2.04d, 2.04h - Choosing distributions and the normal approximation | OCR A-Level Mathematics A

A thousand trials sounds like plenty for a smooth bell curve. Can that still be a poor model? In 2.04h you will choose distributions from the process and the shape; in 2.04d you will match a normal approximation's centre and spread to a binomial count.

Choose a model from the variable and its generating process

Can a count from a thousand trials still be too lopsided for a normal curve? Picture a rare fault: if its probability is only $0.002$ , the expected number in a thousand components is just $2$ . More trials do not, by themselves, make a count nearly symmetric. First find out what is being counted and how the outcomes arise.

You already know

$X\sim\mathrm B(n,p)$ counts successes in $n$ independent trials, each with two outcomes and the same success probability $p$ . A normal variable is continuous, with a symmetric bell-shaped distribution.

The word “success” is a label: a faulty component can be the success we count. For $20$ independently produced components, each faulty with probability $0.1$ , a fault count can have a binomial model. A measurement of their lengths is a different variable and needs a different justification.

Probability model

A probability model assigns probabilities to the possible values of a specified random variable. Its assumptions describe how those values arise. For example, a binomial model describes a count under fixed-trial assumptions; it does not describe every set of $20$ observations merely because there are $20$ of them.

For a continuous measurement, inspect its histogram as well as its physical range. A roughly symmetric, unimodal bell shape supports a normal model, provided impossible-value tails are negligible for the task. A strongly skewed waiting time or two-peak mixture gives a reason to reject one normal curve. Selecting one of six equally likely numbered tokens instead gives a discrete uniform model: an integer-valued variable is not automatically binomial.

Worked example 1: justify the count model

Question. Twenty components are produced independently. Each has probability $0.1$ of being faulty. Let $X$ be the number faulty. State a suitable model and justify it.

Plan. Match the count and each production condition to the binomial assumptions.

\begin{aligned} n&=20 &&\text{fixed number of trials}\\ p&=0.1 &&\text{same fault probability}\\ X&\sim\mathrm B(20,0.1) &&\text{independent binary trials} \end{aligned}

Why this choice? The variable counts faults, and the stated process supplies all four binomial conditions.

Answer. $\mathrm B(20,0.1)$ is suitable under the given production assumptions.

Check. The possible counts are $0,1,\ldots,20$ , exactly the model's support.

Worked example 2 changes only independence. The same marginal fault probability is not enough to rescue the model.

Worked example 2: a shared cause changes the model

Question. The same twenty-component inspection now concerns components from one batch. A faulty machine setting tends to make several components faulty together, although each component's fault probability remains $0.1$ . Explain why $\mathrm B(20,0.1)$ may be unsuitable.

Plan. Identify the changed assumption and connect it to the batch mechanism.

\begin{aligned} n&=20 &&\text{still fixed}\\ p&=0.1 &&\text{still common}\\ \text{shared setting}&\longrightarrow\text{linked faults} &&\text{dependence} \end{aligned}

Why this step? Naming dependence earns a useful explanation only when we say what could cause it here.

Answer. The faults may not be independent because the shared machine setting affects several components. The binomial model may misrepresent how often large groups of faults occur.

Check. There are still two outcomes per component. The problem is dependence, not the number of possible outcomes.

A model criticism should point to a particular assumption or shape, rather than just say that “real life varies”.

Watch out

Wrong: “There are $20$ components, so $X\sim\mathrm B(20,0.1)$ .” A fixed count alone does not establish independence or common probability. State what the variable counts and connect the other assumptions to the process.

Now change the source of failure in the binomial assumptions: the trials are independent, but their probabilities vary.

A normal approximation matches binomial centre and spread

If a binomial count has a suitable shape, choose a normal curve with the same mean and variance. You are matching two features, not turning integer counts into measurements.

For $X\sim\mathrm B(60,0.5)$ , the centre is $30$ and the variance is $15$ . The matching continuous model is therefore $Y\sim\mathrm N(30,15)$ . Its standard deviation is $\sqrt{15}$ , not $15$ .

Binomial centre and spread

If $X\sim\mathrm B(n,p)$ , with $q=1-p$ , then

\mu=np,\qquad \sigma^2=npq=np(1-p).

These are the binomial mean and variance. Use them as the normal approximation's mean and variance when an approximation is suitable.

In the formula sheet.

The sheet supplies the two moments; choosing a matching normal distribution uses them. Write normal notation as $\mathrm N(\mu,\sigma^2)$ , while a calculator's normal-probability function usually asks for $\sigma$ .

Worked example 3: match the parameters

Question. A binomial variable has distribution $\mathrm B(60,0.4)$ . State the matching normal model and its standard deviation to three significant figures.

Plan. Calculate $np$ and $np(1-p)$ , then square-root the variance.

\begin{aligned} \mu&=60(0.4)=24 &&\text{match the mean}\\ \sigma^2&=60(0.4)(0.6)=14.4 &&\text{match the variance}\\ \sigma&=\sqrt{14.4}=3.79473\ldots &&\text{standard deviation} \end{aligned}

Why take a square root? The second normal parameter is variance, while standard deviation measures spread in the same units as the count.

Answer. $Y\sim\mathrm N(24,14.4)$ , with standard deviation $3.79$ .

Check. $(3.79473\ldots)^2=14.4$ ; the mean $24$ lies within the possible count range.

Worked example 4 changes only the number of trials, from $60$ to $120$ . Doubling it doubles mean and variance, but does not double standard deviation.

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2.04d, 2.04h - Choosing distributions and the normal approximation

Choose a model from the variable and its generating process

A normal approximation matches binomial centre and spread

Continue this lesson.

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