5.06b, 5.06c - Fitting a theoretical distribution | OCR A-Level Further Mathematics A

Turn a proposed probability model into the frequencies it predicts for a sample. Given ratios and discrete uniform models provide the starting point in 5.06b; 5.06c extends the same idea to binomial, geometric, Poisson and continuous distributions, including a distribution specified by its density.

Given proportions and discrete uniform models

Can a model predict $31.478$ waiting times in a group when the observed number must be a whole number? Yes: an expected frequency is an average prediction, not an instruction to find a fraction of an observation. By the end, you will obtain such counts directly from continuous probabilities.

You already know

A probability distribution assigns total probability $1$ . A binomial variable counts successes in a fixed number of independent equal-probability trials; a geometric variable counts trials up to and including the first success. Expected frequencies and observed frequencies have different jobs.

Imagine repeating a sample of size $N$ many times under a fixed model. If an observation has probability $p_i$ of landing in class $i$ , the long-run average count in that class is $Np_i$ . This is the bridge from any distribution to a frequency table.

Expected frequency from a model

For $N$ independent observations with class probabilities $p_i$ ,

E_i=Np_i,\qquad \sum_i p_i=1,\qquad \sum_i E_i=N

The classes must be mutually exclusive and cover the model's entire possible range.
Not in the formula booklet: learn it.

A proposed ratio provides probabilities after dividing each weight by the sum of all weights. It need not match the observed proportions: matching the model to its own hypothesis is the point.

Worked example 1: a specified ratio

Question. A model claims that three categories occur in the ratio $2:3:5$ . Find their expected frequencies in $80$ observations.

Plan. Normalise the ratio, then multiply the probabilities by $80$ .

\begin{aligned} 2+3+5&=10 &&\text{total weight}\\ (p_1,p_2,p_3)&=(2/10,3/10,5/10) &&\text{convert weights to probabilities}\\ (E_1,E_2,E_3)&=80(2/10,3/10,5/10) &&\text{multiply by sample size}\\ &=(16,24,40) &&\text{evaluate} \end{aligned}

Why this step? A ratio weight is not itself a probability: the probabilities must sum to $1$ .

Answer. The expected frequencies are $16,24,40$ .

Check. They sum to $80$ and retain the ratio $2:3:5$ .

Worked example 2 changes the proposed weights to equal weights. This is the discrete uniform case; equal probabilities are a model assumption, not a consequence of having named categories.

Worked example 2: a discrete uniform hypothesis

Question. A model assigns equal probabilities to the five values $1,2,3,4,5$ . Find the expected frequencies in $80$ observations.

Plan. Give each of the five possible values probability $1/5$ .

\begin{aligned} p_r&=\frac15\quad(r=1,2,3,4,5) &&\text{discrete uniform model}\\ E_r&=80\times\frac15=16 &&\text{same probability for each value} \end{aligned}

Why this step? There are five possible values, not four intervals between the endpoints.

Answer. The expected table is $(16,16,16,16,16)$ .

Check. $5\times16=80$ ; each expectation is at least $5$ .

The discrete uniform probability $1/n$ for $n$ equally likely values is printed in the formula booklet. Its use to produce counts still requires multiplying by the sample size.

Watch out

Wrong: use observed proportions $O_i/N$ as the hypothesis probabilities. That gives $E_i=O_i$ by construction and makes every discrepancy zero. Use the proposed model, estimating only parameters the question requires.

A fit is a prediction from the stated model; whether the data support that prediction is the next lesson's question.

Fitting binomial models

A binomial model predicts one expected frequency for each possible success count, using the trial count and success probability.

Recall $\mathrm{P}(X=r)=\binom nr p^r(1-p)^{n-r}$ for $X\sim\mathrm{B}(n,p)$ ; this probability and $\mathrm{E}(X)=np$ are in the formula booklet. Here $n$ is the number of trials in one observation, whereas $N$ is the number of observed groups. Confusing them changes the entire model.

If $p$ is specified independently of the data, retain it. If the question asks you to fit an unspecified binomial probability with known $n$ , match the model mean $np$ to the sample mean $\bar x$ .

Fitted binomial probability

For $N$ observed success counts from groups of $n$ trials, estimate an unknown $p$ by

\widehat p=\frac{\bar x}{n}=\frac{\sum_r rO_r}{nN}

Then use $\mathrm{B}(n,\widehat p)$ to calculate expected frequencies. This assumes the binomial model is appropriate.
Not in the formula booklet: learn it.

Record that a parameter was estimated: it will matter when counting degrees of freedom for a goodness of fit test.

Worked example 3: a specified success probability

Question. For each of $80$ independent groups of three trials, the model is $X\sim\mathrm{B}(3,1/2)$ . Find the expected frequencies for $X=0,1,2,3$ .

Plan. Calculate the four binomial probabilities, then multiply each by $80$ .

\begin{aligned} \mathrm{P}(X=r)&=\binom3r(1/2)^3 &&\text{given model}\\ (p_0,p_1,p_2,p_3)&=\frac18(1,3,3,1) &&\text{binomial coefficients}\\ (E_0,E_1,E_2,E_3)&=80\times\frac18(1,3,3,1) &&\text{convert probabilities to counts}\\ &=(10,30,30,10) &&\text{evaluate} \end{aligned}

Why this step? The sample size is $80$ , while the binomial trial count is $3$ : they enter different parts of the calculation.

Answer. $10,30,30,10$ .

Check. They sum to $80$ and are symmetric because $p=1/2$ .

Worked example 4 changes only whether the success probability is known. The sample now supplies its estimate, even though it happens to produce the same fitted probability.

