability of negative values. For waiting times close to zero with substantial spread, that feature can be implausible.

Self-check 2

The heights $H$ cm of adult plants of one variety are modelled by a Normal distribution with mean $74$ cm and variance $36\text{ cm}^2$. State the model and its standard deviation.

Answer

$$\boxed{H\sim\operatorname{N}(74,36)}, \qquad \boxed{\sigma=\sqrt{36}=6\text{ cm}}.$$

Choosing between the models

Worked example 5: one context, two distributions

The mass of a manufactured tablet is Normally distributed with mean $500$ mg and standard deviation $8$ mg. A tablet is underweight if its mass is below $485$ mg. Tablets are produced independently. A packet contains $30$ tablets.

Two different random variables arise.

Let $M$ be the mass of one tablet. Then

$$M\sim\operatorname{N}(500,8^2)=\operatorname{N}(500,64).$$

The probability that one tablet is underweight is

$$\begin{aligned} p &=\operatorname{P}(M<485)\\ &=\operatorname{P}\left(Z<\frac{485-500}{8}\right)\\ &=\operatorname{P}(Z<-1.875)\\ &\approx 0.0304. \end{aligned}$$

Now let $U$ be the number of underweight tablets in a packet. This is a count from $30$ independent trials, each with underweight probability $p\approx0.0304$. Therefore

$$\boxed{U\sim\operatorname{B}(30,0.0304)}.$$

For example, the probability of at least one underweight tablet is

$$\begin{aligned} \operatorname{P}(U\geq1) &=1-\operatorname{P}(U=0)\\ &=1-(1-0.0304)^{30}\\ &\approx\boxed{0.604}. \end{aligned}$$

The individual measurement is Normal; the number satisfying a condition is binomial. This distinction appears frequently in multi-stage questions.

Worked example 6: neither model

A customer repeatedly rolls a fair die until a six appears. Let $R$ be the number of rolls required.

$R$ is discrete, but the number of trials is not fixed in advance. Its possible values are

$$1,2,3,\ldots,$$

with no finite upper bound. It is not binomial. It is not continuous, so it is not Normal either. Therefore

$$\boxed{\text{neither model is suitable}.}$$

You are not required to force every variable into one of the two familiar distributions.

Using evidence to assess a Normal model

When raw data, a histogram or a summary is supplied, inspect it before accepting Normality.

A Normal model becomes more credible when:

• the histogram is roughly symmetric and unimodal;
• the mean and median are close;
• observations are reasonably balanced on either side of the mean;
• there are no strong outliers or physical bounds close to the main body of data.

It becomes less credible when the data have a long tail, several peaks, substantial outliers, or a large pile-up at a boundary such as zero.

Worked example 7: judging from summaries

Two samples of delivery times have the following summaries.

Sample	Mean	Median	Lower quartile	Upper quartile	Maximum
A	$31.2$	$31.0$	$27.4$	$35.1$	$43.0$
B	$31.2$	$24.6$	$18.3$	$36.5$	$118.0$

Sample A has similar mean and median, and its quartiles are approximately balanced around the centre. These facts are consistent with a Normal model, although they do not prove it.

Sample B has a mean much larger than its median and an unusually large maximum. This suggests strong positive skew, so a Normal model is doubtful.

Model conclusions should be measured

Finite data cannot prove that a population is Normal. Write that the evidence supports, is consistent with, or casts doubt on a Normal model.

Common misconceptions

”The question gives a mean and standard deviation, so it must be Normal”

Every distribution with finite spread has a mean and standard deviation. These values do not determine the shape.

”Every discrete variable is binomial”

A binomial variable specifically counts successes in a fixed number of independent trials with constant $p$. Shoe size is discrete but not binomial. The number of attempts until a first success is also not binomial.

”Every measurement is Normal”

Measurements can be skewed, uniform, multimodal or bounded in ways that make a Normal model poor.

”Independent means mutually exclusive”

Independent events may occur together. Mutually exclusive events cannot occur together. Repeated trials in a binomial model must be independent, not mutually exclusive.

”The parameters are just numbers to copy”

Interpret them. In $X\sim\operatorname{B}(n,p)$, $n$ is the number of trials and $p$ is the probability of the outcome counted by $X$. In $Y\sim\operatorname{N}(\mu,\sigma^2)$, the second parameter is variance.

Mixed self-check

For each variable, choose a binomial model, a Normal model, or neither. Give parameters where possible.

1. $X$ is the number of heads in $25$ independent tosses of a coin for which $\operatorname{P}(\text{head})=0.4$.
2. $Y$ is the diameter in mm of a bearing, stated to be Normally distributed with mean $12.0$ mm and standard deviation $0.03$ mm.
3. $W$ is the number of red counters obtained in $10$ draws without replacement from a bag containing $6$ red and $14$ blue counters.
4. $Q$ is a customer’s expenditure, which has many values near zero and a long right tail.

Answers

1. The binomial conditions are stated, so

   $$\boxed{X\sim\operatorname{B}(25,0.4)}.$$

2. This is a continuous measurement and Normality is stated. Convert standard deviation to variance:

   $$\boxed{Y\sim\operatorname{N}(12.0,0.03^2)} =\boxed{\operatorname{N}(12.0,0.0009)}.$$

3. Neither. Without replacement, the trials are dependent and the red probability changes.

4. Neither of these models is justified. $Q$ is continuous, but the strong right skew contradicts the symmetry of a Normal distribution.

Exam-ready checklist

Before committing to a distribution, ask:

$$\begin{array}{ll} \text{Binomial:} & \text{fixed }n,\text{ two outcomes, independent trials, constant }p?\\[4pt] \text{Normal:} & \text{continuous, approximately symmetric and bell-shaped?} \end{array}$$

Then state the model completely:

$$X\sim\operatorname{B}(n,p) \qquad\text{or}\qquad X\sim\operatorname{N}(\mu,\sigma^2).$$

If an assumption fails, say exactly which one. If evidence is insufficient, do not invent certainty.

Next steps

• Practise exact and cumulative calculations in the binomial distribution.
• Standardise continuous variables and find probabilities in the Normal distribution.
• Apply model choice when testing claims in binomial hypothesis tests and Normal hypothesis tests.