Worked example 4: estimate the success probability

Question. In $80$ groups of three trials, the observed frequencies for $0,1,2,3$ successes are $9,33,27,11$ . Fit a binomial distribution with unknown success probability and find the expected frequencies.

Plan. Count all observed successes, estimate $p$ , then use the fitted binomial distribution.

\begin{aligned} \sum_r rO_r&=0(9)+1(33)+2(27)+3(11)=120 &&\text{total successes}\\ \widehat p&=\frac{120}{3\times80}=\frac12 &&\text{successes divided by trials}\\ E_r&=80\binom3r(1/2)^3 &&\text{fit the model}\\ (E_0,E_1,E_2,E_3)&=(10,30,30,10) &&\text{evaluate} \end{aligned}

Why this step? The observed mean estimates the model's mean; no individual frequency is forced to match its observed count.

Answer. The fitted model is $\mathrm{B}(3,1/2)$ , giving $10,30,30,10$ . One model parameter was estimated from these data.

Check. The fitted expected total number of successes is $1(30)+2(30)+3(10)=120$ , matching the observed total.

The numerical fit does not establish independence of the trials or a constant success probability. Those remain modelling assumptions to check in context.

Watch out

Wrong: $\widehat p=\bar x$ for a binomial success count. Right: $\widehat p=\bar x/n$ . Also, grouped data with an open-ended class do not reveal its exact success total; do not invent class values to estimate $p$ .

Use exact totals when they are available, and keep the fitted probability unrounded while finding expectations.

Fitting geometric models and tails

A geometric model has infinitely many possible trial counts, so a finite frequency table must include the remaining tail.

Recall that OCR's geometric variable counts the successful trial: its support is $1,2,3,\ldots$ . For $X\sim\mathrm{Geo}(p)$ , the booklet gives $\mathrm{P}(X=r)=(1-p)^{r-1}p$ , $\mathrm{P}(X>r)=(1-p)^r$ and $\mathrm{E}(X)=1/p$ . There is no zero class with positive probability.

For an unspecified $p$ , match $1/p$ to the observed mean. The resulting fitting rule $\widehat p=1/\bar x=N/\sum rO_r$ is not printed in the formula booklet. If the exact total is unavailable because the tail is grouped, use any mean or parameter supplied; do not replace “ $4$ or more” by exactly $4$ .

Worked example 5: retain the infinite tail

Question. Fit the given model $X\sim\mathrm{Geo}(1/2)$ to $80$ observations in the classes $1,2,3,\ge4$ .

Plan. Calculate the first three point probabilities and use a tail probability for the final class.

\begin{aligned} (p_1,p_2,p_3)&=(1/2,1/4,1/8) &&\text{point probabilities}\\ \mathrm{P}(X\ge4)&=\mathrm{P}(X>3)=(1/2)^3=1/8 &&\text{all remaining trials}\\ (E_1,E_2,E_3,E_{\ge4})&=80(1/2,1/4,1/8,1/8) &&\text{multiply by }80\\ &=(40,20,10,10) &&\text{evaluate} \end{aligned}

Why this step? The final class includes $4,5,6,\ldots$ , rather than just the event $X=4$ .

Answer. $40,20,10,10$ .

Check. The probabilities sum to $1$ , and the expected frequencies sum to $80$ .

Worked example 6 changes only the source of the parameter: an observed total supplies it instead of a fixed hypothesis value.

Worked example 6: estimate a geometric parameter

Question. The exact sum of $80$ observed trial counts is $200$ . Fit a geometric model and calculate expectations for $1,2,3,\ge4$ .

Plan. Estimate $p$ from the sample mean, then retain the entire tail.

\begin{aligned} \bar x&=200/80=2.5 &&\text{sample mean}\\ \widehat p&=1/2.5=0.4 &&\text{match the model mean}\\ (p_1,p_2,p_3,p_{\ge4})&=(0.4,0.6(0.4),0.6^2(0.4),0.6^3) &&\text{geometric probabilities}\\ (E_1,E_2,E_3,E_{\ge4})&=(32,19.2,11.52,17.28) &&\text{multiply by }80 \end{aligned}

Why this step? Lower success probability produces a longer average wait; using $p=\bar x$ would not even give a valid probability.

Answer. The fitted model is $\mathrm{Geo}(0.4)$ , with expectations $32,19.2,11.52,17.28$ . One parameter was estimated.

Check. The expectations sum to $80$ , and the fitted mean is $1/0.4=2.5$ .

An observed zero trial count would be impossible under this convention. That is a support mismatch, not a small discrepancy to hide by combining the zero class with another one.

Watch out

Wrong: $\mathrm{P}(X\ge4)=0.6^4$ when $p=0.4$ . Right: the first three trials must fail, so $\mathrm{P}(X\ge4)=0.6^3$ ; trial $4$ is allowed to be the success.

Write the final class as a probability event before using a calculator.

Fitting Poisson models

A Poisson fit uses its mean count for the actual observation interval, then converts count probabilities into expected frequencies.

Recall $\mathrm{P}(X=r)=e^{-m}m^r/r!$ and $\mathrm{E}(X)=m$ for $X\sim\mathrm{Po}(m)$ ; both are in the formula booklet. To fit an unspecified $m$ , set $\widehat m=\bar x$ . This estimation rule is not printed in the booklet. Use an exact count total or a supplied mean, rather than inventing values in an open tail class.

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5.06b, 5.06c - Fitting a theoretical distribution

Given proportions and discrete uniform models

Fitting binomial models

Fitting geometric models and tails

Fitting Poisson models

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